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Corollary 9.5.19. If \( p \geq 5 \) and \( k \) is even then\n\n\[ p{B}_{k} \equiv {S}_{k}\left( p\right) \left( {\;\operatorname{mod}\;{p}^{3}}\right) ,\]\n\nexcept if \( \left( {p - 1}\right) \mid \left( {k - 2}\right) \) and \( p \nmid k\left( {k - 1}\right) \), in which case the congruence is only modulo \( {p}^{2}... | Proof. Using again the expression for \( {S}_{k}\left( p\right) \) used at the beginning of the proof of Proposition 9.5.18, the Clausen-von Staudt theorem, and \( {B}_{k - 1} = 0 \) when \( k \geq 4 \) is even, we deduce that\n\n\[ {S}_{k}\left( p\right) = p{B}_{k} + \frac{k\left( {k - 1}\right) {p}^{3}}{6}{B}_{k - 2}... | Yes |
Proposition 9.5.20 (Voronoi). For any even \( k \geq 2 \) and for all coprime integers \( a \) and \( n \) in \( {\mathbb{Z}}_{ > 0} \) we have\n\n\[ \left( {{a}^{k} - 1}\right) {N}_{k} \equiv k{a}^{k - 1}{D}_{k}\mathop{\sum }\limits_{{m = 1}}^{{n - 1}}{m}^{k - 1}\left\lfloor \frac{ma}{n}\right\rfloor \left( {\;\operat... | Proof. For \( 1 \leq m \leq n - 1 \), write \( {ma} = {q}_{m}n + {r}_{m} \) with \( 0 \leq {r}_{m} < n \), so \( {q}_{m} = \lfloor {ma}/n\rfloor \) . By the binomial theorem we have\n\n\[ {\left( ma\right) }^{k} \equiv {r}_{m}^{k} + {kn}{q}_{m}{r}_{m}^{k - 1} \equiv {r}_{m}^{k} + {kn}{\left( ma\right) }^{k - 1}\left\lf... | Yes |
Corollary 9.5.21. Let \( p \) be a prime such that \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) and \( p > 3 \). Then \[ 2\left( {2 - \left( \frac{2}{p}\right) }\right) {B}_{\left( {p + 1}\right) /2} \equiv - \mathop{\sum }\limits_{{m = 1}}^{{\left( {p - 1}\right) /2}}\left( \frac{m}{p}\right) \left( {\;\ope... | Proof. We set \( k = \left( {p + 1}\right) /2, a = 2 \), and \( n = p \) in the above proposition. Since for any \( a \) we have \( {a}^{\left( {p - 1}\right) /2} \equiv \left( \frac{a}{p}\right) \left( {\;\operatorname{mod}\;p}\right) \), we obtain \[ \left( {2\left( \frac{2}{p}\right) - 1}\right) {N}_{\left( {p + 1}\... | Yes |
Corollary 9.5.22. Let \( p \) be a prime such that \( p \equiv 3\left( {\;\operatorname{mod}\;4}\right) \) and \( p > 3 \) . If we denote by \( h\left( {-p}\right) \) the class number of the imaginary quadratic field \( \mathbb{Q}\left( \sqrt{-p}\right) \) , then\n\n\[ h\left( {-p}\right) \equiv - 2{B}_{\left( {p + 1}\... | Proof. The classical Dirichlet class number formula gives for any fundamental discriminant \( D < - 4 \) the identity\n\n\[ \left( {2 - \left( \frac{D}{2}\right) }\right) h\left( D\right) = \mathop{\sum }\limits_{{m = 1}}^{{\lfloor \left| D\right| /2\rfloor }}\left( \frac{D}{m}\right) . \]\n\nWhen \( D = - p \) with \(... | Yes |
Proposition 9.5.23 (J. Adams). If \( \left( {p - 1}\right) \nmid k \) then \( {B}_{k}/k \) is p-integral. | Proof. By Theorem 9.5.14, we already know that \( {B}_{k} \) is \( p \) -integral. Write \( k = {p}^{e}{k}_{0} \) with \( p \nmid {k}_{0} \) . Choosing \( n = {p}^{e} \) in Proposition 9.5.20, we see that \( \left( {{a}^{k} - 1}\right) {N}_{k} \equiv 0\left( {\;\operatorname{mod}\;{p}^{e}}\right) \) . Take for \( a \) ... | Yes |
Corollary 9.5.25. If \( k \) and \( {k}^{\prime } \) are even with \( \min \left( {k,{k}^{\prime }}\right) \geq e + 1 \), and \( p \) is a prime such that \( \left( {p - 1}\right) \nmid k \) and \( {k}^{\prime } \equiv k\left( {{\;\operatorname{mod}\;\phi }\left( {p}^{e}\right) }\right) \), then \( {B}_{{k}^{\prime }}/... | Proof. Clear from the above theorem since \( {B}_{k}/k \) is \( p \) -integral by Proposition 9.5.23. | No |
Lemma 9.5.28 (Hermite). Let \( p \) be a prime number and let \( n \geq 1 \) be an integer. We have the congruence\n\n\[ \mathop{\sum }\limits_{{1 \leq j \leq \left( {n - 1}\right) /\left( {p - 1}\right) }}\left( \begin{matrix} n \\ \left( {p - 1}\right) j \end{matrix}\right) \equiv 0\left( {\;\operatorname{mod}\;p}\ri... | Proof. By Lemma 2.5.1 we know that for \( m \geq 1 \) we have \( \mathop{\sum }\limits_{{a \in {\mathbb{F}}_{p}}}{a}^{m} = \) \( - {\delta }_{m, p - 1} \), where \( {\delta }_{m, p - 1} = 1 \) if \( \left( {p - 1}\right) \mid m \) and \( {\delta }_{m, p - 1} = 0 \) otherwise, while \( \mathop{\sum }\limits_{{a \in {\ma... | Yes |
Proposition 9.6.2. We have the functional equation\n\n\\[ \n\\zeta \\left( {s, x + 1}\\right) = \\zeta \\left( {s, x}\\right) - {x}^{-s}, \n\\]\n\nthe (partial) differential equation\n\n\\[ \n\\frac{\\partial \\zeta \\left( {s, x}\\right) }{\\partial x} = - {s\\zeta }\\left( {s + 1, x}\\right) \n\\]\n\nthe asymptotic f... | Proof. The first formula corresponds to changing \\( n \\) into \\( n + 1 \\), and the next two formulas are immediate by normal convergence of the series and its derivative. By definition we have \\( \\zeta \\left( {s,1}\\right) = \\zeta \\left( s\\right) \\) . Finally,\n\n\\[ \n\\zeta \\left( {s,1/2}\\right) = {2}^{s... | Yes |
Corollary 9.6.3. We have the following series expansion valid for \( \\left| y\\right| < x \) :\n\n\[ \n\\zeta \\left( {s, x + y}\\right) = \\mathop{\\sum }\\limits_{{k \\geq 0}}{\\left( -1\\right) }^{k}\\left( \\begin{matrix} s + k - 1 \\\\ k \\end{matrix}\\right) {y}^{k}\\zeta \\left( {s + k, x}\\right) .\n\] | Proof. Denote by RHS the right-hand side of the first formula, so that by definition of \( \\zeta \\left( {s, x}\\right) \) and of \( \\left( \\begin{array}{l} a \\\\ k \\end{array}\\right) \) we have\n\n\[ \n\\mathrm{{RHS}} = \\mathop{\\sum }\\limits_{{k \\geq 0}}\\left( \\begin{matrix} - s \\\\ k \\end{matrix}\\right... | Yes |
Lemma 9.6.4. For \( \left| y\right| < x \) we have the formulas\n\n\[ \zeta \left( {s, x}\right) = \frac{\zeta \left( {s - 1, x}\right) - \zeta \left( {s - 1, x + y}\right) }{y\left( {s - 1}\right) } + \mathop{\sum }\limits_{{k \geq 1}}{\left( -1\right) }^{k - 1}\left( \begin{matrix} s + k - 1 \\ k \end{matrix}\right) ... | Proof. The first formula is a simple rearrangement of terms of the first formula of the corollary. The second and third formulas follow by changing \( y \) into \( - y \) in the first formula and computing the sum and the difference. | No |
Corollary 9.6.5. We have the following formulas, valid for \( x > 1 \), except for the fourth, which is also valid for \( x > 1/2 \) :\n\n\[ \zeta \left( {s, x}\right) = \frac{{x}^{1 - s}}{s - 1} + \mathop{\sum }\limits_{{k \geq 1}}{\left( -1\right) }^{k - 1}\left( \begin{matrix} s + k - 1 \\ k \end{matrix}\right) \fra... | Proof. This follows by taking \( y = 1 \) and \( y = - 1 \) in the first formula of the lemma, \( y = 1 \) and \( y = 1/2 \) in the second formula, and \( y = 1 \) in the third formula. | Yes |
Proposition 9.6.6. The parameter \( x \in {\mathbb{R}}_{ > 0} \) being fixed, the function \( \zeta \left( {s, x}\right) \) (hence in particular the function \( \zeta \left( s\right) \) ) can be analytically continued to the whole complex plane to a meromorphic function with a single pole, at \( s = 1 \) , which is sim... | Proof. To prove this proposition, we can use any of the formulas of the above corollary. First note that since \( \zeta \left( {s, x + 1}\right) = \zeta \left( {s, x}\right) - {x}^{-s} \), it is enough to prove analytic continuation when \( x > 1 \) . In that case, since by Proposition 9.6.2 we know that \( \zeta \left... | Yes |
