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Theorem 63. If \( {f}_{1}\left( x\right) ,\ldots ,{f}_{n}\left( x\right) \) have the same sign on some \( {V}^{ * }\left( a\right) \) and if \( {f}_{2}\left( x\right) ,\ldots \) , \( {f}_{n}\left( x\right) \) are infinitesimal (infinite) of the same or higher (lower) order than \( {f}_{1}\left( x\right) \), then\n\n\[ ... | Proof. We are to show that\n\n\[ \n\underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) + {f}_{2}\left( x\right) + \ldots + {f}_{n}\left( x\right) }{{f}_{1}\left( x\right) } = k \neq 0.\n\]\n\nBy hypothesis,\n\n\[ \n\underset{x \doteq a}{L}\frac{{f}_{2}\left( x\right) }{{f}_{1}\left( x\right) } = {k}_{2},\underset{x \d... | Yes |
Theorem 64. If \( {f}_{3}\left( x\right) \) and \( {f}_{4}\left( x\right) \) are infinitesimals with respect to \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \), then\n\n\[ \underset{x \doteq a}{L}\frac{\left\{ {{f}_{1}\left( x\right) + {f}_{3}\left( x\right) }\right\} \cdot \left\{ {{f}_{2}\left( x\right)... | Proof.\n\n\[ \underset{x \doteq a}{L}\frac{\left\{ {{f}_{1}\left( x\right) + {f}_{3}\left( x\right) }\right\} \cdot \left\{ {{f}_{2}\left( x\right) + {f}_{4}\left( x\right) }\right\} }{{f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) }\n\n\[ = \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) \cdot {f}_{2}\left( ... | Yes |
Theorem 65. If as \( x \doteq a,{\varepsilon }_{1}\left( x\right) \) is an infinitesimal with respect to \( {f}_{1}\left( x\right) \) and \( {\varepsilon }_{2}\left( x\right) \) with respect to \( {f}_{2}\left( x\right) \), then the values approached by\n\n\[ \frac{{f}_{1}\left( x\right) + {\varepsilon }_{1}\left( x\ri... | Proof. This follows from the identity\n\n\[ \frac{{f}_{1}\left( x\right) + {\varepsilon }_{1}\left( x\right) }{{f}_{2}\left( x\right) + {\varepsilon }_{2}\left( x\right) } = \frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } \cdot \frac{\left( 1 + \frac{{\varepsilon }_{1}\left( x\right) }{{f}_{1}\left( x\right) }\... | Yes |
Theorem 66. If \( \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{\phi }_{1}\left( x\right) } = \underset{x \doteq a}{L}\frac{{f}_{2}\left( x\right) }{{\phi }_{2}\left( x\right) } = k \), and if \( \underset{x \doteq a}{L}\frac{{\phi }_{1}\left( x\right) }{{\phi }_{2}\left( x\right) } = l \) is finite, then \[ ... | Proof. \[ \frac{{f}_{1}\left( x\right) + {f}_{2}\left( x\right) }{{\phi }_{1}\left( x\right) + {\phi }_{2}\left( x\right) } - \frac{{f}_{2}\left( x\right) }{{\phi }_{2}\left( x\right) } = \frac{{f}_{1}\left( x\right) {\phi }_{2}\left( x\right) - {f}_{2}\left( x\right) {\phi }_{1}\left( x\right) }{{\phi }_{2}\left( x\ri... | Yes |
Theorem 67. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are both infinitesimals as \( x \doteq a \), then a necessary and sufficient condition that\n\n\[ \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = k\;\left( {k\text{ finite and not zero }}\right) \]\n\nis that ... | Proof. (1) The condition is necessary. Since \( \underset{x \doteq a}{L}\frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = k \) ,\n\n\[ \frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) } = k + {\varepsilon }^{\prime }\left( x\right) \]\n\nor \( {f}_{1}\left( x\right) = {f}_{2}\left( x\right) \cdot k + {f}_{... | Yes |
Theorem 68. If \( f\left( x\right) \) and \( \phi \left( x\right) \), defined on some \( V\left( {+\infty }\right) \), are both infinitesimal as \( x \) approaches \( + \infty \), and if for some positive number \( h,\phi \left( {x + h}\right) \) is always less than \( \phi \left( x\right) \) and\n\n\[ \underset{x \dot... | Proof. Let \( {V}_{1}\left( k\right) \) and \( {V}_{2}\left( k\right) \) be a pair of vicinities of \( k \) such that \( {V}_{2}\left( k\right) \) is entirely within \( {V}_{1}\left( k\right) \). By hypothesis there exists an \( h \) and an \( {X}_{2} \) such that if \( x > {X}_{2} \), \n\n\[ \frac{f\left( {x + h}\righ... | Yes |
Theorem 71. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are of the same rank as \( x \) approaches a, then \( c \cdot {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are of the same rank, \( c \) being any constant not zero. | Proof. By hypothesis for some positive number \( M \) ,\n\n\[ \left| \frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) }\right| < M\text{ and }\left| \frac{{f}_{2}\left( x\right) }{{f}_{1}\left( x\right) }\right| < M \]\n\nhence\n\n\[ \left| \frac{c \cdot {f}_{1}\left( x\right) }{{f}_{2}\left( x\right) }\right| < M... | Yes |
Theorem 72. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are of the same rank and \( {f}_{2}\left( x\right) \) and \( {f}_{3}\left( x\right) \) are of the same rank as \( x \) approaches \( a \), then \( {f}_{1}\left( x\right) \) and \( {f}_{3}\left( x\right) \) are of the same rank as \( x \) appro... | Proof. By hypothesis,\n\n\[ \left| \frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) }\right| < {M}_{1}\text{ and }\left| \frac{{f}_{2}\left( x\right) }{{f}_{3}\left( x\right) }\right| < {M}_{2} \]\n\nin some neighborhood of \( x = a \) . Therefore\n\n\[ \left| \frac{{f}_{1}\left( x\right) }{{f}_{2}\left( x\right) ... | Yes |
Theorem 73. If \( {f}_{1}\left( x\right) \) is infinitesimal (infinite) and does not vanish on some \( {V}^{ * }\left( a\right) \), and if \( {f}_{2}\left( x\right) \) and \( {f}_{3}\left( x\right) \) are infinitesimal (infinite) of the same rank as \( x \) approaches \( a \), then \( {f}_{1}\left( x\right) \cdot {f}_{... | Proof. Since \( \left| \frac{{f}_{1}\left( x\right) }{{f}_{3}\left( x\right) }\right| \) is bounded as \( x \) approaches \( a \), it follows by Theorem 33 that\n\n\[ \n{\underline{L}}_{x \doteq a}\frac{{f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) }{{f}_{3}\left( x\right) } = 0 \n\]\n\nwhich proves the first par... | Yes |
Theorem 74. The derivative of a constant is zero. More precisely: If there exists a neighborhood of \( {x}_{1} \) such that for every value of \( x \) on this neighborhood \( f\left( x\right) = f\left( {x}_{1}\right) \), then \( {f}^{\prime }\left( {x}_{1}\right) = 0 \) . | Proof. In the neighborhood specified \( \frac{f\left( x\right) - f\left( {x}_{1}\right) }{x - {x}_{1}} = 0 \) for every value of \( x \) . | Yes |
Theorem 75. When for two functions \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) the derived functions \( {f}_{1}^{\prime }\left( x\right) \) and \( {f}_{2}^{\prime }\left( x\right) \) exist at \( {x}_{1} \) it follows that, except in the indeterminate case \( \infty - \infty \) ,\n\n(a) If \( {f}_{3}\l... | Proof. By definition and the theorems of Chapter IV (which exclude the case \( \infty - \infty \) ),\n\n(a)\n\n\[ {f}_{1}^{\prime }\left( {x}_{1}\right) + {f}_{2}^{\prime }\left( {x}_{1}\right) = \underset{x \doteq {x}_{1}}{L}\frac{{f}_{1}\left( x\right) - {f}_{1}\left( {x}_{1}\right) }{x - {x}_{1}} + \underset{x \dote... | Yes |
