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f8515bde6dc16861210d9ac25344de2e7afd12c2	therefore: ∫ c z z 2 + 6	therefore: ∫ c z z 2 + 6 z + 8 d z = 2 ∫ 0 ∞ x x 2 + 6 x + 8 d x. {\displaystyle \int _{c}{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz=2\int _{0}^{\infty }{\frac {\sqrt {x}}{x^{2}+6x+8}}\,dx.}	wikipedia
9c681e884218240de3a3c91e9cd70b01515152fe	it can be shown that the integrals over	it can be shown that the integrals over Γ and γ both tend to zero as ε → 0 and r → ∞, by an estimation argument above, that leaves two terms. now since z = e, on the contour outside the branch cut, we have gained 2 π in argument along γ. (by euler's identity, e represents the unit vector, which therefore has π as its l...	wikipedia
66bd3a28449f34ca4f1d3d438c9f261bb461bc24	it can be shown that the integrals over	^{\varepsilon }{\frac {e^{{\frac {1}{2}}(\log \|z\|+i\arg {z})}}{z^{2}+6z+8}}\,dz\\&=\int _{r}^{\varepsilon }{\frac {e^{{\frac {1}{2}}\log \|z\|}e^{{\frac {1}{2}}(2\pi i)}}{z^{2}+6z+8}}\,dz\\&=\int _{r}^{\varepsilon }{\frac {e^{{\frac {1}{2}}\log \|z\|}e^{\pi i}}{z^{2}+6z+8}}\,dz\\&=\int _{r}^{\varepsilon }{\frac {-{\sqrt {z...	wikipedia
22b8b60445310faa556a67ff3aa006f87271982f	let γ be the small circle of radius	let γ be the small circle of radius ε, Γ the larger, with radius r, then ∫ c = ∫ ε r + ∫ Γ + ∫ r ε + ∫ γ. {\displaystyle \int _{c}=\int _{\varepsilon }^{r}+\int _{\gamma }+\int _{r}^{\varepsilon }+\int _{\gamma }.}	wikipedia
5a6b77c5c47629e42e9f04d7346bb5be06859146	note that z = −2 and z =	note that z = −2 and z = −4 are inside the big circle. these are the two remaining poles, derivable by factoring the denominator of the integrand. the branch point at z = 0 was avoided by detouring around the origin.	wikipedia
97aa72d5612eb29eb68334e60fec285558d758d0	then, we use the so-called keyhole contour, which	then, we use the so-called keyhole contour, which consists of a small circle about the origin of radius ε say, extending to a line segment parallel and close to the positive real axis but not touching it, to an almost full circle, returning to a line segment parallel, close, and below the positive real axis in the nega...	wikipedia
666b53a35933d7dc5659a41dfdba66c7937e7ba4	we can use the cauchy integral formula or	we can use the cauchy integral formula or residue theorem again to obtain the relevant residues. however, the important thing to note is that z = e, so z has a branch cut. this affects our choice of the contour c. normally the logarithm branch cut is defined as the negative real axis, however, this makes the calculatio...	wikipedia
2ec1a0d802dbc317f2c3f093ddaa793cb23c692f	we can begin by formulating the complex integral	we can begin by formulating the complex integral ∫ c z z 2 + 6 z + 8 d z = i. {\displaystyle \int _{c}{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz=i.}	wikipedia
334d408345d292eac4544f7e5ce03dd88354879a	consider the real integral ∫ 0 ∞ x	consider the real integral ∫ 0 ∞ x x 2 + 6 x + 8 d x. {\displaystyle \int _{0}^{\infty }{\frac {\sqrt {x}}{x^{2}+6x+8}}\,dx.}	wikipedia
432709bd6a6a4bafe1dcce9ec8bd5a02ad6e21d5	the poles of this function are at 1	the poles of this function are at 1 ± √ 2 and −1 ± √ 2. of these, 1 + √ 2 and −1 − √ 2 are outside the unit circle (shown in red, not to scale), whereas 1 − √ 2 and −1 + √ 2 are inside the unit circle (shown in blue). the corresponding residues are both equal to − i √ 2 / 16, so that the value of the integral is i = 2 ...	wikipedia