Corollary 9.6.9. As \( x \rightarrow \infty \) we have:\n\n(1) For \( \Re \left( s\right) \geq 1 \) and \( s \neq 1 \) ,\n\n\[ \zeta \left( {s, x}\right) = \frac{{x}^{1 - s}}{s - 1} + O\left( {x}^{-s}\right) . \]\n\n(2) For \( \Re \left( s\right) < 1 \) ,\n\n\[ \zeta \left( {s, x}\right) = - \frac{{x}^{1 - s}}{1 - s} +... | Proof. Clear by Proposition 9.6.7. | No |
Corollary 9.6.10. If \( k \in {\mathbb{Z}}_{ \geq 1} \) we have\n\n\[ \zeta \left( {1 - k, x}\right) = - \frac{{B}_{k}\left( x\right) }{k} \]\n\nand in particular \( \zeta \left( {1 - k}\right) = - {B}_{k}/k - {\delta }_{k,1} \) . | Proof. Setting \( \alpha = k - 1 \) in Proposition 9.6.7, we find that for \( n \geq k \) ,\n\n\[ - {k\zeta }\left( {1 - k, x}\right) = {x}^{k} + \mathop{\sum }\limits_{{j = 1}}^{n}\left( \begin{array}{l} k \\ j \end{array}\right) {B}_{j}{x}^{k - j} = {B}_{k}\left( x\right) . \] | Yes |
Proposition 9.6.11. As \( x \rightarrow 0 \) we have\n\n\[ \zeta \left( {s, x}\right) = \left\{ \begin{array}{ll} {x}^{-s} + \zeta \left( s\right) + o\left( 1\right) & \text{ if }\Re \left( s\right) \geq 0, \\ 1/2 + o\left( 1\right) & \text{ if }s = 0, \\ \zeta \left( s\right) + o\left( 1\right) & \text{ if }\Re \left(... | Proof. For \( s \neq - {2k} \) with \( k \in {\mathbb{Z}}_{ \geq 1} \) this immediately follows from\n\n\[ \zeta \left( {s, x}\right) = {x}^{-s} + \zeta \left( {s, x + 1}\right) = {x}^{-s} + \zeta \left( s\right) + o\left( 1\right) . \]\n\nFor \( s = - {2k} \), by the above corollary we have \( \zeta \left( {-{2k}, x}\... | No |
Proposition 9.6.12. We have the duplication formula\n\n\[ \zeta \left( {s, x}\right) + \zeta \left( {s, x + \frac{1}{2}}\right) = {2}^{s}\zeta \left( {s,{2x}}\right) \] | Proof. Follows from an easy rearrangement of terms and left to the reader (Exercise 64). | No |
Proposition 9.6.14. For all \( x \in {\mathbb{R}}_{ > 0} \) we have \( \Gamma \left( {x + 1}\right) = {x\Gamma }\left( x\right) \) and when \( n \in {\mathbb{Z}}_{ \geq 1} \) we have \( \Gamma \left( n\right) = \left( {n - 1}\right) ! \) . | Proof. Since \( \zeta \left( {s, x + 1}\right) = \zeta \left( {s, x}\right) - {x}^{-s} \) we obtain the first formula by derivation with respect to \( x \) . The second follows by induction since \( \log \left( {\Gamma \left( 1\right) }\right) = {\zeta }^{\prime }\left( {0,1}\right) - {\zeta }^{\prime }\left( {0,1}\rig... | No |
Proposition 9.6.15. (1) Let \( u \in {\mathbb{R}}_{ > 0} \) . For \( \left| x\right| < u \) we have\n\n\[ \log \left( {\Gamma \left( {x + u}\right) }\right) = \log \left( {\Gamma \left( u\right) }\right) + \psi \left( u\right) x + \mathop{\sum }\limits_{{k \geq 2}}{\left( -1\right) }^{k}\frac{\zeta \left( {k, u}\right)... | Proof. This follows by differentiating with respect to \( s \) the first and second formulas of Corollary 9.6.3, and using the fact that around \( s = 1 \) we have \( \zeta \left( {s, u}\right) = 1/\left( {s - 1}\right) - \psi \left( u\right) + O\left( {s - 1}\right) \), and in particular \( \zeta \left( s\right) = 1/\... | Yes |
Proposition 9.6.16. For \( x > 0 \) we have for any \( k \geq 1 \) ,\n\n\[ \n{\zeta }^{\prime }\left( {0, x}\right) = \left( {x - \frac{1}{2}}\right) \log \left( x\right) - x + \mathop{\sum }\limits_{{j = 1}}^{{k - 1}}\frac{{B}_{j + 1}}{j\left( {j + 1}\right) {x}^{j}} - \frac{1}{k}{\int }_{0}^{\infty }\frac{{B}_{k}\lef... | Proof. This follows by derivation after a short computation from the formula for \( \zeta \left( {-\alpha, x}\right) \) given in Proposition 9.6.7. | No |
For all \( x \in \mathbb{C} \smallsetminus {\mathbb{Z}}_{ \leq 0} \) set\n\n\[ \n{u}_{N}\left( x\right) = \frac{{N}^{x - 1}N!}{x\left( {x + 1}\right) \cdots \left( {x + N - 1}\right) }.\n\]\n\nThen for all \( x \in {\mathbb{R}}_{ > 0} \) we have\n\n\[ \n\Gamma \left( x\right) = \mathop{\lim }\limits_{{N \rightarrow \in... | Proof. By differentiating the first formula of Proposition 9.6.7 a short computation gives\n\n\[ \n\mathop{\sum }\limits_{{m = 0}}^{{N - 1}}\log \left( {m + x}\right) = - {\zeta }^{\prime }\left( {0, x}\right) + \left( {N + x - \frac{1}{2}}\right) \log \left( {N + x}\right) \n\]\n\n\[ \n- \left( {N + x}\right) - {\int ... | No |
Proposition 9.6.19. The above limit exists and defines a meromorphic function on \( \mathbb{C} \) that generalizes the real gamma function defined above for \( s \in {\mathbb{R}}_{ > 0} \). It has no zeros in \( \mathbb{C} \), and it has simple poles on \( {\mathbb{Z}}_{ \leq 0} \), the residue at \( s = - k \) being e... | Proof. Since\n\n\[ \n{u}_{N + 1}\left( s\right) /{u}_{N}\left( s\right) = {\left( 1 + 1/N\right) }^{s}/\left( {1 + s/N}\right) = 1 + O\left( {1/{N}^{2}}\right) , \n\] \n\nit is clear that the limit exists and that it converges uniformly on any compact subset of \( \mathbb{C} \smallsetminus {\mathbb{Z}}_{ \leq 0} \). It... | Yes |
Corollary 9.6.20. There exists a unique holomorphic function \( \log \Gamma \left( s\right) \) defined on the simply connected set \( \mathbb{C} \smallsetminus {\mathbb{R}}_{ \leq 0} \) such that \( \exp \left( {\operatorname{Log}\Gamma \left( s\right) }\right) = \Gamma \left( s\right) \) and \( \log \Gamma \left( 1\ri... | Proof. Let \( \Omega = \mathbb{C} \smallsetminus {\mathbb{R}}_{ \leq 0} \) . Since \( \Gamma \left( s\right) \) has no zeros or poles on \( \Omega \) and since \( \Omega \) is simply connected, it follows that there exists a holomorphic function \( \log \left( {\Gamma \left( s\right) }\right) \) defined on \( \Omega \)... | Yes |
\[ {\int }_{-\infty }^{\infty }{e}^{-{t}^{2}}{dt} = \sqrt{\pi } \] | Proof. There are several classical proofs of this result, and we give two. The first is using polar coordinates. Set \( {I}_{N} = {\int }_{-N}^{N}{e}^{-{t}^{2}}{dt} \) . Then \( {I}_{N}^{2} = \) \( {\int }_{S\left( N\right) }{e}^{-\left( {{t}^{2} + {u}^{2}}\right) }{dtdu} \), where \( S\left( N\right) \) is the square ... | Yes |
Proposition 9.6.22 (Stirling’s formula). As \( n \rightarrow \infty \) we have\n\n\[ n! \sim {n}^{n}{e}^{-n}\sqrt{2\pi n} \]\n\nor equivalently,\n\n\[ \log \left( {n!}\right) = \left( {n + \frac{1}{2}}\right) \log \left( n\right) - n + \frac{1}{2}\log \left( {2\pi }\right) + o\left( 1\right) . \] | Proof. Once again there are several classical proofs. Certainly the most classical is as follows: if we set \( {u}_{n} = \log \left( {n!/\left( {{n}^{n}{e}^{-n}\sqrt{n}}\right) }\right) \) then\n\n\[ {u}_{n + 1} - {u}_{n} = 1 - \left( {n + \frac{1}{2}}\right) \log \left( {1 + \frac{1}{n}}\right) \sim - \frac{1}{{12}{n}... | Yes |
Corollary 9.6.23. For any \( m \geq 1 \) we have\n\n\[ \log \Gamma \left( s\right) = \left( {s - \frac{1}{2}}\right) \log \left( s\right) - s + \frac{\log \left( {2\pi }\right) }{2} + \mathop{\sum }\limits_{{k = 1}}^{m}\frac{{B}_{2k}}{{2k}\left( {{2k} - 1}\right) {s}^{{2k} - 1}} \]\n\n\[ - \frac{1}{{2m} + 1}{\int }_{0}... | Proof. Clear from Stirling's formula and Proposition 9.6.16. | No |