Theorem 76. If \( x > 0 \), then \( \frac{d}{dx}{x}^{k} = k \cdot {x}^{k - 1} \) . | (a) If \( k \) is a positive integer, we have\n\n\[ \underset{x \doteq {x}_{1}}{L}\frac{{x}^{k} - {x}_{1}^{k}}{x - {x}_{1}} = \underset{x \doteq {x}_{1}}{L}\left\{ {{x}^{k - 1} + {x}^{k - 2} \cdot {x}_{1} + \ldots + {x}^{k} \cdot {x}_{1}^{k - 2} + {x}_{1}^{k - 1}}\right\} \]\n\n\[ = k \cdot {x}_{1}^{k - 1}\text{.} \] | Yes |
Theorem 77. \( \frac{d}{dx}{\log }_{a}x = \frac{1}{x} \cdot {\log }_{a}e \) . | Proof.\n\n\[ \frac{{\log }_{a}\left( {x + {\Delta x}}\right) - {\log }_{a}x}{\Delta x} = \frac{1}{\Delta x}{\log }_{a}\frac{x + {\Delta x}}{x} \]\n\n\[ = \frac{1}{x} \cdot {\log }_{a}{\left( 1 + \frac{\Delta x}{x}\right) }^{\frac{x}{\Delta x}}. \]\n\nBut, by Theorem 57,\n\n\[ \underset{{\Delta x} \doteq 0}{L}{\left( 1 ... | Yes |
Theorem 78. If \( {f}_{1}^{\prime }\left( x\right) \) exists and if there is a \( V\left( {x}_{1}\right) \) upon which \( {f}_{1}\left( x\right) \) is continuous and possesses a single-valued inverse \( x = {f}_{2}\left( y\right) \), then \( {f}_{2}\left( y\right) \) is differentiable and\n\n\[ \n{f}_{1}^{\prime }\left... | Proof. To prove this theorem we observe that\n\n\[ \n{f}_{1}^{\prime }\left( {x}_{1}\right) = \underset{x \doteq {x}_{1}}{L}\frac{{f}_{1}\left( x\right) - {f}_{1}\left( {x}_{1}\right) }{x - {x}_{1}} = \underset{x \doteq {x}_{1}}{L}\frac{1}{\frac{x - {x}_{1}}{{f}_{1}\left( x\right) - {f}_{1}\left( {x}_{1}\right) }}.\n\]... | Yes |
\[ \frac{d}{dx}{a}^{x} = {a}^{x}\log a \] | Proof. Let\n\n\[ y = {a}^{x} \]\n\ntherefore\n\n\[ \log y = x \cdot \log a \]\n\nand, by Theorem 77,\n\n\[ \frac{\frac{dy}{dx}}{y} = \log a \]\n\nwhence\n\n\[ \frac{dy}{dx} = y \cdot \log a = {a}^{x}\log a. \] | Yes |
Theorem 81. When for two functions \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \)\n\n\[ d{f}_{1}\left( x\right) = {f}_{1}^{\prime }\left( x\right) \cdot {dx}\text{ and }d{f}_{2}\left( x\right) = {f}_{2}\left( x\right) \cdot {dx}\text{ at }{x}_{1}, \]\n\nit follows that\n\n(a) If \( {f}_{3}\left( x\right)... | (a) If \( {f}_{3}\left( x\right) = {f}_{1}\left( x\right) + {f}_{2}\left( x\right) \), then\n\n\[ d{f}_{3}\left( {x}_{1}\right) = \left\{ {{f}_{1}^{\prime }\left( {x}_{1}\right) + {f}_{2}^{\prime }\left( {x}_{1}\right) }\right\} {dx} \]\n\n\[ = d{f}_{1}\left( {x}_{1}\right) + d{f}_{2}\left( {x}_{1}\right) . \] | Yes |
Theorem 82. If \( f\left( x\right) \) has a unique and finite derivative at \( x = {x}_{1} \), then \( f\left( x\right) \) is continuous at \( {x}_{1} \) . | Proof. The proof depends upon the evident fact that if \( f\left( x\right) - f\left( {x}_{1}\right) \) approach anything but zero as \( x \) approaches \( {x}_{1} \), then one of the values approached by\n\n\[ \frac{f\left( x\right) - f\left( {x}_{1}\right) }{x - {x}_{1}} \]\n\nis \( + \infty \) or \( - \infty \) . | No |
Theorem 83. If \( {f}^{\prime }\left( {x}_{1}\right) \) exists and if \( f\left( x\right) \) has a maximum or a minimum at \( x = {x}_{1} \), then \( {f}^{\prime }\left( {x}_{1}\right) = 0 \) . | Proof. In case of a maximum at \( {x}_{1} \), it follows directly from the hypothesis that\n\n\[ \underset{\begin{matrix} {x \doteq {x}_{1}} \\ {x > {x}_{1}} \end{matrix}}{L}\frac{f\left( x\right) - f\left( {x}_{1}\right) }{x - {x}_{1}} \gtreqless 0,\text{ and also }\underset{\begin{matrix} {x \doteq {x}_{1}} \\ {x < {... | Yes |
Theorem 84. If \( f\left( {x}_{1}\right) = f\left( {x}_{2}\right), f\left( x\right) \) being continuous on the interval \( {x}_{1}{x}_{2} \), and if the derivative exists \( {}^{3} \) at every point between \( {x}_{1} \) and \( {x}_{2} \), then there is a value \( \xi \) between \( {x}_{1} \) and \( {x}_{2} \) such tha... | Proof. (a) The function may be a constant between \( {x}_{1} \) and \( {x}_{2} \), in which case \( {f}^{\prime }\left( x\right) = 0 \) for all values of \( x \) between \( {x}_{1} \) and \( {x}_{2} \) by Theorem 74 . \n\n(b) There may be values of the function between \( {x}_{1} \) and \( {x}_{2} \) which are greater ... | Yes |
Theorem 85. If \( f\left( x\right) \) is continuous on the interval \( \overrightarrow{{x}_{1}{x}_{2}} \), and if the derivative exists at every point between \( {x}_{1} \) and \( {x}_{2} \), then there is a value of \( x, x = \xi \), between \( {x}_{1} \) and \( {x}_{2} \) such that \[ {f}^{\prime }\left( \xi \right) ... | Proof. Consider a function \( {f}_{1}\left( x\right) \) such that \[ {f}_{1}\left( x\right) = f\left( x\right) - \left( {x - {x}_{2}}\right) \cdot \frac{f\left( {x}_{1}\right) - f\left( {x}_{2}\right) }{{x}_{1} - {x}_{2}}; \] then \( {f}_{1}\left( {x}_{1}\right) = f\left( {x}_{2}\right) \) and \( {f}_{1}\left( {x}_{2}\... | Yes |
Theorem 86. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are continuous on an interval \( \bar{a}\bar{b} \), and if \( {f}_{1}^{\prime }\left( x\right) \) and \( {f}_{2}^{\prime }\left( x\right) \) exist between \( a \) and \( b,{f}_{2}^{\prime }\left( x\right) \neq \pm \infty \), and \( {f}_{2}^{\p... | Proof. Consider a function\n\n\[ {f}_{3}\left( x\right) = \frac{{f}_{1}\left( a\right) - {f}_{1}\left( b\right) }{{f}_{2}\left( a\right) - {f}_{2}\left( b\right) }\left\{ {{f}_{2}\left( x\right) - {f}_{2}\left( b\right) }\right\} - \left\{ {{f}_{1}\left( x\right) - {f}_{1}\left( b\right) }\right\} . \]\n\nSince \( {f}_... | Yes |
Theorem 87. If the first \( n \) derivatives of the function \( f\left( x\right) \) exist and are finite upon the interval a \( b \), there is a value of \( x,{x}_{n} \) on the interval \( {ab} \) such that\n\n\[ f\left( b\right) = f\left( a\right) + \frac{\left( b - a\right) }{1!}{f}^{\prime }\left( a\right) + \frac{{... | Proof. Let \( {R}_{n} \) be a constant such that\n\n\[ F\left( x\right) = f\left( x\right) - f\left( a\right) - \left( {x - a}\right) {f}^{\prime }\left( a\right) - \frac{{\left( x - a\right) }^{2}}{2!}{f}^{\prime \prime }\left( a\right) - \ldots - \frac{{\left( x - a\right) }^{n - 1}}{\left( {n - 1}\right) !}{f}^{\lef... | Yes |
Theorem 88. If \( {f}^{\left( n\right) }\left( x\right) \) exists and \( \left| {{f}^{\left( n\right) }\left( x\right) }\right| \) is less than a fixed quantity \( M \) for every \( x \) on the interval \( \overrightarrow{a}\overrightarrow{b} \) and for every \( n\left( {n = 1,2,\ldots }\right) \), then | \[ f\left( b\right) = f\left( a\right) + \frac{\left( b - 1\right) }{1!}{f}^{\prime }\left( a\right) + \ldots + \frac{{\left( b - a\right) }^{n}}{n!}{f}^{\left( n\right) }\left( a\right) + \ldots \] | No |
Theorem 89. If on some \( V\left( a\right) \) the first \( n \) derivatives of \( f\left( x\right) \) exist and are finite and on \( {V}^{ * }\left( a\right) {f}^{\left( n + 1\right) }\left( x\right) \) exists and is bounded, \( {}^{7} \) and if\n\n\[ 0 = {f}^{\prime }\left( a\right) = {f}^{\prime \prime }\left( a\righ... | Proof. By Taylor’s theorem, for every \( x \) in the vicinity of \( a \)\n\n\[ f\left( x\right) = f\left( a\right) + {\left( x - a\right) }^{n}{f}^{\left( n\right) }\left( a\right) + {\left( x - a\right) }^{n + 1} \cdot {f}^{\left( n + 1\right) }\left( {\xi }_{x}\right) ,\]\n\nwhere \( {\xi }_{x} \) is between \( x \) ... | Yes |