cdac0d00e40f7fdb369db00dbf7213403ad66b09	the substitution yields 1 4 ∮ \| z	the substitution yields 1 4 ∮ \| z \| = 1 4 i z z 4 − 6 z 2 + 1 d z = ∮ \| z \| = 1 i z z 4 − 6 z 2 + 1 d z. {\displaystyle {\frac {1}{4}}\oint _{\|z\|=1}{\frac {4iz}{z^{4}-6z^{2}+1}}\,dz=\oint _{\|z\|=1}{\frac {iz}{z^{4}-6z^{2}+1}}\,dz.}	wikipedia
4d04540a52f27032c8cb522325f56ab2b8495453	the image at right illustrates this for i	the image at right illustrates this for i = ∫ 0 π 2 1 1 + (sin t) 2 d t, {\displaystyle i=\int _{0}^{\frac {\pi }{2}}{\frac {1}{1+(\sin t)^{2}}}\,dt,} which we now compute. the first step is to recognize that i = 1 4 ∫ 0 2 π 1 1 + (sin t) 2 d t. {\displaystyle i={\frac {1}{4}}\int _{0}^{2\pi }{\frac {1}{1+(\sin t)^{2}}...	wikipedia
e9f07b7a2a3b6382e9d3dd55706cbfda566c6456	this substitution maps the interval to the unit	this substitution maps the interval to the unit circle. furthermore, sin (k t) = e i k t − e − i k t 2 i = z k − z − k 2 i {\displaystyle \sin(kt)={\frac {e^{ikt}-e^{-ikt}}{2i}}={\frac {z^{k}-z^{-k}}{2i}}} and cos (k t) = e i k t + e − i k t 2 = z k + z − k 2 {\displaystyle \cos(kt)={\frac {e^{ikt}+e^{-ikt}}{2}}={\frac...	wikipedia
28c7bb4284dd7a9fd6922088912b428cd2c92794	where p and q are polynomials, i.e. a	where p and q are polynomials, i.e. a rational function in trigonometric terms is being integrated. note that the bounds of integration may as well be π and − π, as in the previous example, or any other pair of endpoints 2 π apart.	wikipedia
a1780dfa53cea19ba414f7eaa32ffbfc9b1bccbe	the above method may be applied to all	the above method may be applied to all integrals of the type ∫ 0 2 π p (sin (t), sin (2 t), …, cos (t), cos (2 t), …) q (sin (t), sin (2 t), …, cos (t), cos (2 t), …) d t {\displaystyle \int _{0}^{2\pi }{\frac {p{\big (}\sin(t),\sin(2t),\ldots,\cos(t),\cos(2t),\ldots {\big)}}{q{\big (}\sin(t),\sin(2t),\ldots,\cos(t),\c...	wikipedia
b1162aed7b4ab43d0a43b9aed3ad6fad63b9a255	the singularities to be considered are at ±	the singularities to be considered are at ± i 3. {\displaystyle {\tfrac {\pm i}{\sqrt {3}}}.} let c be a small circle about i 3, {\displaystyle {\tfrac {i}{\sqrt {3}}},} and c be a small circle about − i 3. {\displaystyle {\tfrac {-i}{\sqrt {3}}}.} then we arrive at the following: − 4 i 3 = − 4 i 3 z = i 3 + 2 π i z = ...	wikipedia
3dce610c8dfc7362c5bf0ae47ce070b282062205	∮ c 1 1 + 3 (1 2	∮ c 1 1 + 3 (1 2 (z + 1 z)) 2 d z i z = ∮ c 1 1 + 3 4 (z + 1 z) 2 1 i z d z = ∮ c − i z + 3 4 z (z + 1 z) 2 d z = − i ∮ c d z z + 3 4 z (z 2 + 2 + 1 z 2) = − i ∮ c d z z + 3 4 (z 3 + 2 z + 1 z) = − i ∮ c d z 3 4 z 3 + 5 2 z + 3 4 z = − i ∮ c 4 3 z 3 + 10 z + 3 z d z = − 4 i ∮ c d z 3 z 3 + 10 z + 3 z = − 4 i ∮ c z 3 z ...	wikipedia
3eed4e1acfe1cf4afa0ac6b755f1a60b5b64c39b	∮ c 1 1 + 3 (1 2	{1}{1+{\frac {3}{4}}\left(z+{\frac {1}{z}}\right)^{2}}}{\frac {1}{iz}}\,dz\\&=\oint _{c}{\frac {-i}{z+{\frac {3}{4}}z\left(z+{\frac {1}{z}}\right)^{2}}}\,dz\\&=-i\oint _{c}{\frac {dz}{z+{\frac {3}{4}}z\left(z^{2}+2+{\frac {1}{z^{2}}}\right)}}\\&=-i\oint _{c}{\frac {dz}{z+{\frac {3}{4}}\left(z^{3}+2z+{\frac {1}{z}}\rig...	wikipedia
97773520c2015bdc7b919b45256d0d1b49ccb8e5	∮ c 1 1 + 3 (1 2	i}{\sqrt {3}}}\right)}}\,dz\\&=-{\frac {4i}{3}}\oint _{c}{\frac {z}{\left(z+{\sqrt {3}}i\right)\left(z-{\sqrt {3}}i\right)\left(z+{\frac {i}{\sqrt {3}}}\right)\left(z-{\frac {i}{\sqrt {3}}}\right)}}\,dz.\end{aligned}}}	wikipedia