Proposition 9.6.24. We have the following expansions:\n\n(1)\n\n\[ \pi \operatorname{cotan}\left( {\pi x}\right) = \frac{1}{x} + {2x}\mathop{\sum }\limits_{{n \geq 1}}\frac{1}{{x}^{2} - {n}^{2}}. \]\n\n(2)\n\n\[ {\left( \frac{\pi }{\sin \left( {\pi x}\right) }\right) }^{2} = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}\f... | Proof. Let \( a \notin \mathbb{Z} \) be a parameter, and define \( f\left( x\right) \) to be the \( {2\pi } \) -periodic function such that \( f\left( x\right) = \cos \left( {ax}\right) \) for \( - \pi \leq x \leq \pi \) . This function is clearly continuous and piecewise differentiable. It is thus everywhere equal to ... | Yes |
Proposition 9.6.25. For all \( s \) such that \( \Re \left( s\right) > 0 \) we have\n\n\[ \log \left( s\right) = {\int }_{0}^{\infty }\frac{{e}^{-t} - {e}^{-{st}}}{t}{dt} \]\n\nwhere the left-hand side is the principal determination of the logarithm. More generally, if \( \Re \left( {s}_{1}\right) > 0 \) and \( \Re \le... | Proof. Let \( I\left( s\right) \) be the first integral above. It is clearly absolutely convergent for \( \Re \left( s\right) > 0 \), and its (for the moment formal) derivative with respect to \( s \) is \( {\int }_{0}^{\infty }{e}^{-{st}}{dt} \), which is normally convergent in the domain \( \Re \left( s\right) \geq \... | Yes |
Proposition 9.6.26 (Hadamard product). We have\n\n\[ \Gamma \left( {s + 1}\right) = {e}^{-{\gamma s}}\mathop{\prod }\limits_{{n \geq 1}}\frac{{e}^{s/n}}{1 + s/n}, \]\n\nwhere \( \gamma = {0.57721}\ldots \) is Euler’s constant. | Proof. If we divide the numerator and the denominator of \( {u}_{n}\left( s\right) \) by \( n! = \) \( 1 \cdot 2\cdots n \) we obtain\n\n\[ {u}_{n}\left( {s + 1}\right) = \frac{{n}^{s}}{\mathop{\prod }\limits_{{1 \leq k \leq n}}\left( {1 + s/k}\right) } = {n}^{s}{e}^{-s{H}_{n}}\mathop{\prod }\limits_{{1 \leq k \leq n}}... | Yes |
Corollary 9.6.28. (1) As \( x \rightarrow \infty \) in \( \mathbb{R} \) we have \( \Gamma \left( x\right) \sim {x}^{x - 1/2}{e}^{-x}{\left( 2\pi \right) }^{1/2} \). | Proof. The first statement is clear. | No |
Proposition 9.6.29. We have the following integral representation, valid for \( \Re \left( s\right) > - 1 \) :\n\n\[ \log \Gamma \left( {s + 1}\right) = {\int }_{0}^{\infty }\left( {s\frac{{e}^{-t}}{t} - \frac{1 - {e}^{-{st}}}{t\left( {{e}^{t} - 1}\right) }}\right) {dt}. \] | Proof. By Corollary 9.6.20, as \( n \rightarrow \infty \) we have \( \log \Gamma \left( {s + 1}\right) = s\log \left( n\right) - \) \( \mathop{\sum }\limits_{{1 \leq k \leq n}}\log \left( {\left( {s + k}\right) /k}\right) + o\left( 1\right) \), so by Proposition 9.6.25 we have\n\n\[ \log \Gamma \left( {s + 1}\right) = ... | Yes |
Corollary 9.6.30. We have the following integral representations:\n\n\[ \gamma = {\int }_{0}^{\infty }\left( {\frac{1}{{e}^{t} - 1} - \frac{{e}^{-t}}{t}}\right) {dt} \]\n\nand for \( k \geq 2 \)\n\n\[ \zeta \left( k\right) = \frac{1}{\left( {k - 1}\right) !}{\int }_{0}^{\infty }\frac{{t}^{k - 1}}{{e}^{t} - 1}{dt} \] | Proof. This follows by expanding in powers of \( s \) the integrand of the proposition and comparing with Proposition 9.6.15. | No |
Corollary 9.6.31. We have the integral representation\n\n\[ \n\frac{1}{2}\log \left( {2\pi }\right) = {\int }_{0}^{\infty }\left( {\frac{1}{t} - \frac{{e}^{-t}}{2} - \frac{1}{{e}^{t} - 1}}\right) \frac{dt}{t}.\n\] | Proof. Integrating the first formula of Proposition 9.6.25 we obtain\n\n\[ \ns\log \left( s\right) - s = {\int }_{0}^{\infty }\left( {s\frac{{e}^{-t}}{t} - \frac{1 - {e}^{-{st}}}{{t}^{2}}}\right) {dt}.\n\]\n\nSubtracting this and \( \log \left( s\right) /2 \) given by Proposition 9.6.25 from the integral representation... | No |
Corollary 9.6.32. For \( \Re \left( s\right) > 0 \) we have\n\n\[ \log \Gamma \left( s\right) = \left( {s - \frac{1}{2}}\right) \log \left( s\right) - s + \frac{1}{2}\log \left( {2\pi }\right) + {\int }_{0}^{\infty }\frac{{e}^{-{st}}}{t}\left( {\frac{1}{{e}^{t} - 1} - \frac{1}{t} + \frac{1}{2}}\right) {dt}. \] | Proof. Immediate from Propositions 9.6.29 and the integral representations of \( s\log \left( s\right) - s,\log \left( s\right) \), and \( \log \left( {2\pi }\right) /2 \) given above. | No |
Proposition 9.6.33. We have the duplication formula\n\n\[ \Gamma \left( s\right) \Gamma \left( {s + 1/2}\right) = {2}^{1 - {2s}}{\pi }^{1/2}\Gamma \left( {2s}\right) \] | Proof. Differentiating with respect to \( s \) the formula of Proposition 9.6.12, setting \( s = 0 \), and using Definition 9.6.13 gives\n\n\[ \mathop{\sum }\limits_{{0 \leq j < N}}\log \left( {\Gamma \left( {x + j/N}\right) }\right) = \log \left( {\Gamma \left( {Nx}\right) }\right) - \left( {N - 1}\right) {\zeta }^{\p... | Yes |
Proposition 9.6.34. We have the reflection formula\n\n\[ \Gamma \left( s\right) \Gamma \left( {1 - s}\right) = \frac{\pi }{\sin \left( {\pi s}\right) }.\] | Proof. By Propositions 9.6.26 and 9.6.24 we have\n\n\[ \Gamma \left( {1 + s}\right) \Gamma \left( {1 - s}\right) = \mathop{\prod }\limits_{{n \geq 1}}\left( {1 - {s}^{2}/{n}^{2}}\right) = {\pi s}/\sin \left( {\pi s}\right) ,\]\n\nso the result follows since \( \Gamma \left( {1 + s}\right) = {s\Gamma }\left( s\right) \)... | Yes |
Proposition 9.6.35. For \( \Re \left( s\right) > 0 \) we have\n\n\[ \Gamma \left( s\right) = {\int }_{0}^{\infty }{t}^{s}{e}^{-t}\frac{dt}{t} \] | Proof. Recall that for any function \( f \in {C}^{\infty }\left( \left\lbrack {0,1}\right\rbrack \right) \) we have Taylor’s formula (which is trivially proved by integration by parts)\n\n\[ f\left( 1\right) = \mathop{\sum }\limits_{{0 \leq k \leq n}}\frac{{f}^{\left( k\right) }\left( 0\right) }{k!} + \frac{1}{n!}{\int... | Yes |
Corollary 9.6.36. Let \( a \) and \( s \) be two complex numbers such that \( \Re \left( a\right) > 0 \) and \( \Re \left( s\right) > 0 \), or \( \Re \left( a\right) = 0, a \neq 0 \), and \( 0 < \Re \left( s\right) < 1 \) . Then\n\n\[{\int }_{0}^{\infty }{t}^{s}{e}^{-{at}}\frac{dt}{t} = {a}^{-s}\Gamma \left( s\right)\]... | Proof. The integral converges (absolutely) at \( t = 0 \) if and only if \( \Re \left( s\right) > 0 \) . When \( \Re \left( a\right) > 0 \) it is clear that the integral converges absolutely at infinity. When \( \Re \left( a\right) = 0 \) and \( a \neq 0 \), then \( f\left( t\right) = {e}^{-{at}} \) is such that \( {\i... | Yes |
Corollary 9.6.37. For \( 0 < \Re \left( s\right) < 1 \) we have\n\n\[{\int }_{0}^{\infty }{t}^{s - 1}\cos \left( t\right) {dt} = \cos \left( \frac{\pi s}{2}\right) \Gamma \left( s\right) \;\text{ and }\;{\int }_{0}^{\infty }{t}^{s - 1}\sin \left( t\right) {dt} = \sin \left( \frac{\pi s}{2}\right) \Gamma \left( s\right)... | Proof. Clear. | No |
Proposition 9.6.38. We have\n\n\[ \n{\int }_{0}^{\infty }\frac{\sin \left( {xt}\right) }{t}{dt} = \frac{\pi }{2}\operatorname{sign}\left( x\right) \n\]\n\nand for \( a > 0 \) ,\n\n\[ \n{\int }_{0}^{a}\frac{\cos \left( {xt}\right) - 1}{t}{dt} + {\int }_{a}^{\infty }\frac{\cos \left( {xt}\right) }{t}{dt} = - \left( {\gam... | Proof. By integration by parts we have for \( s < 1 \) ,\n\n\[ \n{\int }_{1}^{\infty }{t}^{s}{e}^{ixt}\frac{dt}{t} = - \frac{{e}^{ix}}{ix} + \frac{1 - s}{ix}{\int }_{1}^{\infty }\frac{{e}^{ixt}}{{t}^{2 - s}}{dt}. \n\]\n\nThis last integral is absolutely convergent, so that\n\n\[ \n\mathop{\lim }\limits_{{s \rightarrow ... | Yes |