Theorem 90. If \( f\left( x\right) \) and \( \phi \left( x\right) \) are continuous and differentiable and \( \phi \left( x\right) \) is monotonic and \( {\phi }^{\prime }\left( x\right) \neq 0 \) and \( {\phi }^{\prime }\left( x\right) \neq \infty \) and\n\n(1) if \( \underset{x \doteq \infty }{L}f\left( x\right) = 0 ... | Proof. For every positive \( h \) we have, by the second mean-value theorem,\n\n\[ \frac{f\left( {x + h}\right) - f\left( x\right) }{\phi \left( {x + h}\right) - \phi \left( x\right) } = \frac{{f}^{\prime }\left( {\xi }_{x}\right) }{{\phi }^{\prime }\left( {\xi }_{x}\right) } \]\n\nwhere \( {\xi }_{x} \) lies between \... | Yes |
Theorem 91. If \( f\left( x\right) \) and \( \phi \left( x\right) \) are continuous and differentiable on some \( {V}^{ * }\left( a\right) \) and \( f\left( x\right) \) is bounded on every finite interval, while \( \phi \left( x\right) \) is monotonic and (1) \( \underset{x \doteq a}{L}f\left( x\right) = 0,\underset{x ... | Proof. If \( L\frac{{f}^{\prime }\left( x\right) }{{\phi }^{\prime }\left( x\right) } \) exists, the limit exists when the approach is only on values of \( x > a \) . Consider only such values of \( x \) . Then if \[ z = \frac{1}{x - a}, f\left( x\right) = f\left( {a + \frac{1}{z}}\right) = F\left( z\right) \] and \[ \... | Yes |
Theorem 92. If \( f\left( x\right) \) is continuous and \( {f}^{\prime }\left( x\right) \) exists for every \( x \) on an interval \( \overset{\overleftrightarrow{} }{ab} \), then \( {f}^{\prime }\left( x\right) \) takes on every value between any two of its values. | Proof. Consider any two values of \( {f}^{\prime }\left( x\right) ,{f}^{\prime }\left( {x}_{1}\right) \), and \( {f}^{\prime }\left( {x}_{2}\right) \) on the interval \( \overrightarrow{ab} \) . Consider, further, the function \( \frac{f\left( x\right) - f\left( {x}_{1}\right) }{x - {x}_{1}} \) on the interval between ... | Yes |
Theorem 93. If the derivative exists at every point on an interval, including its endpoints, it does not follow that the derivative is continuous or that it takes on its upper and lower bounds. | Proof. This is shown by the following example.\n\nThe curve shall lie between the \( x \) -axis and the parabola \( y = \frac{1}{2}{x}^{2} \) . The straight lines of slopes \( 1,1\frac{1}{2},1\frac{3}{4},\ldots ,1 + \frac{{2}^{n} - 1}{{2}^{n}}\ldots \) through the points \( \left( {\frac{1}{2},0}\right) ,\left( {\frac{... | Yes |
Theorem 94. If \( {f}^{\prime }\left( x\right) \) exists and is equal to zero for every value of \( x \) on the interval \( \overrightarrow{a}\overrightarrow{b} \) , then \( f\left( x\right) \) is a constant on that interval. | Proof. By Theorem 82, \( f\left( x\right) \) is continuous. Suppose \( f\left( x\right) \) not a constant, so that for two values of \( x,{x}_{1} \), and \( {x}_{2}, f\left( {x}_{1}\right) \neq f\left( {x}_{2}\right) \), then, by Theorem 85, there is a value of \( x, x = \xi \) between \( {x}_{1} \) and \( {x}_{2} \) s... | Yes |
Theorem 95. If \( {f}^{\prime }\left( x\right) \) exists and is positive for every value of \( x \) on the interval \( \overrightarrow{a}\overrightarrow{b} \), then \( f\left( x\right) \) is monotonic increasing on this interval. If \( {f}^{\prime }\left( x\right) \) is negative for every value of \( x \) on this inter... | Proof. If \( {f}^{\prime }\left( x\right) \) is positive for every value of \( x \), then it follows from Theorem 85, provided that \( f\left( x\right) \) is continuous, that the function is monotonic increasing, for if there were two values of \( x,{x}_{1} \) and \( {x}_{2} \), such that \( f\left( {x}_{1}\right) \geq... | Yes |
Theorem 96. If a function \( f\left( x\right) \) is monotonic increasing on an interval \( \overrightarrow{a}\overrightarrow{b} \), and if \( {f}^{\prime }\left( x\right) \) exists for every value of \( x \) on this interval, then there is no point on the interval for which \( {f}^{\prime }\left( x\right) \) is negativ... | Proof. If \( {f}^{\prime }\left( x\right) \) is negative for some value of \( x \), say \( {x}_{1} \), then\n\n\[ \underset{x \doteq {x}_{1}}{L}\frac{f\left( x\right) - f\left( {x}_{1}\right) }{x - {x}_{1}} = C,\text{ a negative number,}\]\n\nwhence there is a neighborhood of \( {x}_{1} \) on which \( f\left( x\right) ... | Yes |
Theorem 99. Every non-oscillating bounded function is integrable. | Proof. The proof runs, as in the preceding theorem, to the paragraph following (4). Let \( D \) and \( d \) be the upper and lower bounds of \( f\left( x\right) .\delta \), being arbitrary, can be so chosen that \( \delta = \frac{\varepsilon }{D - d} \) . Then\n\n\[ \n{O}_{{\pi }_{B}} = \mathop{\sum }\limits_{{k = 1}}^... | Yes |
Theorem 100. If \( f\left( x\right) \) is a constant, \( C \), then\n\n\[{\int }_{a}^{b}{Cdx} = C\left( {b - a}\right)\] | Proof. The function \( f\left( x\right) = C \) is integrable either according to Theorem 98 or Theorem 99 . Hence\n\n\[{\int }_{a}^{b}{Cdx} = {\iint }_{n \doteq \infty }\mathop{\sum }\limits_{{k = 1}}^{n}C\frac{b - a}{n} = {\iint }_{n \doteq \infty }n \cdot C \cdot \frac{b - a}{n} = C\left( {b - a}\right) .\] | Yes |
\[ {\int }_{a}^{b}{e}^{x}{dx} = {e}^{b} - {e}^{a} \] | Let\n\n\[ {S}_{\Delta x} = {e}^{a}{\Delta x} + {e}^{a + {\Delta x}} \cdot {\Delta x} + {e}^{a + {2\Delta x}} \cdot {\Delta x} + \ldots + {e}^{a + \left( {n - 1}\right) {\Delta x}} \cdot {\Delta x} \]\n\n\[ = {e}^{a} \cdot {\Delta x}\left\lbrack {1 + {e}^{\Delta x} + {e}^{2\Delta x} + \ldots + {e}^{\left( {n - 1}\right)... | Yes |
Theorem 102. In all cases where \( m \) is a whole number \( \neq - 1 \), and if \( a > 0, b > 0 \) for every value of \( m \neq - 1 \) ,\n\n\[ \n{\int }_{a}^{b}{x}^{m}{dx} = \frac{{b}^{m + 1} - {a}^{m + 1}}{m + 1}.\n\] | Proof.\n\n\[ \n{\int }_{a}^{b}{x}^{m}{dx} = \underset{q \doteq 1}{L}a\left( {q - 1}\right) \mathop{\sum }\limits_{{k = 0}}^{{n - 1}}{q}^{k}{\left( a{q}^{k}\right) }^{m} \n\]\n\n\[ \n= {a}^{m + 1}{\int }_{q \doteq 1}\left( {q - 1}\right) \left\lbrack {1 + \left( {q}^{m + 1}\right) + {\left( {q}^{m + 1}\right) }^{2} + \l... | Yes |
Theorem 103.\n\[ \n{\\int }_{a}^{b}\\frac{1}{x}{dx} = \\log b - \\log a,\\left( {0 < a < b}\\right) .\n\] | Proof. By equation (1) in the last theorem, since \( {q}^{m + 1} = {q}^{0} = 1 \) ,\n\n\[ \n{\\int }_{a}^{b}\\frac{1}{x}{dx} = \\underset{n \\doteq \\infty }{L}n\\left( {q - 1}\\right) \n\]\n\nbut \( n = \\frac{\\log \\left( \\frac{b}{a}\\right) }{\\log q} \), hence\n\n\[ \n{\\int }_{a}^{b}\\frac{1}{x}{dx} = \\underset... | Yes |