aabf3c551fd4a3dcfab3c652fa19ac001a15df26	we seek to make a substitution of z	we seek to make a substitution of z = e. now, recall cos t = 1 2 (e i t + e − i t) = 1 2 (z + 1 z) {\displaystyle \cos t={\frac {1}{2}}\left(e^{it}+e^{-it}\right)={\frac {1}{2}}\left(z+{\frac {1}{z}}\right)} and d z d t = i z, d t = d z i z. {\displaystyle {\frac {dz}{dt}}=iz,\ dt={\frac {dz}{iz}}.}	wikipedia
f5e4cfb37577b717074941c09846cdaf5e140f1d	as an example, consider ∫ − π π	as an example, consider ∫ − π π 1 1 + 3 (cos t) 2 d t. {\displaystyle \int _{-\pi }^{\pi }{\frac {1}{1+3(\cos t)^{2}}}\,dt.}	wikipedia
89c2759efe7f5a2bd3f9d9ca11d601f5190b7dcd	a similar argument with an arc that winds	a similar argument with an arc that winds around − i rather than i shows that if t < 0 then ∫ − ∞ ∞ e i t x x 2 + 1 d x = π e t, {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx=\pi e^{t},} and finally we have this: ∫ − ∞ ∞ e i t x x 2 + 1 d x = π e − \| t \|. {\displaystyle \int _{-\infty }^{\inft...	wikipedia
241fa1398a445cee5488510dd41f0f27d916e0d4	therefore, if t > 0 then ∫ −	therefore, if t > 0 then ∫ − ∞ ∞ e i t x x 2 + 1 d x = π e − t. {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx=\pi e^{-t}.}	wikipedia
514668684b64d405524c88060159dfd423d3cc7f	according to jordan's lemma, if t > 0	according to jordan's lemma, if t > 0 then ∫ arc e i t z z 2 + 1 d z → 0 as a → ∞. {\displaystyle \int _{\text{arc}}{\frac {e^{itz}}{z^{2}+1}}\,dz\rightarrow 0{\mbox{ as }}a\rightarrow \infty.}	wikipedia
9cab6cf1767d3e9358dd2090d4085b90b54dbce7	the contour c may be split into a	the contour c may be split into a "straight" part and a curved arc, so that ∫ straight + ∫ arc = π e − t, {\displaystyle \int _{\text{straight}}+\int _{\text{arc}}=\pi e^{-t},} and thus ∫ − a a = π e − t − ∫ arc. {\displaystyle \int _{-a}^{a}=\pi e^{-t}-\int _{\text{arc}}.}	wikipedia
e143f6037e3d94a29fe0f67aad51a6ee2502e9ac	according to the residue theorem, then, we have	according to the residue theorem, then, we have ∫ c f (z) d z = 2 π i res z = i f (z) = 2 π i e − t 2 i = π e − t. {\displaystyle \int _{c}f(z)\,dz=2\pi i\operatorname {res} _{z=i}f(z)=2\pi i{\frac {e^{-t}}{2i}}=\pi e^{-t}.}	wikipedia
df663405f59ca6f6569d1ecf4da15b2dce95cb70	since e is an entire function (having no	since e is an entire function (having no singularities at any point in the complex plane), this function has singularities only where the denominator z + 1 is zero. since z + 1 = (z + i)(z − i), that happens only where z = i or z = − i. only one of those points is in the region bounded by this contour. the residue of f...	wikipedia
6e90417b638dd7def021566150961797c3986687	(which arises in probability theory as a scalar	(which arises in probability theory as a scalar multiple of the characteristic function of the cauchy distribution) resists the techniques of elementary calculus. we will evaluate it by expressing it as a limit of contour integrals along the contour c that goes along the real line from − a to a and then counterclockwis...	wikipedia
34754ba0f02038ab84b1702cf93c6b0f70ee0887	the integral ∫ − ∞ ∞ e i	the integral ∫ − ∞ ∞ e i t x x 2 + 1 d x {\displaystyle \int _{-\infty }^{\infty }{\frac {e^{itx}}{x^{2}+1}}\,dx}	wikipedia
f072dcb393a37950987e7283252de1c927a82660	as an aside, a question can arise whether	as an aside, a question can arise whether we do not take the semicircle to include the other singularity, enclosing − i. to have the integral along the real axis moving in the correct direction, the contour must travel clockwise, i.e., in a negative direction, reversing the sign of the integral overall.	wikipedia