Proposition 9.6.39. For \( \Re \left( a\right) > 0 \) and \( \Re \left( b\right) > 0 \) define the beta function \( B\left( {a, b}\right) {by} \n\n\[ \nB\left( {a, b}\right) = {\int }_{0}^{1}{t}^{a - 1}{\left( 1 - t\right) }^{b - 1}{dt}. \n\] \n\nThen \n\[ \nB\left( {a, b}\right) = \frac{\Gamma \left( a\right) \Gamma \... | Proof. By Proposition 9.6.35 we have \n\n\[ \n\Gamma \left( a\right) \Gamma \left( b\right) = {\int }_{0}^{\infty }{\int }_{0}^{\infty }{t}^{a - 1}{u}^{b - 1}{e}^{-\left( {t + u}\right) }{dtdu}, \n\] \n\nso that setting \( v = t + u \) we obtain \n\n\[ \n\Gamma \left( a\right) \Gamma \left( b\right) = {\int }_{0}^{\inf... | Yes |
Corollary 9.6.40. (1) For \( 0 < \Re \left( a\right) < 2\Re \left( b\right) \) we have\n\n\[ \n{\int }_{0}^{\infty }\frac{{t}^{a - 1}}{{\left( 1 + {t}^{2}\right) }^{b}}{dt} = \frac{1}{2}B\left( {a/2, b - a/2}\right) = \frac{\Gamma \left( {a/2}\right) \Gamma \left( {b - a/2}\right) }{{2\Gamma }\left( b\right) }.\n\] | Proof. If we set \( u = {t}^{2}/\left( {1 + {t}^{2}}\right) \) then \( 1 + {t}^{2} = 1/\left( {1 - u}\right), t = {\left( u/\left( 1 - u\right) \right) }^{1/2} \) , and \( {dt} = {du}/\left( {2{\left( 1 - u\right) }^{3/2}{u}^{1/2}}\right) \), so\n\n\[ \n{\int }_{0}^{\infty }\frac{{t}^{a - 1}}{{\left( 1 + {t}^{2}\right)... | Yes |
Proposition 9.6.41. Let \( k \in {\mathbb{Z}}_{ \geq 0} \). (1) We have \[ \psi \left( s\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}\left( {\log \left( N\right) - \mathop{\sum }\limits_{{n = 0}}^{N}\frac{1}{n + s}}\right) \] | \[ = - \gamma + \left( {s - 1}\right) \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{\left( {n + 1}\right) \left( {n + s}\right) } = - \gamma + \mathop{\sum }\limits_{{n = 0}}^{\infty }\left( {\frac{1}{n + 1} - \frac{1}{n + s}}\right) . \] | Yes |
Proposition 9.6.43. We have for \( \Re \left( s\right) > - 1 \) ,\n\n\[ \n\psi \left( {s + 1}\right) = {\int }_{0}^{\infty }\left( {\frac{{e}^{-t}}{t} - \frac{{e}^{-{st}}}{{e}^{t} - 1}}\right) {dt} \n\]\n\n\[ \n= - \gamma + {\int }_{0}^{\infty }\frac{1 - {e}^{-{st}}}{{e}^{t} - 1}{dt} = - \gamma + {\int }_{0}^{1}\frac{1... | Proofs. Immediate and left to the reader (Exercise 77). | No |
Proposition 9.6.45. Let \( f \) be a rational function, and write\n\n\[ f\left( x\right) = C\mathop{\prod }\limits_{{\alpha \text{ zero or pole }}}{\left( x - \alpha \right) }^{v\left( \alpha \right) },\]\n\nwhere \( \alpha \) runs through the zeros and poles of \( f, v\left( \alpha \right) \in \mathbb{Z} \) is the ord... | Proof. Again immediate and left to the reader (Exercise 78). | No |
Proposition 9.6.46. Assume that \( 0 < r < m \) are integers, and set as usual \( {\zeta }_{m} = \exp \left( {{2i\pi }/m}\right) \) . We have\n\n\[ \psi \left( \frac{r}{m}\right) = - \gamma - \log \left( m\right) + \mathop{\sum }\limits_{{1 \leq k \leq m - 1}}{\zeta }_{m}^{-{rk}}\log \left( {1 - {\zeta }_{m}^{k}}\right... | Proof. By Proposition 9.6.41 we have\n\n\[ \psi \left( {r/m}\right) = - \gamma + \mathop{\sum }\limits_{{n \geq 0}}\left( {\frac{1}{n + 1} - \frac{m}{{mn} + r}}\right) . \]\n\nBy Abel's theorem on the continuity of power series on their circle of convergence, since \( 1/\left( {n + 1}\right) - m/\left( {{mn} + r}\right... | Yes |
Proposition 9.6.47. We have\n\n\[ \mathop{\sum }\limits_{{1 \leq r \leq m}}\psi \left( \frac{r}{m}\right) {e}^{{2i\pi ar}/m} = \left\{ \begin{array}{ll} m\log \left( \left| {2\sin \left( \frac{\pi a}{m}\right) }\right| \right) + {im\pi }\left( {\left\{ \frac{a}{m}\right\} - \frac{1}{2}}\right) & \text{ if }m \nmid a, \... | Proof. Left to the reader (Exercise 103). | No |
Corollary 9.6.51. For \( x \in \mathbb{R} \smallsetminus \mathbb{Z} \) and \( \Re \left( s\right) < 1 \) we have\n\n\[ \zeta \left( {s,\{ x\} }\right) = 2{\left( 2\pi \right) }^{s - 1}\Gamma \left( {1 - s}\right) \mathop{\sum }\limits_{{n \geq 1}}\frac{\sin \left( {{2\pi nx} + {s\pi }/2}\right) }{{n}^{1 - s}}. \] | Proof. The variable \( s \) being fixed, the periodic function \( \zeta \left( {s,\{ x\} }\right) \) is a piecewise \( {C}^{\infty } \) function with simple discontinuities at the integers, so the corollary follows by a simple computation from the proposition and the fundamental theorem on Fourier series, which implies... | No |
Corollary 9.6.52. For \( x \in \mathbb{R} \smallsetminus \mathbb{Z} \) and \( \Re \left( s\right) > 0 \) we have\n\n\[ \mathop{\sum }\limits_{{n \geq 1}}\frac{\cos \left( {2\pi nx}\right) }{{n}^{s}} = \frac{{\left( 2\pi \right) }^{s}}{{4\Gamma }\left( s\right) \cos \left( {{s\pi }/2}\right) }\left( {\zeta \left( {1 - s... | Proof. Immediate and left to the reader. | No |
For all \( x \in \mathbb{R} \smallsetminus \mathbb{Z} \) the Fourier expansion of \( \log \left( {\Gamma \left( {\{ x\} }\right) }\right) \) is given by\n\n\[ \log \left( {\Gamma \left( {\{ x\} }\right) }\right) = \frac{1}{2}\log \left( {2\pi }\right) + \frac{1}{2}\mathop{\sum }\limits_{{n \geq 1}}\frac{\cos \left( {2\... | Proof. Using for instance Abel summation, we note that for fixed \( x \notin \) \( \mathbb{Z} \), the series \( \mathop{\sum }\limits_{{n \geq 1}}\log \left( n\right) {e}^{2i\pi nx}{n}^{-s} \) is uniformly convergent in any compact subset of the right half-plane \( \Re \left( s\right) > 0 \) . It follows that we can di... | Yes |
Theorem 9.7.1 (Monotone convergence theorem). Let \( X \subset \mathbb{R} \), and let \( {f}_{n} \) be a sequence of measurable functions on \( X \) such that \( f\left( x\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( x\right) \) exists for every \( x \in X \) . If for all \( x \in X \) we have ... | \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{X}{f}_{n}\left( X\right) {dx} = {\int }_{X}f\left( x\right) {dx}. \] | No |
Theorem 9.7.2 (Dominated convergence theorem). Let \( X \subset \mathbb{R} \), and let \( {f}_{n} \) be a sequence of measurable functions on \( X \) such that \( f\left( x\right) = \mathop{\lim }\limits_{{n \rightarrow infty }}{f}_{n}\left( x\right) \) exists for every \( x \in X \) . If there exists a function \( g \... | \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{X}\left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| {dx} = 0\;\text{ and }\;\mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{X}{f}_{n}\left( x\right) {dx} = {\int }_{X}f\left( x\right) {dx}. \] | Yes |
Theorem 9.7.5. (1) (Inversion formula.) Assume that both \( f \) and \( \mathcal{F}\left( f\right) \) are in \( {L}^{1}\left( \mathbb{R}\right) \) . Then\n\n\[ f\left( x\right) = {\int }_{-\infty }^{\infty }{e}^{2i\pi xt}\mathcal{F}\left( f\right) \left( t\right) {dt} \]\n\nfor all \( x \), where \( f \) is continuous.... | Proof. Set for \( T > 0 \),\n\n\[ {f}_{T}\left( x\right) = {\int }_{-T}^{T}\left( {1 - \frac{\left| t\right| }{T}}\right) {e}^{2i\pi xt}\mathcal{F}\left( f\right) \left( t\right) {dt}. \]\n\nReplacing \( \mathcal{F}\left( f\right) \left( t\right) \) by its expression we find after an easy computation that\n\n\[ {f}_{T}... | No |