Theorem 104. If on an interval \( \overset{\overleftrightarrow{} }{ab} \) two functions \( f\left( x\right) \) and \( F\left( x\right) \) have the property that for every two values of \( x,{x}_{1} \) and \( {x}_{2} \), where \( a < {x}_{1} < {x}_{2} < b \) ,\n\n\[ f\left( {x}_{1}\right) \left( {{x}_{2} - {x}_{1}}\righ... | Proof. We consider first the case\n\n\[ f\left( {x}_{1}\right) \left( {{x}_{2} - {x}_{1}}\right) \leqq F\left( {x}_{2}\right) - F\left( {x}_{1}\right) \leqq f\left( {x}_{2}\right) \left( {{x}_{2} - {x}_{1}}\right) . \]\n\nThis gives\n\n\[ f\left( {x}_{1}\right) \leqq \frac{F\left( {x}_{2}\right) - F\left( {x}_{1}\right... | Yes |
Theorem 105. If \( a < b < c \), and if a bounded function \( f\left( x\right) \) is integrable from a to \( c \), then it is integrable from a to \( b \) and from \( b \) to \( c \) . | Proof. Suppose \( f\left( x\right) \) not integrable from \( a \) to \( b \), then by the definition of a limit (see Chap. II.) there must be a set of values of \( {}_{a}^{b}{S}_{\delta },\left\lbrack {{}_{a}^{b}{S}_{\delta }^{\prime }}\right\rbrack \), such that \( {}_{\delta }^{L}{}_{ \doteq 0}^{b}{}_{a}^{{S}_{\delta... | Yes |
Theorem 106. If \( a < b < c \) and if a bounded function \( f\left( x\right) \) is integrable from a to \( b \) and from \( b \) to \( c \), then \( f\left( x\right) \) is integrable from a to \( c \) and \( {\int }_{a}^{c}f\left( x\right) {dx} = {\int }_{a}^{b}f\left( x\right) {dx} + {\int }_{b}^{c}f\left( x\right) {... | Proof. Since \( {\int }_{a}^{b}f\left( x\right) {dx} \) and \( {\int }_{b}^{c}f\left( x\right) {dx} \) exist, we know by Theorem 26 that for every \( \varepsilon \) there exists a \( {\delta }_{c}^{\prime } \) such that for \( {}_{a}^{b}{S}_{\delta } \) where \( \delta \leqq {\delta }_{\varepsilon } \) ,\n\n\[ \left| {... | Yes |
Theorem 107. Provided both integrals exist, \( {}^{3} \) and \( a < b \) , \[ {\int }_{a}^{b}\left| {f\left( x\right) }\right| {dx} \geqq \left| {{\int }_{a}^{b}f\left( x\right) {dx}}\right| . \] | Proof. \[ \sum \left| {f\left( {\xi }_{k}\right) }\right| {\Delta }_{k}x \geqq \left| {\sum f\left( {\xi }_{k}\right) {\Delta }_{k}x}\right| . \] Hence for every \( {S}_{\delta }\left| {f\left( x\right) }\right| \) there is a smaller or equal \( {S}_{\delta }f\left( x\right) \), the \( \delta \) ’s being the same. Henc... | Yes |
Theorem 108. If \( {\int }_{a}^{b}f\left( x\right) {dx} \) exists, then \( {\int }_{b}^{a}f\left( x\right) {dx} \) exists and\n\n\[{\int }_{a}^{b}f\left( x\right) {dx} = - {\int }_{b}^{a}f\left( x\right) {dx}.\] | Proof. This is a consequence of the theorem (Corollary 1 Theorem 27) that\n\n\[ \underset{x \doteq a}{L}\left( {-f\left( x\right) }\right) = - \underset{x \doteq a}{L}f\left( x\right) \]\n\nfor to every \( S \) used in defining \( {\int }_{a}^{b}f\left( x\right) {dx} \) corresponds a sum equal to \( - S \) which is use... | Yes |
Theorem 111. If \( C \) is any constant and if \( f\left( x\right) \) is integrable on \( \overrightarrow{ab} \), then \( {Cf}\left( x\right) \) is integrable on \( {ab} \) and\n\n\[{\int }_{a}^{b}{Cf}\left( x\right) {dx} = C{\int }_{a}^{b}f\left( x\right) {dx}.\] | Proof.\n\n\[{S}_{\delta } = \mathop{\sum }\limits_{{k = 1}}^{n}f\left( {\xi }_{k}\right) {\Delta }_{k}x\]\n\n\( {}^{4} \) First stated formally by H. Lebesgue, Leçons sur l’Intégration, Chapter VII, page 98.\n\nis an \( {S}_{\delta } \) of the set which defines \( {\int }_{a}^{b}f\left( x\right) {dx} \) and\n\n\[{S}_{\... | No |
Theorem 112. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are any two functions each integrable on the interval \( \overrightarrow{a}\overrightarrow{b} \), then \( f\left( x\right) = {f}_{1}\left( x\right) \pm {f}_{2}\left( x\right) \) is integrable on \( \overrightarrow{a}\overrightarrow{b} \) and\... | Proof. The proof depends directly upon the theorem that if \( \underset{x = a}{L}{\phi }_{1}\left( x\right) = {b}_{1} \), and \( \underset{x = a}{L}{\phi }_{2}\left( x\right) = \) \( {b}_{2} \), then \( \underset{x \doteq a}{L}\left( {{\phi }_{1}\left( x\right) \pm {\phi }_{2}\left( x\right) }\right) = {b}_{1} \pm {b}_... | Yes |
Theorem 113. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are integrable on \( \overset{\overleftrightarrow{} }{ab} \) and such that for every value of x on \( \overrightarrow{ab}{f}_{1}\left( x\right) \geqq {f}_{2}\left( x\right) \), then\n\n\[{\int }_{a}^{b}{f}_{1}\left( x\right) {dx} \geqq {\int ... | Proof. Since \( {S}_{1} \) is always greater than or equal to \( {S}_{2} \), then, by Theorem 34, \( \underset{\delta = 0}{L}{S}_{1} \geqq \underset{\delta \doteq 0}{L}{S}_{2} \) , which proves the theorem. | No |
Theorem 114. (Maximum-Minimum Theorem.) If\n\n(1) the product \( {f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) \) and the factor \( {f}_{1}\left( x\right) \) are integrable on \( \bar{a}\bar{b} \) ,\n\n(2) \( {f}_{1}\left( x\right) \) is always positive or always negative on \( \overrightarrow{a}\overrightarrow{b... | Proof. By Theorem 111,\n\n\[ M \cdot {\int }_{a}^{b}{f}_{1}\left( x\right) {dx} = {\int }_{a}^{b}M \cdot {f}_{1}\left( x\right) {dx} \]\n\nand\n\n\[ m \cdot {\int }_{a}^{b}{f}_{1}\left( x\right) {dx} = {\int }_{a}^{b}m \cdot {f}_{1}\left( x\right) {dx}. \]\n\nBut in case \( {f}_{1}\left( x\right) \) is always positive,... | Yes |
Theorem 117. In case the integral of \( f\\left( x\\right) \) exists on the interval \( \\bar{a}\\bar{b} \) , | \[ {}_{a}^{b}{Mf}\\left( x\\right) = \\frac{{\\int }_{a}^{b}f\\left( x\\right) {dx}}{b - a}. \] | Yes |
Theorem 118. If \( f\left( x\right) \) is integrable on an interval \( \overrightarrow{ab} \), and if \( x \) is any point of \( \overrightarrow{ab} \) , \( {\int }_{a}^{x}f\left( x\right) {dx} \) is a continuous function of \( x \) . | Proof. \( {\int }_{a}^{x}f\left( x\right) {dx} \) exists, by Theorem 105, and by the definition of a continuous function we need only to show that\n\n\[ \underset{{x}^{\prime } \doteq x}{L}\left( {{\int }_{a}^{{x}^{\prime }}f\left( x\right) {dx} - {\int }_{a}^{x}f\left( x\right) {dx}}\right) = 0. \]\n\nBy the theorems ... | Yes |
Theorem 119. If \( f\left( x\right) \) is continuous on an interval \( \overrightarrow{a},\overrightarrow{b},{\int }_{a}^{x}f\left( x\right) {dx}\left( {a < x < b}\right) \) possesses a derivative with respect to \( x \) such that\n\n\[ \frac{d}{dx}{\int }_{a}^{x}f\left( x\right) {dx} = f\left( x\right) \] | Proof. By the preceding theorem \( {\int }_{a}^{x}f\left( x\right) {dx} \) is continuous. To form the derivative we investigate the expression\n\n\[ \frac{{\int }_{a}^{{x}^{\prime }}f\left( x\right) {dx} - {\int }_{a}^{x}f\left( x\right) {dx}}{{x}^{\prime } - x} = \frac{{\int }_{x}^{{x}^{\prime }}f\left( x\right) {dx}}... | Yes |
