{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sinh a x} {e^{b x} + 1} \rd x 	| r = \frac 1 2 \int_0^\infty \frac {e^{-b x} \paren {e^{a x} - e^{-a x} } } {1 - \paren {-e^{-b x} } } \rd x 	| c = {{Defof|Hyperbolic Sine}} }} {{eqn	| r = \frac 1 2 \int_0^\infty \paren {e^{\paren {a - b} x} - e^{-\paren {a + b} x} } \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n e^{-b n x} } \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^\infty \paren {e^{\paren {a - \paren {n + 1} b} x} - e^{-\paren {a + \paren {n + 1} b} x} } \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\intlimits {-\frac {e^{\paren {a - \paren {n + 1} b} x} } {\paren {n + 1} b - a} } 0 \infty - \intlimits {-\frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} } 0 \infty} 	| c = Primitive of $e^{a x}$ }} {{end-eqn}} Note that as $b > a$, we have that $a - b < 0$. As $b > 0$, we therefore have $a - \paren {n + 1} b < 0$ for all positive integer $n$. We also have that as $a + \paren {n + 1} b > 0$, that $-\paren {a + \paren {n + 1} b} < 0$. So, by Exponential Tends to Zero and Infinity:  :$\displaystyle \lim_{x \mathop \to \infty} \frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} = 0$ and:  :$\displaystyle \lim_{x \mathop \to \infty} \frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} = 0$ We therefore have: {{begin-eqn}} {{eqn	| l = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\intlimits {-\frac {e^{\paren {a - \paren {n + 1} b} x} } {\paren {n + 1} b - a} } 0 \infty - \intlimits {-\frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} } 0 \infty} 	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac 1 {\paren {n + 1} b - a} - \frac 1 {\paren {n + 1} b + a} } 	| c = Exponential of Zero }} {{eqn	| r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 a} {\paren {n + 1}^2 b^2 - a^2} 	| c = Difference of Two Squares }} {{eqn	| r = \frac 1 b \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\paren {\frac a b} } {n^2 - \paren {\frac a b}^2} 	| c = shifting the index, extracting factors }} {{eqn	| r = \frac 1 b \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} }} {{end-eqn}} We have by Mittag-Leffler Expansion for Cosecant Function: :$\displaystyle \pi \map \csc {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 - n^2}$  Setting $z = \dfrac a b$ we have:  :$\displaystyle \pi \map \csc {\frac {\pi a} b} = \frac b a + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$ Rearranging gives:  :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} = \frac \pi 2 \map \csc {\frac {\pi a} b} - \frac b {2 a}$ Therefore:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sinh a x} {e^{b x} + 1} \rd x 	| r = \frac 1 b \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2} }} {{eqn	| r = \frac \pi {2 b} \map \csc {\frac {\pi a} b} - \frac 1 {2 a} }} {{end-eqn}} {{qed}}