Proposition 9.7.7. Assume that \( f \) is continuous on \( \rbrack 0,\infty \lbrack \), that \( f\left( t\right) = \) \( O\left( {t}^{-\alpha }\right) \) for some \( \alpha \in \mathbb{R} \) as \( t \rightarrow 0 \), and that \( f\left( t\right) \) tends to 0 faster than any power of \( t \) as \( t \rightarrow \infty ... | Proof. (1). By our assumptions on \( f \) the integral converges absolutely in a neighborhood of infinity, and in any compact interval not containing 0 . Since \( \left| {{t}^{s - 1}f\left( t\right) }\right| = O\left( {t}^{\Re \left( s\right) - \alpha - 1}\right) \), the integral converges absolutely also at 0 when \( ... | Yes |
Proposition 9.7.9. Assume that \( f \) is piecewise continuous on \( \rbrack 0,\infty \lbrack \), that \( f\left( t\right) = O\left( {t}^{-\alpha }\right) \) for some \( \alpha < 1 \) as \( t \rightarrow 0 \), and that \( {e}^{-{at}}f\left( t\right) \) tends to 0 for all \( a > 0 \) as \( t \rightarrow \infty \) .\n\n(... | Proof. Thanks to the assumptions made on \( f \) ,(1) and the first formula of (2) are clear, and the second is immediate by integration by parts. For (3) we make the change of variable \( u = {e}^{-t} \), which gives for \( x > 0 \) ,\n\n\[ \mathcal{L}\left( f\right) \left( x\right) = {\int }_{0}^{1}{u}^{x - 1}f\left(... | Yes |
Proposition 9.8.1. For \( \nu \in \mathbb{C} \), let \( {E}_{\nu } \) be the differential equation\n\n\[ \n{y}^{\prime \prime } + \frac{{y}^{\prime }}{x} + \left( {1 - \frac{{\nu }^{2}}{{x}^{2}}}\right) y = 0.\n\]\n\n(1) When \( \nu \notin \mathbb{Z} \), a basis of the space of solutions of \( {E}_{\nu } \) is given by... | Proof. This is a classical undergraduate exercise, so we only give a sketch (see Exercise 110). If \( \nu \notin \mathbb{Z} \), we can set \( y = {x}^{\pm \nu }\mathop{\sum }\limits_{{k \geq 0}}{a}_{k}{x}^{k} \) with \( {a}_{0} \neq \) 0 . The differential equation gives \( {a}_{1} = 0 \) and a simple recurrence for \(... | No |
Proposition 9.8.2. For \( \nu \in \mathbb{C} \), let \( {F}_{\nu } \) be the differential equation\n\n\[ \n{y}^{\prime \prime } + \frac{{y}^{\prime }}{x} - \left( {1 + \frac{{\nu }^{2}}{{x}^{2}}}\right) y = 0.\n\]\n\n(1) When \( \nu \notin \mathbb{Z} \), a basis of the space of solutions of \( {F}_{\nu } \) is given by... | Proof. Exactly the same proof as the preceding proposition. Note that \( {T}_{\pm \nu }\left( x\right) = {S}_{\pm \nu }\left( {ix}\right) . \) | No |
Proposition 9.8.4. We have\n\n\[ \n{J}_{\nu - 1}\left( x\right) + {J}_{\nu + 1}\left( x\right) = \frac{2\nu }{x}{J}_{\nu }\left( x\right) ,\;{J}_{\nu - 1}\left( x\right) - {J}_{\nu + 1}\left( x\right) = 2{J}_{\nu }^{\prime }\left( x\right) ,\n\]\n\n\[ \n{Y}_{\nu - 1}\left( x\right) + {Y}_{\nu + 1}\left( x\right) = \fra... | Proof. Immediate from the series expansions and the definitions of \( Y \) and \( K \), using \( \Gamma \left( {\nu + k + 1}\right) = \left( {\nu + k}\right) \Gamma \left( {\nu + k}\right) \), and left to the reader (Exercise 116). | No |
Proposition 9.8.5. When \( \nu \in \left( {1/2}\right) + \mathbb{Z} \) the four Bessel functions are elementary functions. More precisely:\n\n(1) We have\n\n\[ \n{J}_{1/2}\left( x\right) = {Y}_{-1/2}\left( x\right) = \sqrt{\frac{2}{\pi x}}\sin \left( x\right) ,\;{J}_{-1/2}\left( x\right) = - {Y}_{1/2}\left( x\right) = ... | Proof. The formulas for \( {J}_{\pm 1/2}\left( x\right) \) and \( {I}_{\pm 1/2}\left( x\right) \) follow immediately from the power series expansion, using the formula\n\n\[ \n\Gamma \left( {k + 3/2}\right) = \left( {{2k} + 1}\right) !\sqrt{\pi }/\left( {k!{2}^{{2k} + 1}}\right) ,\n\]\n\nwhich is an immediate consequen... | "No" |
Proposition 9.8.8. For \( 0 < \Re \left( s\right) < 1 \) we have the following Mellin transforms:\n\n\[{\int }_{0}^{\infty }{t}^{s - 1}{J}_{0}\left( t\right) {dt} = \frac{{2}^{s - 1}}{\pi }\sin \left( \frac{\pi s}{2}\right) \Gamma {\left( s/2\right) }^{2},\] | Proof. All of these formulas immediately follow from the corresponding integral representations, so we simply prove the first. Exchanging the orders of integration, which is legal since \( 0 < \Re \left( s\right) < 1 \), and making the change of variable \( y = x\sin \left( t\right) \), we have\n\n\[{\int }_{0}^{\infty... | Yes |
Theorem 9.8.9. For \( x > 0 \) and \( \Re \left( s\right) > 1/2 \) we have\n\n\[{\int }_{0}^{\infty }\frac{\cos \left( {xt}\right) }{{\left( {t}^{2} + 1\right) }^{s}}{dt} = \frac{{\pi }^{1/2}{\left( x/2\right) }^{s - 1/2}}{\Gamma \left( s\right) }{K}_{s - 1/2}\left( x\right) .\] | Proof. Set \( {I}_{s}\left( x\right) = {\int }_{0}^{\infty }\cos \left( {xt}\right) /{\left( {t}^{2} + 1\right) }^{s}{dt} \) . Using the integral definition of \( \Gamma \left( s\right) \) we have by Fubini’s theorem\n\n\[ \Gamma \left( s\right) {I}_{s}\left( x\right) = {\int }_{0}^{\infty }\cos \left( {xt}\right) \lef... | Yes |
Proposition 10.1.1. Let \( a\left( n\right), b\left( n\right) \), and \( c\left( n\right) \) be three arithmetic functions. The following two conditions are equivalent:\n\n(1)\n\n\[ L\left( {c, s}\right) = L\left( {a, s}\right) L\left( {b, s}\right) .\n\]\n\n(2) For all \( n \geq 1 \) ,\n\n\[ c\left( n\right) = \mathop... | Proof. We simply write\n\n\[ L\left( {a, s}\right) L\left( {b, s}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{a\left( n\right) }{{n}^{s}}\mathop{\sum }\limits_{{m = 1}}^{\infty }\frac{b\left( m\right) }{{m}^{s}} = \mathop{\sum }\limits_{{n, m \geq 1}}\frac{a\left( n\right) b\left( m\right) }{{\left( nm\righ... | Yes |
Proposition 10.1.5 (Möbius inversion formula). Let \( a \) and \( b \) be arithmetic functions.\n\n(1) (First form.) Assume that \( b\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}a\left( d\right) \). Then\n\n\[ a\left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) b\left( {n/d}\right) = \mathop... | Proof. By definition, \( b \) is the arithmetic convolution of \( a \) with \( \mathbf{1} \) ; hence \( L\left( {b, s}\right) = L\left( {a, s}\right) \zeta \left( s\right) \), so that \( L\left( {a, s}\right) = L\left( {b, s}\right) /\zeta \left( s\right) = L\left( {\mu, s}\right) L\left( {b, s}\right) \), and (1) foll... | Yes |
Proposition 10.1.6 (Evaluation of Ramanujan sums). We have\n\n\[ R\left( {m, a}\right) = \mathop{\sum }\limits_{\substack{{x{\;\operatorname{mod}\;m}} \\ {\gcd \left( {x, m}\right) = 1} }}\exp \left( {{2i\pi ax}/m}\right) = \mathop{\sum }\limits_{{d \mid \gcd \left( {m, a}\right) }}\mu \left( {m/d}\right) d. \] | Proof. Using the above argument, setting \( x = {dy} \) we have\n\n\[ R\left( {m, a}\right) = \mathop{\sum }\limits_{{d \mid m}}\mu \left( d\right) \mathop{\sum }\limits_{{y{\;\operatorname{mod}\;m}/d}}\exp \left( {{2i\pi ay}/\left( {m/d}\right) }\right) . \]\n\nThe inner sum is now an honest geometric series that vani... | Yes |