Theorem 120. If \( f\left( x\right) \) is any continuous function on the interval \( \overrightarrow{ab} \), and \( F\left( x\right) \) any function on this interval such that\n\n\[ \frac{d}{dx}F\left( x\right) = f\left( x\right) \]\n\nthen \( F\left( x\right) \) differs from \( {\int }_{a}^{x}f\left( x\right) {dx} \) ... | Proof. Let \( F\left( x\right) = {\int }_{a}^{x}f\left( x\right) {dx} + \phi \left( x\right) \) .\n\nSince \( F\left( x\right) \) and \( {\int }_{a}^{x}f\left( x\right) {dx} \) are both differentiable,\n\n\[ \frac{d}{dx}F\left( x\right) = \frac{d}{dx}\left( {{\int }_{a}^{x}f\left( x\right) {dx} + \phi \left( x\right) }... | Yes |
Theorem 121. If \( f\left( x\right) \) is a continuous function on an interval \( \overset{\overleftrightarrow{} }{ab} \) and \( F\left( x\right) \) is such that\n\n\[ \frac{d}{dx}F\left( x\right) = f\left( x\right) \]\n\nthen\n\n\[ {\int }_{a}^{b}f\left( x\right) {dx} = F\left( b\right) - F\left( a\right) \] | Proof. By the last theorem,\n\n\[ {\int }_{a}^{x}f\left( x\right) {dx} = F\left( x\right) + c. \]\n\nBut\n\n\[ 0 = {\int }_{a}^{a}f\left( x\right) {dx} = F\left( a\right) + c. \]\n\nTherefore\n\n\[ - F\left( a\right) = c. \]\n\nWhence\n\n\[ {\int }_{a}^{b}f\left( x\right) {dx} = F\left( b\right) + c = F\left( b\right) ... | Yes |
Theorem 122. (Integration by parts.)\n\n\[ \n{\int }_{a}^{b}{f}_{1}\left( x\right) \cdot {f}_{2}^{\prime }\left( x\right) {dx} = {\left\lbrack {f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) \right\rbrack }_{a}^{b} - {\int }_{a}^{b}{f}_{2}\left( x\right) \cdot {f}_{1}^{\prime }\left( x\right) {dx}, \]\n\nprovided \... | Proof. By Theorem 75,\n\n\[ \n\frac{d}{dx}\left( {{f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) }\right) = {f}_{1}\left( x\right) \cdot {f}_{2}^{\prime }\left( x\right) + {f}_{2}\left( x\right) \cdot {f}_{1}^{\prime }\left( x\right) . \]\n\nTherefore\n\n\[ \n{\int }_{a}^{b}\frac{d}{dx}\left( {{f}_{1}\left( x\righ... | Yes |
Theorem 123. (Integration by substitution.) If \( y = \phi \left( x\right) \) has a continuous derivative at every point of \( \bar{a}\bar{b} \) and \( f\left( y\right) \) is continuous for all values taken by \( y = \phi \left( x\right) \) as \( x \) varies from a to \( b \) ,\n\n\[ \n{\int }_{A}^{B}f\left( y\right) {... | Proof. By Theorem 120 and by Theorem 79,\n\n\[ \n{\int }_{A}^{\phi \left( x\right) }f\left( y\right) {dy} = {\int }_{a}^{x}\frac{d}{dx}\left( {{\int }_{A}^{\phi \left( x\right) }f\left( y\right) {dy}}\right) {dx} + C = {\int }_{a}^{x}f\left( y\right) \frac{dy}{dx} \cdot {dx} + C, \n\] \n\n\( C \) being an arbitrary con... | Yes |
Theorem 124.\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} = {\int }_{A}^{B}f\left( {\phi \left( y\right) }\right) \frac{dx}{dy}{dy} \]\n\nwhere \( x = \phi \left( y\right) \) and \( a = \phi \left( A\right), b = \phi \left( B\right) \) ; provided that both integrals exist, and that \( \phi \left( y\right) \) is non-oscil... | Proof.\n\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} = \underset{n \doteq \infty }{L}\mathop{\sum }\limits_{{k = 1}}^{n}f\left( {\xi }_{k}\right) {\Delta }_{k}x \]\n\n(1)\n\nwhenever the least upper bound of \( {\Delta }_{k}x \) for each \( n \) approaches zero as \( n \) approaches \( + \infty \) . Now let \( {\Delta y... | Yes |
Theorem 126. A necessary and sufficient condition that a function \( f\left( x\right) \), defined, single-valued, and bounded on an interval \( \overrightarrow{a}\overrightarrow{b} \), is integrable is that\n\n\[ \underset{\begin{matrix} {\delta \doteq 0} \end{matrix}}{L}{O}_{\delta } = 0 \] | Proof. The condition is necessary.\n\nBy Theorem 125 the integrability of \( f\left( x\right) \) implies \( \underline{B}{O}_{\pi } = 0 \) . Hence for every \( \varepsilon \) there exists a partition \( \pi \) such that\n\n\[ {O}_{\pi } < \varepsilon \]\n\nBy Lemma 4 there exists a \( {\delta }_{\varepsilon } \) such t... | Yes |
Theorem 127. A necessary and sufficient condition that a function, defined, single-valued, and bounded on an interval \( \overset{\overleftrightarrow{} }{ab} \), shall be integrable on that interval is that for every pair of positive numbers \( \sigma \) and \( \lambda \) there exists a partition \( \pi \) such that th... | Proof. The condition is necessary.\n\nIf for a given pair of positive numbers \( \sigma \) and \( \lambda \) there exists no \( \pi \) such as is required by the theorem, then \( {O}_{\pi } > \sigma \cdot \lambda \) for every \( \pi \), which is contrary to the conclusion of Theorem 125 that\n\n\[ \underline{B}{O}_{\pi... | Yes |
Theorem 128. A necessary and sufficient condition for the integrability of a function \( f\left( x\right) \) on an interval ’ \( a \) ’ is that for every \( \sigma > 0 \) the set of points \( \left\lbrack {x}_{\sigma }\right\rbrack \) at which the oscillation of \( f\left( x\right) \) is greater than or equal to \( \si... | Proof. If at every point of an interval \( \overrightarrow{cd} \) the oscillation of \( f\left( x\right) \) is less than \( \sigma \), then about each point of \( \overrightarrow{cd} \) there is a segment upon which the oscillation is less than \( \sigma \), and hence by Theorem 11, Chapter II, there is a partition of ... | Yes |
Theorem 129. A perfect set of points is not numerably infinite. | Proof. Suppose the theorem not true. Then there exists a sequence of points \( \left\{ {x}_{n}\right\} \) containing every point of a perfect set \( \left\lbrack x\right\rbrack \) . Let \( {P}_{1} \) be any point of \( \left\lbrack x\right\rbrack \), and \( \overline{{a}_{1}{b}_{1}} \) a segment containing \( {P}_{1} \... | Yes |
Theorem 130. A numerably infinite set of sets of points each of content zero cannot contain every point of any interval. | Proof. Let the set of sets be ordered into a sequence \( \left\{ {\left\lbrack x\right\rbrack }_{n}\right\} \) . We show that on every segment \( \overline{ab} \) there is at least one point not of \( \left\{ {\left\lbrack x\right\rbrack }_{n}\right\} \) . Since \( {\left\lbrack x\right\rbrack }_{1} \) is of content ze... | Yes |
Theorem 131. The points of discontinuity of an integrable function form at most a set consisting of a numerable set of sets, each of content zero. | Proof. Let \( {\sigma }_{1},{\sigma }_{2},{\sigma }_{3},\ldots \) be any set of numbers such that\n\n\[{\sigma }_{n} > {\sigma }_{n + 1}\]\n\nand\n\n\[\underset{n \doteq \infty }{L}{\sigma }_{n} = 0\]\n\nBy Theorem 128 the set of points \( \left\lbrack {x}_{{\sigma }_{n}}\right\rbrack \) at which the oscillation of \( ... | Yes |
Theorem 132. If a function \( f\left( x\right) \) is integrable on an interval \( \overrightarrow{ab} \), then it is continuous at a set of points which is everywhere dense on \( {ab} \) . | Proof. If the theorem fails to hold, then there exists an interval \( \overrightarrow{a}\overrightarrow{b} \) on which the function is discontinuous at every point. By Theorem 131 an integrable function is discontinuous at most on a numerably infinite set of sets each of content zero, and by Theorem 130 such sets of se... | Yes |
