Theorem: Definite Integral to Infinity of Sine of a x over Hyperbolic Sine of b x

:$\ds \int_0^\infty \frac {\sin a x} {\sinh b x} \rd x = \frac \pi {2 b} \tanh \frac {a \pi} {2 b}$

Proof:
{{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 = Euler's Sine Identity, {{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: 
:$\ds \pi \map \tanh {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac z {4 z^2 + \paren {2 n + 1}^2}$
where $z$ is not a half-integer multiple of $i$. 
Setting $z = \dfrac a {2 b}$ gives: 
:$\ds \pi \map \tanh {\frac {a \pi} {2 b} } = 8 \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a {2 b} } } {4 \paren {\frac a {2 b} }^2 + \paren {2 n + 1}^2}$
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}}