Proposition 10.1.8. A function a is multiplicative if and only if \( L\left( {a, s}\right) \) has a formal Euler product, i.e., can be written formally as\n\n\[ L\left( {a, s}\right) = \mathop{\prod }\limits_{p}{L}_{p}\left( {a, s}\right) ,\text{ where }{L}_{p}\left( {a, s}\right) = 1 + \mathop{\sum }\limits_{{k = 1}}^... | Proof. If we expand formally the product \( {L}_{p}\left( {a, s}\right) \), we obtain a formal Dirichlet series of the form \( \mathop{\sum }\limits_{{n \geq 1}}b\left( n\right) /{n}^{s} \), where\n\n\[ b\left( n\right) = \mathop{\prod }\limits_{{{p}^{k}\parallel n}}a\left( {p}^{k}\right) . \]\n\nThe notation \( {p}^{k... | Yes |
Corollary 10.1.9. If \( a \) and \( b \) are multiplicative functions, then so is the arithmetic convolution of \( a \) and \( b \), and if \( a \) is invertible, then its inverse is also multiplicative. | Proof. Clear from the above interpretation of multiplicativity in terms of formal Euler products. Of course this can also be proved directly. | No |
Proposition 10.1.11(1) We have\n\n\[ \phi \left( n\right) = n\mathop{\prod }\limits_{{p \mid n}}\left( {1 - \frac{1}{p}}\right) \]\n\nwhere the product is over primes dividing \( n \) . | Proof. By expanding the formal power series we have\n\n\[ \frac{1 - 1/{p}^{s}}{1 - p/{p}^{s}} = 1 + \frac{p - 1}{{p}^{s}} + \frac{{p}^{2} - p}{{p}^{2s}} + \cdots = 1 + \mathop{\sum }\limits_{{k \geq 1}}\frac{{p}^{k}\left( {1 - 1/p}\right) }{{p}^{ks}}; \]\n\nhence for \( k \geq 1 \) we have \( \phi \left( {p}^{k}\right)... | Yes |
Corollary 10.1.12. We have\n\n\[ \phi \left( n\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( d\right) \left( {n/d}\right) = \mathop{\sum }\limits_{{d \mid n}}\mu \left( {n/d}\right) d. \] | Proof. Simply apply the Möbius inversion formula to (2). | No |
Proposition 10.1.14. (1) We have \( \Lambda \left( n\right) = 0 \) if \( n \) is not a prime power, and \( \Lambda \left( n\right) = \log \left( p\right) \) if \( n = {p}^{k} \) is a power of a prime \( p \) with \( k \geq 1 \) . (2) We have\n\n\[ \mathop{\sum }\limits_{{d \mid n}}\Lambda \left( d\right) = \log \left( ... | Proof. (1) is immediate since\n\n\[ - \frac{{\zeta }^{\prime }\left( s\right) }{\zeta \left( s\right) } = \mathop{\sum }\limits_{{p\text{ prime }}}\frac{\log \left( p\right) }{{p}^{s}\left( {1 - 1/{p}^{s}}\right) } = \mathop{\sum }\limits_{{p\text{ prime }}}\mathop{\sum }\limits_{{k \geq 1}}\frac{\log \left( p\right) }... | Yes |
Proposition 10.1.15. Let \( f\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}a\left( n\right) /{n}^{s} \) be a Dirichlet series. There exists a \( \sigma \in \left\lbrack {-\infty , + \infty }\right\rbrack \) (i.e., a real number or \( \pm \infty \) ), called the abscissa of absolute convergence of the series, such... | Proof. Assume that \( f\left( {s}_{0}\right) \) converges absolutely for some \( {s}_{0} \in \mathbb{C} \) . Then\n\nsince\n\[ \frac{\left| a\left( n\right) \right| }{\left| {n}^{s}\right| } = \frac{\left| a\left( n\right) \right| }{\left| {n}^{{s}_{0}}\right| }\left| {n}^{{s}_{0} - s}\right| = \frac{\left| a\left( n\r... | Yes |
Proposition 10.1.16. Let \( f\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}/{n}^{s} \) be a Dirichlet series with nonnegative coefficients, i.e., such that \( {a}_{n} \geq 0 \) for all \( n \), let \( {\sigma }_{0} \) be the abscissa of convergence of \( f \), and assume that \( {\sigma }_{0} \neq \pm \inf... | Proof. Assume the contrary. Then for \( \varepsilon > 0 \) sufficiently small, \( f \) is holomorphic in a circle centered at \( {\sigma }_{0} + 1 \) of radius \( 1 + {2\varepsilon } \), so inside this circle it is equal to the sum of its power series expansion\n\n\[ f\left( s\right) = \mathop{\sum }\limits_{{k \geq 0}... | Yes |
Corollary 10.1.17. Let \( f\left( s\right) = \mathop{\sum }\limits_{{n \geq 1}}{a}_{n}/{n}^{s} \) be a Dirichlet series with nonnegative coefficients and abscissa of convergence different from \( + \infty \) . If \( f\left( s\right) \) can be holomorphically continued to \( \Re \left( s\right) > \sigma \) then the seri... | Proof. Indeed, if \( {\sigma }_{0} < \infty \) is the abscissa of convergence of \( f\left( s\right) \) then either \( {\sigma }_{0} = - \infty \) and there is nothing to prove, or \( {\sigma }_{0} \neq \pm \infty \) and by the above proposition we know that \( f\left( s\right) \) has a singularity at \( s = {\sigma }_... | Yes |
Lemma 10.2.1. Let \( \chi \) be a character modulo \( m \), let \( f \mid m \) be the conductor of \( \chi \), and \( {\chi }_{f} \) the corresponding primitive character modulo \( f \) . We have\n\n\[ L\left( {\chi, s}\right) = L\left( {{\chi }_{f, s}\right) \mathop{\prod }\limits_{{p \mid m}}\left( {1 - \frac{{\chi }... | Proof. Clear. | No |
Proposition 10.2.2. Let \( f \) be a \( {C}^{\infty } \) function on \( \lbrack 0,\infty \lbrack \) tending rapidly to 0 at infinity, and for \( \Re \left( s\right) > 0 \) define\n\n\[ L\left( {f, s}\right) = \frac{1}{\Gamma \left( s\right) }{\int }_{0}^{\infty }f\left( t\right) {t}^{s}\frac{dt}{t}. \]\n\n(1) For any \... | Proof. Assume first that all the derivatives of \( f \) also tend rapidly to 0 at infinity. Integrating by parts the definition of \( L\left( {f, s}\right) \) and using the fact that \( f \) tends rapidly to 0 at infinity, we obtain\n\n\[ L\left( {f, s}\right) = - \frac{1}{{s\Gamma }\left( s\right) }{\int }_{0}^{\infty... | Yes |
Corollary 10.2.3. Let \( \chi \) be any arithmetic function of period dividing \( m \) , and recall that \( {B}_{0}\left( \chi \right) = {s}_{0}\left( \chi \right) /m = \left( {\mathop{\sum }\limits_{{0 \leq r < m}}\chi \left( r\right) }\right) /m \) .\n\n(1) For any \( k \in {\mathbb{Z}}_{ \geq 0} \) and \( \Re \left(... | Proof. (1), (2), and (3). The integral definition of the gamma function immediately implies that \n\n\[ {\int }_{0}^{\infty }{e}^{-{nt}}{t}^{s}\frac{dt}{t} = \frac{\Gamma \left( s\right) }{{n}^{s}} \] \n\nIt follows by absolute convergence that for \( \Re \left( s\right) > 1 \) , \n\n\[ \Gamma \left( s\right) L\left( {... | Yes |
Proposition 10.2.4. Let \( \chi \) be a primitive Dirichlet character modulo \( m \), let \( W\left( \chi \right) \) be the root number given by Definition 2.2.25, let \( k \in {\mathbb{Z}}_{ \geq 1} \) be such that \( \chi \left( {-1}\right) = {\left( -1\right) }^{k} \), and let \( e = 0 \) or 1 be such that \( k \equ... | Proof. Applying Proposition 9.4.14 to \( \bar{\chi } \) and \( x = 0 \) (with \( k \) and \( n \) exchanged) we obtain \[ {B}_{k}\left( \bar{\chi }\right) = - \frac{{m}^{k - 1}k!}{{\left( 2i\pi \right) }^{k}}\mathop{\sum }\limits_{{n \in \mathbb{Z}, n \neq 0}}\frac{\tau \left( {\bar{\chi }, n}\right) }{{n}^{k}}. \] Thi... | Yes |
Proposition 10.2.5. Let \( \chi \) be any arithmetic function of period dividing \( m \) . (1) We have \[ L\left( {\chi, s}\right) = \frac{1}{{m}^{s}}\mathop{\sum }\limits_{{1 \leq r \leq m}}\chi \left( r\right) \zeta \left( {s, r/m}\right) . \] | Proof. Since \[ L\left( {\chi, s}\right) = \mathop{\sum }\limits_{{n \geq 1}}\frac{\chi \left( n\right) }{{n}^{s}} = \mathop{\sum }\limits_{{1 \leq r \leq m}}\chi \left( r\right) \mathop{\sum }\limits_{{q \geq 0}}\frac{1}{{\left( qm + r\right) }^{s}} = \frac{1}{{m}^{s}}\mathop{\sum }\limits_{{1 \leq r \leq m}}\chi \lef... | Yes |