Theorem 133. If\n\n\\[ \n{\\int }_{a}^{X}f\\left( x\\right) {dx} = 0 \n\\]\n\nfor every \\( X \\) of \\( \\overrightarrow{ab} \\), then \\( f\\left( x\\right) = 0 \\) on a set of points everywhere dense on \\( \\overrightarrow{ab} \\), and for every \\( \\sigma > 0 \\) the points where \\( \\left| {f\\left( x\\right) }... | Proof. At every point \\( X \\) where \\( f\\left( x\\right) \\) is continuous, according to the corollary of Theorem 119,\n\n\\[ \n\\frac{d}{dX}{\\int }_{a}^{X}f\\left( x\\right) {dx} = f\\left( X\\right) = 0, \n\\]\n\nsince \\( {\\int }_{a}^{X}f\\left( x\\right) {dx} \\) is a constant. The points of continuity of \\(... | Yes |
Theorem 134. If\n\n\[ \n{\int }_{a}^{X}f\left( x\right) {dx} = {\int }_{a}^{X}\phi \left( x\right) {dx} \n\]\n\nfor every \( X \) of \( \overrightarrow{ab} \), then \( f\left( x\right) = \phi \left( x\right) \) on a set of points everywhere dense on \( \overrightarrow{ab} \), and for every \( \sigma > 0 \) the points w... | Proof. Apply the theorem above to \( f\left( x\right) - \phi \left( x\right) \). | No |
Theorem 135. If \( f\left( x\right) \) is integrable from a to \( b \), then \( \left| {f\left( x\right) }\right| \) is integrable from a to \( b{.}^{10} \) | Proof. Since\n\n\[ 0 \leqq {O}_{\pi }\left| {f\left( x\right) }\right| \leqq {O}_{\pi }f\left( x\right) \]\n\nit follows that \( \underline{B}{O}_{\pi }f\left( x\right) = 0 \) implies \( \underline{B}{O}_{\pi }\left| {f\left( x\right) }\right| = 0 \), and hence the integrability of \( f\left( x\right) \) implies the in... | Yes |
Theorem 136. If \( f\left( x\right) \) and \( \phi \left( x\right) \) are both integrable on an interval \( \overrightarrow{ab} \), then\n\n\[ f\left( x\right) \cdot \phi \left( x\right) \]\n\n(1)\n\nis integrable on \( \overset{\overleftrightarrow{} }{ab} \) ; and, provided there is a constant \( m > 0 \) such that \(... | Proof. Since \( f\left( x\right) \) and \( \phi \left( x\right) \) are both integrable on \( \overrightarrow{ab} \), it follows that for every pair of positive numbers \( \sigma \) and \( \lambda \) there is a partition \( {\pi }_{1} \) for \( f\left( x\right) \) and a partition \( {\pi }_{2} \) for \( \phi \left( x\ri... | Yes |
Theorem 137. If \( f\left( x\right) \) is an integrable function on an interval \( \overrightarrow{ab} \), and if \( \phi \left( y\right) \) is a continuous function on an interval \( \underline{B}f\bar{B}f \), where \( \underline{B}f \) and \( \bar{B}f \) are the lower and upper bounds respectively of \( f\left( x\rig... | Proof. By Theorem 48 there exists for every \( \sigma > 0 \) a \( {\delta }_{\sigma } \) such that for \( \left| {{y}_{1} - {y}_{2}}\right| < {\delta }_{\sigma } \), \[ \left| {\phi \left( {y}_{1}\right) - \phi \left( {y}_{2}\right) }\right| < \sigma \] (1) Since \( f\left( x\right) \) is integrable on \( \overrightarr... | Yes |
Theorem 138. If \( {\int }_{x}^{b}f\left( x\right) {dx} \) exists for every \( x, a < x < b \), then a necessary and sufficient condition that \[ \underset{x \doteq a}{L}{\int }_{x}^{b}f\left( x\right) {dx} \] shall exist and be finite is that for every \( \varepsilon \) there exists a \( {V}_{\varepsilon }^{ * }\left(... | Proof. This theorem is a special case of Theorem 27, since, by Theorem 110, \[ {\int }_{{x}_{1}}^{{x}_{2}}f\left( x\right) {dx} = {\int }_{{x}_{1}}^{b}f\left( x\right) {dx} - {\int }_{{x}_{2}}^{b}f\left( x\right) {dx}. \] | Yes |
Theorem 139. If \( {\int }_{x}^{b}f\left( x\right) {dx} \) exists for every \( x, a < x < b \), and if\n\n\[ \n{\int }_{x \doteq a}{\int }_{x}^{b}\left| {f\left( x\right) }\right| {dx}\n\]\n\nis finite, then\n\n\[ \n{L}_{x \doteq a}{\int }_{x}^{b}f\left( x\right) {dx}\n\]\n\nexists and is finite. \( {}^{2} \) | Proof. By the necessary condition of Theorem 138 there is a \( {V}_{\varepsilon }^{ * }\left( a\right) \) corresponding to any preassigned \( \varepsilon \) such that for any two values of \( x,{x}_{1} \) and \( {x}_{2} \), which lie on the segment \( \overrightarrow{ab} \) and on \( {V}_{\varepsilon }^{ * }\left( a\ri... | Yes |
Theorem 140. If \( {\int }_{x}^{b}f\left( x\right) {dx} \) exists for every \( x \) on the segment \( \overline{ab} \), and if \( {\left( x - a\right) }^{k}f\left( x\right) \) is bounded on \( {V}^{ * }\left( a\right) \) for some value of \( k,0 < k < 1 \), then\n\n\[ \underset{x \doteq a}{L}{\int }_{x}^{b}f\left( x\ri... | Proof. By hypothesis \( {\left( x - a\right) }^{k}\left| {f\left( x\right) }\right| \leqq M \), i.e.,\n\n\[ \left| {f\left( x\right) }\right| \leqq \frac{M}{{\left( x - a\right) }^{k}} \]\n\nwhere \( M \) may be taken greater than one. The proof of the theorem consists in showing that for every \( \varepsilon \) there ... | Yes |
Theorem 141. If for any positive number \( m \) and for any \( k \geqq 1 \) there exists a \( {V}^{ * }\left( a\right) \) on which \( f\left( x\right) \) does not change sign, and on which \( {\left( x - a\right) }^{k}f\left( x\right) > m \) for every \( x \), then\n\n\[ \n{L}_{x \doteq a}{\int }_{x}^{b}f\left( x\right... | Proof. (1) In case\n\n\[ \n{\int }_{x}^{b}f\left( x\right) {dx} \n\]\n\nfails to exist for some value of \( x \) between \( a \) and \( b \) ,\n\n\[ \n{\int }_{x \doteq a}{\int }_{x}^{b}f\left( x\right) {dx} \n\]\n\nfails to exist because the limitand function does not exist.\n\n(2) If\n\n\[ \n{\int }_{x}^{b}f\left( x\... | Yes |
Theorem 142. If\n\n\[ \n{L}_{x \doteq a}{\int }_{x}^{b}f\left( x\right) {dx} \]\n\nexists and is finite and if \( f\left( x\right) \) approaches infinity monotonically as \( x \doteq a \) on some \( {V}^{ * }\left( a\right) \) , then\n\n\[ \n\underset{x \doteq a}{L}\left( {x - a}\right) \cdot f\left( x\right) = 0 \]\n\... | Proof. By means of Theorem 138 it follows from the hypothesis that for every \( \varepsilon \) there exists a \( {V}_{\varepsilon }^{ * }\left( a\right) \) within \( {V}^{ * }\left( a\right) \) such that for every \( {x}_{1} \) and \( {x}_{2} \) on \( {ab} \), and also on \( {V}_{\varepsilon }^{ * }\left( a\right) \) ,... | Yes |
Theorem 145. If for a function \( {f}_{1}\left( x\right) \) which does not change sign in the neighborhood of \( x = a \) there exists a monotonic function \( {f}_{2}\left( x\right) \) infinite of the same rank as \( {f}_{1}\left( x\right) \) as \( x \) approaches \( a,{\int }_{x}^{b}{f}_{1}\left( x\right) {dx} \) and ... | Proof. By hypothesis\n\n\[ {\underset{ \doteq }{L}}_{x}{\int }_{x}^{b}{f}_{1}\left( x\right) {dx} \]\n\nexists and is finite. Hence, by Theorem 143,\n\n\[ {L}_{x \doteq a}{\int }_{x}^{b}{f}_{2}\left( x\right) {dx} \]\n\nexists and is finite. Therefore, by Theorem 142,\n\n\[ \underset{x \doteq a}{L}\left( {x - a}\right)... | Yes |
Theorem 146. If\n\n\[ \n{\int }_{a}^{x}f\left( x\right) {dx} \]\n\nexists for every \( x, a < x \), then a necessary and sufficient condition that\n\n\[ \n\underset{x \doteq \infty }{L}{\int }_{a}^{x}f\left( x\right) {dx} \]\n\nexists and is finite, is that for every \( \varepsilon \) there exists a \( {D}_{\varepsilon... | Proof. The theorem is a direct consequence of Theorems 105 and 27. | No |
