{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin a x} {\sinh b x} \rd x 	| r = \frac 1 i \int_0^\infty \frac {e^{-b x} \paren {e^{i a x} - e^{-i a x} } } {1 - e^{-2 b x} } \rd x 	| c = Sine Exponential Formulation, {{Defof|Hyperbolic Sine}} }} {{eqn	| r = \frac 1 i \int_0^\infty \paren {e^{\paren {i a - b} x} - e^{-\paren {i a + b} x} } \paren {\sum_{n \mathop = 0}^\infty e^{-2 n b x} } \rd x 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \frac 1 i \sum_{n \mathop = 0}^\infty \int_0^\infty \paren {e^{\paren {i a - \paren {2 n + 1} b} x} - e^{-\paren {i a + \paren {2 n + 1} b} x} } \rd x 	| c = Fubini's Theorem }} {{eqn	| r = \frac 1 i \sum_{n \mathop = 0}^\infty \paren {\intlimits {-\frac {e^{\paren {i a - \paren {2 n + 1} b} x} } {\paren {2 n + 1} b - i a} } 0 \infty - \intlimits {-\frac {e^{-\paren {i a + \paren {2 n + 1} b} x} } {\paren {2 n + 1} b + i a} } 0 \infty} 	| c = Primitive of $e^{a x}$ }} {{end-eqn}} We have, as $b, n > 0$: {{begin-eqn}} {{eqn	| l = \size {\lim_{x \mathop \to \infty} e^{\paren {i a - \paren {2 n + 1} b} x} } 	| r = \lim_{x \mathop \to \infty} \size {e^{\paren {i a - \paren {2 n + 1} b} x} } 	| c = Modulus of Limit }} {{eqn	| r = \lim_{x \mathop \to \infty} \size {e^{i a x} } \size {e^{-\paren {2 n + 1} b x} } 	| c = Exponential of Sum }} {{eqn	| r = \lim_{x \mathop \to \infty} e^{-\paren {2 n + 1} b x} }} {{eqn	| r = 0 	| c = Exponential Tends to Zero and Infinity }} {{end-eqn}} We similarly have: {{begin-eqn}} {{eqn	| l = \size {\lim_{x \mathop \to \infty} e^{-\paren {i a + \paren {2 n + 1} b} x} } 	| r = \lim_{x \mathop \to \infty} \size {e^{-\paren {i a + \paren {2 n + 1} b} x} } 	| c = Modulus of Limit }} {{eqn	| r = \lim_{x \mathop \to \infty} \size {e^{-i a x} } \size {e^{-\paren {2 n + 1} b x} } 	| c = Exponential of Sum }} {{eqn	| r = \lim_{x \mathop \to \infty} e^{-\paren {2 n + 1} b x} }} {{eqn	| r = 0 	| c = Exponential Tends to Zero and Infinity }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \frac 1 i \sum_{n \mathop = 0}^\infty \paren {\intlimits {-\frac {e^{\paren {i a - \paren {2 n + 1} b} x} } {\paren {2 n + 1} b - i a} } 0 \infty - \intlimits {-\frac {e^{-\paren {i a + \paren {2 n + 1} b} x} } {\paren {2 n + 1} b + i a} } 0 \infty} 	| r = \frac 1 i \sum_{n \mathop = 0}^\infty \paren {\frac 1 {\paren {2 n + 1} b - i a} - \frac 1 {\paren {2 n + 1} b + i a} } 	| c = Exponential of Zero }} {{eqn	| r = \frac 1 i \sum_{n \mathop = 0}^\infty \frac {2 i a} {a^2 + \paren {2 n + 1}^2 b^2} 	| c = Difference of Two Squares }} {{eqn	| r = \frac 2 b \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 + \paren {2 n + 1}^2} }} {{eqn	| r = \frac 4 b \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a {2 b} } } {4 \paren {\frac a {2 b} }^2 + \paren {2 n + 1}^2} }} {{end-eqn}} By Mittag-Leffler Expansion for Hyperbolic Tangent Function, we have:  :$\displaystyle \pi \map {\tanh} {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac z {4 z^2 + \paren {2 n + 1}^2}$ where $z$ is not a half-integer multiple of $i$.  Setting $z = \dfrac a {2 b}$ gives:  :$\displaystyle \pi \map {\tanh} {\frac {a \pi} {2 b} } = 8 \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a {2 b} } } {4 \paren {\frac a {2 b} }^2 + \paren {2 n + 1}^2}$ Therefore:  {{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\sin a x} {\sinh b x} \rd x 	| r = \frac 4 b \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a {2 b} } } {4 \paren {\frac a {2 b} }^2 + \paren {2 n + 1}^2} }} {{eqn	| r = \frac \pi {2 b} \map {\tanh} {\frac {a \pi} {2 b} } }} {{end-eqn}} {{qed}}