Theorem 10.2.6. For \( \Re \left( s\right) > 1 \) set \( Z\left( {s, x}\right) = \mathop{\sum }\limits_{{n \geq 1}}{e}^{2i\pi nx}/{n}^{s} \) . (1) For \( \Re \left( s\right) > 1 \) we have the functional equation \[ \zeta \left( {1 - s,\{ x\} }\right) = {\left( 2\pi \right) }^{-s}\Gamma \left( s\right) \left( {{e}^{-{i... | Proof. Statement (1) is a simple rephrasing of Corollary 9.6.51 seen in the previous chapter. | No |
Lemma 10.2.9. Assume that \( \Im \left( a\right) \geq 0 \), and that when \( \Im \left( a\right) = 0 \) we have \( a \neq 0 \) and \( \Im \left( b\right) = 0 \) . Then\n\n\[ \n{\int }_{-\infty }^{+\infty }\exp \left( {{i\pi }\left( {a{t}^{2} + {bt} + c}\right) }\right) {dt} = {\left( \frac{i}{a}\right) }^{1/2}\exp \lef... | Proof. Note that when \( \Im \left( a\right) = 0 \), we need the condition \( a \neq 0 \) and \( \Im \left( b\right) = 0 \) to ensure the convergence of the integral.\n\nSince \( a/i \) is not a negative real number, by Lemma 10.2.8, we have \( {\left( a/i\right) }^{1/2}{\left( i/a\right) }^{1/2} = 1 \) . If we set \( ... | Yes |
Proposition 10.2.10. Let \( \chi \) be a primitive character modulo \( m \) and let \( e = \) \( 0 \) if \( \chi \left( {-1}\right) = 1 \) and \( e = 1 \) if \( \chi \left( {-1}\right) = - 1 \) . For \( \Im \left( \tau \right) > 0 \) and any \( z \in \mathbb{C} \), set\n\n\[ \Theta \left( {\chi ,\tau, z}\right) = \math... | Proof. We have\n\n\[ \Theta \left( {\chi ,\tau, z}\right) = \mathop{\sum }\limits_{{r{\;\operatorname{mod}\;m}}}\chi \left( r\right) \mathop{\sum }\limits_{{k \in \mathbb{Z}}}\exp \left( \frac{{i\pi }{\left( km + r\right) }^{2}\tau + {2i\pi }\left( {{km} + r}\right) z}{m}\right) . \]\n\nSince \( \Im \left( \tau \right)... | Yes |
Corollary 10.2.12. The function \( \theta \left( {\chi ,\tau }\right) \) satisfies the functional equation\n\n\[ \theta \left( {\chi , - \frac{1}{\tau }}\right) = W\left( \chi \right) {\left( \frac{\tau }{i}\right) }^{\left( {{2e} + 1}\right) /2}\theta \left( {\bar{\chi },\tau }\right) ,\] | Proof. If \( e = 0 \), i.e., if \( \chi \) is even, the corollary immediately follows from the proposition by setting \( z = 0 \) . Assume now that \( e = 1 \) . We clearly have\n\n\[ \frac{\partial \Theta }{\partial z}\left( {\chi ,\tau ,0}\right) = \frac{2i\pi }{m}\theta \left( {\chi ,\tau }\right) .\n\nOn the other ... | Yes |
Corollary 10.2.13. Let \( t \) be a real positive number. As \( t \rightarrow + \infty \), then \( \theta \left( {\chi ,{it}}\right) = \chi \left( 0\right) + O\left( {\exp \left( {-{\pi t}/m}\right) }\right) \), and as \( t \rightarrow {0}^{ + } \), then \( \theta \left( {\chi ,{it}}\right) = {t}^{-e - 1/2}(\chi \left(... | Proof. As \( t \rightarrow + \infty \), this follows immediately from the definition. As \( t \rightarrow {0}^{ + } \), we have by the functional equation\n\n\[ \theta \left( {\chi ,{it}}\right) = W\left( \chi \right) {t}^{-e - 1/2}\theta \left( {\bar{\chi }, i/t}\right) = W\left( \chi \right) {t}^{-e - 1/2}\left( {\ch... | Yes |
Theorem 10.2.14 (Functional Equation for Dirichlet \( L \) -Functions).\n\nLet \( \chi \) be a primitive character modulo \( m \), let \( e = 0 \) (respectively \( e = 1 \) ) if \( \chi \) is an even (respectively odd) character, and set\n\n\[ \gamma \left( s\right) = {\pi }^{-s/2}\Gamma \left( {s/2}\right) \;\text{ an... | Proof. Set\n\n\[ I\left( {\chi, s}\right) = {\int }_{0}^{\infty }{t}^{\left( {s + e}\right) /2}\left( {\theta \left( {\chi ,{it}}\right) - \chi \left( 0\right) }\right) \frac{dt}{t}. \]\n\nBy the above corollary, this integral converges (exponentially fast in fact) when \( t \) is large, while when \( t \) is close to ... | Yes |
Corollary 10.2.15. (1) If \( \chi \) is a primitive character modulo \( m \) and \( e = 0 \) or 1 such that \( \chi \left( {-1}\right) = {\left( -1\right) }^{e} \), the functional equation may be rewritten in the form\n\n\[ L\left( {\chi, s}\right) = {\left( -i\right) }^{e}\tau \left( \chi \right) {\left( \frac{2\pi }{... | Proof. (1) immediately follows from the functional equation and the formulas \( \Gamma \left( s\right) \Gamma \left( {1 - s}\right) = \pi /\sin \left( {s\pi }\right) \) and \( \Gamma \left( {s/2}\right) \Gamma \left( {\left( {s + 1}\right) /2}\right) = {\pi }^{1/2}{2}^{1 - s}\Gamma \left( s\right) \) . | Yes |
Theorem 10.2.17. Keep all the above notation and assumptions. For any \( \sigma \) such that \( 0 < \sigma < 1 \) and \( x > 0 \) set\n\n\[ K\left( x\right) = \frac{1}{2i\pi }{\int }_{\Re \left( s\right) = \sigma }\frac{\gamma \left( s\right) }{\gamma \left( {1 - s}\right) }{x}^{-s}{ds}\;\text{ and }\;g\left( x\right) ... | Proof. (1). Recall from Section 9.7.3 that the Mellin transform of \( f \) is defined by \( M\left( f\right) \left( s\right) = {\int }_{0}^{\infty }f\left( t\right) {t}^{s - 1}{dt} \). Since \( f \) is in the Schwartz space, \( M\left( f\right) \left( s\right) \) converges for \( \Re \left( s\right) > 0 \) and can be a... | Yes |
Theorem 10.3.1. Let \( \chi \) be any periodic arithmetic function with period dividing \( m \) and let \( k \geq 1 \) be an integer.\n\n(1) We have\n\n\[ L\left( {\chi ,1 - k}\right) = - \frac{{B}_{k}\left( \chi \right) }{k} - \chi \left( 0\right) {\delta }_{k,1}. \]\n\n(2) If, in addition, \( \chi \) is a primitive c... | Proof. Thanks to the functional equation (for instance from Corollary 10.2.15) it is clear that (2) follows from (1), so it is enough to prove (1). We have already given two proofs of (1), one using Corollary 10.2.3, the other using \( \zeta \left( {s, x}\right) \) in Proposition 10.2.5. We will give a third proof, whi... | Yes |
Corollary 10.3.4. We have \( \zeta \left( 0\right) = - 1/2 \), and for \( k \geq 1 \) we have \( \zeta \left( {-{2k}}\right) = \) \( 0,\zeta \left( {1 - {2k}}\right) = - {B}_{2k}/\left( {2k}\right) \), and \[ \zeta \left( {2k}\right) = {\left( -1\right) }^{k - 1}\frac{{2}^{{2k} - 1}{\pi }^{2k}{B}_{2k}}{\left( {2k}\righ... | Proofs. Left to the reader (Exercise 12). | No |
Proposition 10.3.5. Let \( \chi \) be a character modulo \( m \) . Then (1) | Proof. We have seen in Proposition 10.2.5 that we have\n\n\[ L\left( {\chi, s}\right) = {m}^{-s}\mathop{\sum }\limits_{{1 \leq r \leq m}}\chi \left( r\right) \zeta \left( {s, r/m}\right) ,\]\n\nwhere \( \zeta \left( {s, x}\right) \) is the Hurwitz zeta function. Since by Definition 9.6.13 we have \( {\zeta }^{\prime }\... | Yes |
Corollary 10.3.6. Let \( D > 1 \) be a fundamental discriminant, and set as usual \( {\chi }_{D} = \left( \frac{D}{ \cdot }\right) \). (1) We have \[ L\left( {{\chi }_{D},1}\right) = \frac{2\log \left( {\varepsilon }_{D}\right) }{{D}^{1/2}}, \] where \[ {\varepsilon }_{D} = \mathop{\prod }\limits_{{r = 1}}^{{\lfloor D/... | Proof. The formula of (1) follows immediately from the above proposition since we know that \( W\left( \left( \frac{D}{ \cdot }\right) \right) = 1 \) and \( {\chi }_{D} \), and the fact that \( {\varepsilon }_{D} \) is the fundamental unit of \( K \) follows from Dirichlet’s class number formula (Proposition 3.4.5). No... | Yes |