Theorem 147. If\n\n\\[ \n{\\int }_{a}^{x}f\\left( x\\right) {dx} \n\\]\n\nexists for every \\( x \\) greater than \\( a \\), and if\n\n\\[ \n{\\int }_{x \\doteq \\infty }{\\int }_{a}^{x}\\left| {f\\left( x\\right) }\\right| {dx} \n\\]\n\nis finite, \\( {}^{7} \\) then\n\n\\[ \n{L}_{x \\doteq \\infty }{\\int }_{a}^{x}f\... | Proof. The proof is like that of Theorem 139. | No |
Theorem 148. If\n\n\[ \n{\int }_{a}^{x}f\left( x\right) {dx} \]\n\nexists for every \( x \) greater than \( a \), and if \( {\left( x - a\right) }^{k} \cdot f\left( x\right) \) is bounded as \( x \) approaches infinity for some \( k, k > 1 \), then\n\n\[ \n\underset{x \doteq \infty }{L}{\int }_{a}^{x}f\left( x\right) {... | Proof. If in the proof of Theorem 140 we write \( {D}_{\varepsilon }^{1 - k} = \frac{\varepsilon \left( {1 - k}\right) }{M} \) instead of \( {\delta }_{\varepsilon }^{1 - k} = \frac{\varepsilon \left( {1 - k}\right) }{M} \), and use Theorem 146 instead of 138 , the proof of Theorem 140 will apply to Theorem 148. | No |
Theorem 149. If \( f\left( x\right) \) does not change sign for \( x \) greater than some fixed number \( D \), and if for some positive number \( m \) and some number \( k \leqq 1,\left| {{\left( x - a\right) }^{k} \cdot f\left( x\right) }\right| > m \) for every \( x \) greater than \( D \), then\n\n\[ \underset{x \d... | Proof. By making suitable changes in the proof of Theorem 141 so as to make \( {x}_{1} \) and \( {x}_{2} \) approach infinity instead of \( a \), that proof applies to this theorem. | No |
Theorem 150. If\n\n\[ \underset{x \doteq \infty }{L}{\int }_{a}^{x}f\left( x\right) {dx} \]\n\nexists and is finite, and if \( f\left( x\right) \) is monotonic for all values of \( x \) greater than some fixed number, then\n\n\[ \underset{x \doteq \infty }{L}\left( {x - a}\right) \cdot f\left( x\right) = 0. \] | Proof. By making slight modifications of the proof of Theorem 142, that proof applies to this theorem. | No |
Theorem 152. If\n\n(1) \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are infinitesimals of the same rank as \( x \) approaches infinity, or if \( {f}_{1}\left( x\right) \) is of lower order than \( {f}_{2}\left( x\right) \) ,\n\n(2) \( {\int }_{a}^{x}{f}_{1}\left( x\right) {dx} \) and \( {\int }_{a}^{x}... | Proof like that of Theorem 144. | No |
Theorem 153. If for a function \( {f}_{1}\left( x\right) \) which does not change sign in the neighborhood of \( x = \infty \) there exists a monotonic function \( {f}_{2}\left( x\right) \) such that \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are infinitesimals of the same rank as \( x \) approaches ... | The proof is like that of Theorem 145. | No |
Theorem 154. If \( a < b < c \), and if two of the three simple improper definite integrals\n\n\[{\int }_{a}^{b}f\left( x\right) {dx},\;{\int }_{b}^{c}f\left( x\right) {dx},\;\text{ and }\;{\int }_{a}^{c}f\left( x\right) {dx}\]\n\nexist, then the third exists and\n\n\[{\int }_{S}^{b}f\left( x\right) {dx} + {\int }_{S}^... | Proof. If \( b \) is a point at which the integral exists improperly, and if\n\n\[{\int }_{a}^{b}f\left( x\right) {dx}\text{ and }{\int }_{b}^{c}f\left( x\right) {dx}\]\n\nboth exist, then by the definition of\n\n\[{\int }_{a}^{c}f\left( x\right) {dx}\]\n\nthe latter exists and is equal to the sum of the two former.\n\... | Yes |
Theorem 155. If\n\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} \n\]\n\nexists, then\n\n\[ \n{\int }_{b}^{a}f\left( x\right) {dx} \n\]\n\nexists and\n\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} = - {\int }_{b}^{a}f\left( x\right) {dx}. \n\] | Proof. In case the integral exists improperly only at one point of the interval, then the theorem is an immediate consequence of Theorem 108 and Corollary 1, Theorem 27. (If \( \left. {\underset{x = a}{L}f\left( x\right) = K\text{, then}\underset{x = a}{L}\{ - f\left( x\right) \} = - K\text{.}}\right) \) The theorem in... | No |
Theorem 156. If \( c \) is a constant and if the simple improper definite integral of \( f\left( x\right) \) exists on \( \overrightarrow{ab} \), then the simple improper definite integral of \( c \cdot f\left( x\right) \) exists on \( \overrightarrow{ab} \) and\n\n\[ c{\int }_{a}^{b}f\left( x\right) {dx} = {\int }_{a}... | Proof. The theorem is a direct consequence of Theorems 111 and 34. | No |
Theorem 157. If the simple improper definite integrals of \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) both exist on \( {ab} \), then the simple improper definite integral of \( {f}_{1}\left( x\right) + {f}_{2}\left( x\right) \) and of \( {f}_{1}\left( x\right) - {f}_{2}\left( x\right) \) both exist an... | Proof. The theorem is a direct consequence of Theorems 112 and 34. | No |
Theorem 158. If the simple improper definite integrals of \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) both exist, and if \( {f}_{1}\left( x\right) \geqq {f}_{2}\left( x\right) \), then \[ {\int }_{a}^{b}{f}_{1}\left( x\right) {dx} \geqq {\int }_{S}^{b}{f}_{2}\left( x\right) {dx}. \] | Proof. The theorem is a direct consequence of Theorem 113 and Corollary 2, Theorem 40. | No |
Theorem 159. If\n\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} \n\]\n\nexists, then\n\n\[ \n{\int }_{a}^{x}f\left( x\right) {dx} \n\]\n\nis a continuous function of the limit of integration on the interval \( \bar{a}\bar{b} \) . | Proof. If \( x \) is a point at which the integral exists properly, the theorem is the same as 118. If \( x \) is a point at which the integral exists improperly, then the theorem follows from Theorems 138 and 27. | No |
Theorem 160. If\n\n\\[ \n{\\int }_{a}^{b}f\\left( x\\right) {dx} \n\\]\n\n exists, it does not follow that\n\n\\[ \n{\\int }_{a}^{b}\\left| {f\\left( x\\right) }\\right| {dx} \n\\]\n\nexists. | Proof. Let\n\n\\[ \n{x}_{1},{x}_{2},{x}_{3},\\ldots ,{x}_{n},\\ldots \n\\]\n\nbe an infinite sequence of points on \\( {01}\\mathrm{\\;m} \\) the order indicated from 1 towards 0 such that\n\n\\[ \n{\\int }_{{x}_{n}}^{{x}_{n - 1}}\\frac{dx}{x} = \\frac{1}{n} \n\\]\n\n\nConsider a function \\( f\\left( x\\right) \\) def... | Yes |
Theorem 161. If \( a < b < c \) and if of the integrals\n\n\[{\int }_{b}^{b}{f}_{a}f\left( x\right) {dx},{\int }_{b}^{c}{f}_{b}\left( x\right) {dx},{\int }_{b}^{c}{f}_{a}f\left( x\right) {dx}\]\n\neither\n\n\[\\text{(a)}{\int }_{b}^{b}f\left( x\right) {dx}\\;\\text{and}\\{int }_{b}^{c}f\left( x\right) {dx}\\text{exist,... | Proof. Every set \( I \) of intervals on \( \\overrightarrow{ac} \) may be regarded as composed of a set \( \\bar{I} \) on \( \\overleftarrow{ab} \) and a set \( \\overline{\\bar{I}} \) on \( \\bar{b}\\bar{c} \), while, conversely, every pair of sets \( \\bar{I} \) and \( \\overline{\\bar{I}} \) constitute a set \( I \... | Yes |