7ac68c249f638e0dcfcbb0b1be873ba3ea5407b0	it is clear by inspection that the residue	it is clear by inspection that the residue is − i / 4, so, by the residue theorem, we have ∮ c f (z) d z = ∮ c 1 (z 2 + 1) 2 d z = 2 π i res z = i f (z) = 2 π i (− i 4) = π 2 ◻ {\displaystyle \oint _{c}f(z)\,dz=\oint _{c}{\frac {1}{\left(z^{2}+1\right)^{2}}}\,dz=2\pi i\,\operatorname {res} _{z=i}f(z)=2\pi i\left(-{\fra...	wikipedia
e88d7fd446df5d6f5d5aa1e94bcdd0866b81b775	consider the laurent series of f (z) about	consider the laurent series of f (z) about i, the only singularity we need to consider. we then have f (z) = − 1 4 (z − i) 2 + − i 4 (z − i) + 3 16 + i 8 (z − i) + − 5 64 (z − i) 2 + ⋯ {\displaystyle f(z)={\frac {-1}{4(z-i)^{2}}}+{\frac {-i}{4(z-i)}}+{\frac {3}{16}}+{\frac {i}{8}}(z-i)+{\frac {-5}{64}}(z-i)^{2}+\cdots ...	wikipedia
7ed3ebbe5ff0b2d46d779925e0a4de229fb54078	where m is an upper bound on \|	where m is an upper bound on \| f (z) \| along the arc and l the length of the arc. now, \| ∫ arc f (z) d z \| ≤ a π (a 2 − 1) 2 → 0 as a → ∞. {\displaystyle \left\|\int _{\text{arc}}f(z)\,dz\right\|\leq {\frac {a\pi }{\left(a^{2}-1\right)^{2}}}\to 0{\text{ as }}a\to \infty.} so ∫ − ∞ ∞ 1 (x 2 + 1) 2 d x = ∫ − ∞ ∞ f (z) d z ...	wikipedia
2add1b38edbf2311f67418a93fb619f4a777d133	we need to show that the integral over	we need to show that the integral over the arc of the semicircle tends to zero as a → ∞, using the estimation lemma \| ∫ arc f (z) d z \| ≤ m l {\displaystyle \left\|\int _{\text{arc}}f(z)\,dz\right\|\leq ml}	wikipedia
ace13fb550ffbd17749010c722d68cf1c14e9103	we take the first derivative, in the above	we take the first derivative, in the above steps, because the pole is a second-order pole. that is, (z − i) is taken to the second power, so we employ the first derivative of f (z). if it were (z − i) taken to the third power, we would use the second derivative and divide by 2!, etc. the case of (z − i) to the first po...	wikipedia
d2dbde878d749c9c2c3166893b08e01a8e8c95da	which puts the function in the form for	which puts the function in the form for direct application of the formula. then, by using cauchy's integral formula, ∮ c f (z) d z = ∮ c 1 (z + i) 2 (z − i) 2 d z = 2 π i d d z 1 (z + i) 2 \| z = i = 2 π i z = i = π 2 {\displaystyle \oint _{c}f(z)\,dz=\oint _{c}{\frac {\frac {1}{(z+i)^{2}}}{(z-i)^{2}}}\,dz=2\pi i\,\left...	wikipedia
f41f89e1dc7d295ff81ab72d66d721bc670293c5	since the only singularity in the contour is	since the only singularity in the contour is the one at i, then we can write f (z) = 1 (z + i) 2 (z − i) 2, {\displaystyle f(z)={\frac {\frac {1}{(z+i)^{2}}}{(z-i)^{2}}},}	wikipedia
de31a027d4fa25e63021b9fe25a3329fc0e398b3	furthermore, observe that f (z) = 1 (z	furthermore, observe that f (z) = 1 (z 2 + 1) 2 = 1 (z + i) 2 (z − i) 2. {\displaystyle f(z)={\frac {1}{\left(z^{2}+1\right)^{2}}}={\frac {1}{(z+i)^{2}(z-i)^{2}}}.}	wikipedia
8476c6cd7ce9486f867927f116ecbc6b14a6f34e	note that: ∮ c f (z) d z	note that: ∮ c f (z) d z = ∫ − a a f (z) d z + ∫ arc f (z) d z {\displaystyle \oint _{c}f(z)\,dz=\int _{-a}^{a}f(z)\,dz+\int _{\text{arc}}f(z)\,dz} thus ∫ − a a f (z) d z = ∮ c f (z) d z − ∫ arc f (z) d z {\displaystyle \int _{-a}^{a}f(z)\,dz=\oint _{c}f(z)\,dz-\int _{\text{arc}}f(z)\,dz}	wikipedia
307b813520a3d27c1910dcde712dd037a99baa1c	which has singularities at i and − i.	which has singularities at i and − i. we choose a contour that will enclose the real-valued integral, here a semicircle with boundary diameter on the real line (going from, say, − a to a) will be convenient. call this contour c.	wikipedia