Theorem 10.3.7. Recall that we denote by \( {\sigma }_{k}\left( n\right) \) the sum of the \( k \) th powers of the (positive) divisors of \( n \) . Let \( D \) be the discriminant of a real quadratic field.\n\n(1)\n\n\[ L\left( {\left( \frac{D}{ \cdot }\right) , - 1}\right) = - \frac{1}{5}\mathop{\sum }\limits_{\subst... | and by the functional equation\n\n\[ L\left( {\left( \frac{D}{ \cdot }\right) ,2}\right) = - \frac{2{\pi }^{2}}{{D}^{3/2}}L\left( {\left( \frac{D}{ \cdot }\right) , - 1}\right) .\] | No |
Proposition 10.3.8. Let \( D \neq 1 \) be a fundamental discriminant.\n\n(1) If \( D \neq 5 \) and \( D \neq 8 \) then \( L\left( {\left( \frac{D}{ \cdot }\right) , - 1}\right) \in 2\mathbb{Z} \), and \( L\left( {\left( \frac{5}{ \cdot }\right) , - 1}\right) = - 2/5 \) and \( L\left( {\left( \frac{8}{ \cdot }\right) , ... | Proof. By Corollary 10.3.3, for \( D < 0 \) we have \( L\left( {\left( \frac{D}{ \cdot }\right) , - 1}\right) = L\left( {\left( \frac{D}{ \cdot }\right) , - 3}\right) = \) 0 . For \( D > 0 \) we have\n\n\[ L\left( {\left( \frac{D}{ \cdot }\right) , - 1}\right) = - \frac{1}{2D}\mathop{\sum }\limits_{{r = 1}}^{{D - 1}}\l... | Yes |
Proposition 10.3.10. Let \( D \neq 1 \) be a fundamental discriminant.\n\n(1) If \( D \neq - 3, - 4, - 7 \), and -8 then \( L\left( {\left( \frac{D}{}\right) , - 2}\right) \in 2\mathbb{Z} \), and \( L\left( {\left( \frac{-3}{}\right) , - 2}\right) = \) \( - 2/9, L\left( {\left( \frac{-4}{}\right) , - 2}\right) = - 1/2,... | Proof. Left as an exercise for the reader (Exercise 13). | No |
Theorem 10.3.12. For \( k \geq 2 \) the Fourier series\n\n\[ \n{\mathcal{H}}_{k}\left( \tau \right) = \mathop{\sum }\limits_{{n \geq 0}}{H}_{k}\left( n\right) {q}^{n} \n\] \n\nis a modular form of weight \( k + 1/2 \) on the congruence subgroup \( {\Gamma }_{0}\left( 4\right) \), where as usual \( q = \exp \left( {2i\p... | Since the space of modular forms is finite-dimensional, it is then an easy matter to identify precisely a given form from its first few Fourier coefficients, given a specific basis. | No |
Proposition 10.3.16. Let \( \chi \) be a nontrivial character modulo \( m \) of conductor \( f > 1 \). (1) We have \[ \left| {L\left( {\chi ,1}\right) }\right| \leq \frac{1}{2}\log \left( f\right) + \log \left( {\log \left( f\right) }\right) + \log \left( {d\left( {m/f}\right) }\right) + {2.8}. \] | Proof. This proof is fundamentally based on partial (or Abel) summation. For \( X \geq 0 \) set \( S\left( X\right) = \mathop{\sum }\limits_{{1 \leq n \leq X}}\chi \left( n\right) \). For any integers \( M \) and \( N \) such that \( N \geq M \geq 1 \) we have \[ \mathop{\sum }\limits_{{n = M + 1}}^{N}\frac{\chi \left(... | Yes |
The average of \( L\left( {\chi ,1}\right) \) over all characters modulo \( m \) is equal to \[ - \frac{1}{m}\left( {\psi \left( \frac{1}{m}\right) + \log \left( m\right) + A\left( m\right) }\right) = 1 - \frac{\log \left( m\right) + A\left( m\right) - \gamma }{m} + O\left( \frac{1}{{m}^{2}}\right) . \] | By Proposition 10.2.5 (4) and orthogonality of characters we have \[ \mathop{\sum }\limits_{{\chi \neq {\chi }_{0}}}L\left( {\chi ,1}\right) = - \frac{1}{m}\mathop{\sum }\limits_{{1 \leq r \leq m}}\psi \left( {r/m}\right) \mathop{\sum }\limits_{{\chi \neq {\chi }_{0}}}\chi \left( r\right) \] \[ = - \frac{1}{m}\left( {\... | Yes |
Proposition 10.3.19. (1) For \( s \) around 1 we have\n\n\[ \zeta \left( s\right) = \frac{1}{s - 1} + \mathop{\sum }\limits_{{m \geq 0}}{\left( -1\right) }^{m}\frac{{\gamma }_{m}}{m!}{\left( s - 1\right) }^{m} = \frac{1}{s - 1} + \gamma + O\left( {s - 1}\right) . \] | Proof. (1). Let us restrict to \( s \in \mathbb{R} \), and for \( k \geq 1 \) set\n\n\[ {u}_{k}\left( s\right) = \frac{1}{{k}^{s}} - {\int }_{k - 1}^{k}\frac{dt}{{t}^{s}}. \]\n\nBy Taylor’s formula to order 2 it is clear that as \( s \rightarrow \infty \) we have \( {u}_{k}\left( s\right) \sim \) \( \left( {s/2}\right)... | Yes |
Proposition 10.3.21. Let \( A\left( x\right) = \mathop{\sum }\limits_{{m \geq 1}}a\left( m\right) {x}^{m} \) be a power series with radius of convergence strictly larger than \( 1/2 \) and such that \( A\left( 0\right) = {A}^{\prime }\left( 0\right) = 0 \) (hence with \( a\left( 1\right) = 0 \) ), and set \( S\left( A\... | Proof. Using the Euler product for the zeta function we have for \( n \geq 2 \) ,\n\n\[ \log \left( {{\zeta }_{p > N}\left( n\right) }\right) = \mathop{\sum }\limits_{{p > N}}\mathop{\sum }\limits_{{k \geq 1}}\frac{1}{k{p}^{kn}} = \mathop{\sum }\limits_{{k \geq 1}}\frac{n}{kn}\mathop{\sum }\limits_{{p > N}}\frac{1}{{p}... | Yes |
Corollary 10.3.22. Let \( B\left( x\right) = 1 + \mathop{\sum }\limits_{{m \geq 1}}b\left( m\right) {x}^{m} \) be a power series with radius of convergence strictly greater than \( 1/2 \), and such that \( B\left( 0\right) = 1 \) and \( {B}^{\prime }\left( 0\right) = 0 \) (hence with \( b\left( 1\right) = 0 \) ), and s... | Proof. If we write \( A\left( X\right) = \log \left( {B\left( x\right) }\right) \) we have \( P\left( B\right) = {e}^{S\left( A\right) } \), so the first formula follows from the proposition. For the second we note that by definition \( {nc}\left( n\right) = \mu * {na}\left( n\right) \), where \( * \) is the arithmetic... | Yes |
Proposition 10.4.2. The above series converges for \( \Re \left( s\right) > 1 \), and for any \( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \in {\mathrm{{SL}}}_{2}\left( \mathbb{Z}\right) \) we have\n\n\[ G\left( {\frac{{a\tau } + b}{{c\tau } + d}, s}\right) = G\left( {\tau, s}\right) \]\n\nin other wor... | Proof. Left to the reader (Exercise 55). | No |
Theorem 10.4.3. Let \( \tau = x + {iy} \) with \( y > 0 \) . For \( \Re \left( s\right) > 1 \) we have\n\n\[ G\left( {\tau, s}\right) = \zeta \left( {2s}\right) {y}^{s} + \frac{{\pi }^{1/2}\Gamma \left( {s - 1/2}\right) }{\Gamma \left( s\right) }\zeta \left( {{2s} - 1}\right) {y}^{1 - s} \]\n\n\[ + 2\frac{{\pi }^{s}}{\... | Proof. First a friendly word to the reader: this result is quite technical; however, its proof is instructive and completely straightforward. Set\n\n\[ S\left( {\tau, s}\right) = \mathop{\sum }\limits_{{n \in \mathbb{Z}}}\frac{{y}^{s}}{{\left| \tau + n\right| }^{2s}}. \]\n\nBy the Poisson summation formula (Corollary 2... | Yes |
Corollary 10.4.4. The function \( G\left( {\tau, s}\right) \) has a meromorphic continuation to the whole complex \( s \) -plane, with a single pole, at \( s = 1 \), which is simple with residue \( \pi /2 \) (in particular which is independent of \( \tau \) ). Moreover, if we set\n\n\[ \mathcal{G}\left( {\tau, s}\right... | Proof. Indeed, by Proposition 9.8.6 the function \( {K}_{s - 1/2}\left( z\right) \) is a holomorphic function of \( s \in \mathbb{C} \) that tends to zero faster than any power of \( z \) as \( z \rightarrow \infty \) . It follows from the known analytic continuation of the other functions occurring in the expansion of... | Yes |
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