Theorem 162. If \( {\int }_{a}^{b}f\left( x\right) {dx} \) exists, then \( {\int }_{b}^{a}f\left( x\right) {dx} \) exists and\n\n\[{\int }_{b}^{b}f\left( x\right) {dx} = - {\int }_{b}^{a}f\left( x\right) {dx}.\] | Proof. By Theorem 108, for every \( I \)\n\n\[{\int }_{aI}^{b}f\left( x\right) {dx} = - {\int }_{bI}^{a}f\left( x\right) {dx}\]\n\nwhence\n\n\[{\int }_{b}^{b}f\left( x\right) {dx} = - {\int }_{b}^{a}f\left( x\right) {dx}.\] | Yes |
Theorem 163. If \( {\int }_{b}^{b}f\left( x\right) {dx} \) exists, then \( {\int }_{b}^{b}c \cdot f\left( x\right) {dx} \) exists and | Proof. This is a direct consequence of Theorems 111 and 34. | No |
Theorem 164. If \( {\int }_{a{P}_{0}}^{b}{f}_{1}\left( x\right) {dx} \) and \( {\int }_{a{P}_{0}}^{b}{f}_{2}\left( x\right) {dx} \) both exist, then \( {\int }_{a}^{b}\left( {{f}_{1}\left( x\right) \pm {f}_{2}\left( x\right) }\right) {dx} \) exists and \[ {\int }_{b}^{b}{f}_{1}\left( x\right) {dx} \pm {\int }_{b}^{b}{f... | Proof. This is a direct consequence of Theorems 112 and 34. | No |
Theorem 165. If \( {f}_{1}\left( x\right) \geqq {f}_{2}\left( x\right) \), then\n\n\[{\int }_{a}^{b}{f}_{1}\left( x\right) {dx} \geqq {\int }_{b}^{b}{f}_{2}\left( x\right) {dx}\]\n\nprovided these integrals exist. | Proof. By Theorems 113 and 40. | No |
Theorem 166. If \( {\int }_{a}^{b}{f}_{1}\left( x\right) {dx} \) and \( {\int }_{a}^{b}{f}_{2}\left( x\right) {dx} \) both exist,\n\n\[{\int }_{b}^{b}{f}_{1}\left( x\right) \cdot {f}_{2}\left( x\right) {dx}\]\ndoes not in general exist. | Proof. Let \( {f}_{1}\left( x\right) = {f}_{2}\left( x\right) = \frac{1}{\sqrt{x}} \) . In this case the hypothesis of the theorem is verified but the product, \( \frac{1}{x} \), fails to be integrable on the interval 0 1 . | Yes |
Theorem 167. \( {\int }_{a}^{x}f\left( x\right) {dx} \) is a continuous function of \( x \) . | Proof. If \( x \) is a point at which the integral exists properly, the continuity follows by Theorem 118. If \( x \) is a point of the set \( {P}_{0} \), then, by Theorem 26, we need to show that for every \( \varepsilon \) there is a \( {\delta }_{\varepsilon } \), such that for every interval \( \overrightarrow{{a}^... | Yes |
Theorem 168. If \( f\left( x\right) \) is integrable with respect to \( {P}_{0} \), and if \( {P}_{1} \) is a set of points of content zero, then \( f\left( x\right) \) is integrable with respect to the set \( {P}_{2} \) consisting of all points in \( {P}_{0} \) and in \( {P}_{1} \) and\n\n\[{\int }_{b}^{b}f\left( x\ri... | Proof. Obviously the set \( {P}_{2} \) is of content zero. Any set of intervals \( I \) not containing a point of \( {P}_{2} \) is also a set \( \bar{I} \) not containing a point of \( {P}_{0} \) . Hence any value approached by \( {\int }_{a\bar{I}}^{b}f\left( x\right) {dx} \) as \( m\left( \bar{I}\right) \) approaches... | Yes |
Theorem 169. If \( {f}_{1}\left( x\right) \) is integrable with respect to \( {P}_{1} \) and \( {f}_{2}\left( x\right) \) is integrable with respect to \( {P}_{2} \), then \( {f}_{1}\left( x\right) \pm {f}_{2}\left( x\right) \) is integrable with respect to the set, \( {P}_{3} \), of all points in \( {P}_{1} \) and \( ... | Proof. By Theorem 168 each of the functions \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) is integrable with respect to \( {P}_{3} \), and\n\n\[{\int }_{b}^{b}{f}_{1}\left( x\right) {dx} = {\int }_{b}^{b}{f}_{1}\left( x\right) {dx}\]\n\nand\n\n\[{\int }_{b}^{b}{f}_{2}\left( x\right) {dx} = {\int }_{b}^{... | Yes |
Theorem 171. For every function \( {f}_{1}\left( x\right) \) defined on the interval \( \overrightarrow{ab} \) there exists a function \( {f}_{2}\left( x\right) \) such that\n\n(1) \( {f}_{2}\left( x\right) \) is continuous and does not change sign on a certain neighborhood of \( x \doteq a \) .\n\n(2) \( {\iint }_{x \... | Proof. Let \( {x}_{1}^{\prime },{x}_{2}^{\prime },\ldots ,{x}_{n}^{\prime },\ldots \) be a set of points of the interval \( \overset{\overleftrightarrow{} }{ab} \) dense only at \( a \) . Let \( {B}_{1},{B}_{2},{B}_{3},\ldots ,{B}_{n},\ldots \) be a set of numbers such that\n\n\[ {B}_{n} \cdot n\left| {{f}_{1}\left( {{... | Yes |
Theorem 172. For every function \( {f}_{1}\left( x\right) \) which is unbounded as \( x \) approaches \( \infty \) there exists a non-oscillating function \( {f}_{2}\left( x\right) \) such that\n\n\[ \underset{x \doteq \infty }{L}{\int }_{a}^{x}{f}_{2}\left( x\right) {dx} \]\n\nexists and is finite, while \( \left( {x ... | Proof. Obviously the lemma of Theorem 170 can be stated so as to apply to the case where \( x \) approaches \( \infty \) instead of \( a \) . If then in the proof of Theorem 161 the set of points \( {x}_{1}\ldots {x}_{n}\ldots \) is so taken that\n\n\[ \underset{n \doteq \infty }{L}{x}_{n} = \infty \]\n\ninstead of \( ... | No |
Theorem 173. For every function \( {f}_{1}\left( x\right) \) defined on the interval \( a\overrightarrow{\infty } \) there exists a function \( {f}_{2}\left( x\right) \) such that\n\n(1) \( {f}_{2}\left( x\right) \) is continuous and does not change sign for \( x \) greater than a certain fixed number.\n\n(2)\n\n\[ \n{... | Proof. Such a function \( {f}_{2}\left( x\right) \) may be defined in a manner analogous to that of the proof of Theorem 171. | No |
Proposition 1. If an algebra \( \mathfrak{A} \) contains a solvable ideal \( \mathfrak{I} \), and if \( \overline{\mathfrak{A}} = \mathfrak{A}/\mathfrak{I} \) is solvable, then \( \mathfrak{A} \) is solvable. | Proof: Since (1) is a homomorphism, it follows that \( \overline{{\mathfrak{A}}^{2}} = {\overline{\mathfrak{A}}}^{2} \) and that \( \overline{{\mathfrak{A}}^{\left( i\right) }} = {\overline{\mathfrak{A}}}^{\left( i\right) } \) . Then \( {\overline{\mathfrak{A}}}^{\left( r\right) } = 0 \) implies \( \overline{{\mathfrak... | Yes |
Proposition 2. If \( \mathfrak{B} \) and \( \mathfrak{C} \) are solvable ideals of an algebra \( \mathfrak{A} \) , then \( \mathfrak{B} + \mathfrak{C} \) is a solvable ideal of \( \mathfrak{A} \) . Hence, if \( \mathfrak{A} \) is finite-dimensional, \( \mathfrak{A} \) has a unique maximal solvable ideal \( \mathfrak{N}... | Proof: \( \mathfrak{B} + \mathfrak{C} \) is an ideal because \( \mathfrak{B} \) and \( \mathfrak{C} \) are ideals. By the second isomorphism theorem \( \left( {\mathfrak{B} + \mathfrak{C}}\right) /\mathfrak{C} \cong \mathfrak{B}/\left( {\mathfrak{B} \cap \mathfrak{C}}\right) \) . But \( \mathfrak{B}/\left( {\mathfrak{B... | Yes |
Theorem 3. An ideal \( \mathfrak{B} \) of an algebra \( \mathfrak{A} \) is nilpotent if and only if the (associative) subalgebra \( {\mathfrak{B}}^{ * } \) of \( \mathfrak{M}\left( \mathfrak{A}\right) \) is nilpotent. | Proof: Suppose that every product of \( t \) elements of \( \mathfrak{B} \), no matter how associated, is 0 . Then the same is true for any product of more than \( t \) elements of \( \mathfrak{B} \) . Let \( T = {T}_{1}\cdots {T}_{t} \) be any product of \( t \) elements of \( {\mathfrak{B}}^{ * } \) . Then \( T \) is... | Yes |
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