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

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Theorem

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

where:

$a$ and $b$ are positive real numbers
$\tanh$ denotes the hyperbolic cotangent function.


Proof

\(\ds \int_0^\infty \frac {\sin a x} {\sinh b x} \rd x\) \(=\) \(\ds \frac 1 i \int_0^\infty \frac {e^{-b x} \paren {e^{i a x} - e^{-i a x} } } {1 - e^{-2 b x} } \rd x\) Euler's Sine Identity, Definition of Hyperbolic Sine
\(\ds \) \(=\) \(\ds \frac 1 i \int_0^\infty \paren {e^{\paren {i a - b} x} - e^{-\paren {i a + b} x} } \paren {\sum_{n \mathop = 0}^\infty e^{-2 n b x} } \rd x\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \int_0^\infty \paren {e^{\paren {i a - \paren {2 n + 1} b} x} - e^{-\paren {i a + \paren {2 n + 1} b} x} } \rd x\) Fubini's Theorem
\(\ds \) \(=\) \(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \paren {\intlimits {-\frac {e^{\paren {i a - \paren {2 n + 1} b} x} } {\paren {2 n + 1} b - i a} } 0 \infty - \intlimits {-\frac {e^{-\paren {i a + \paren {2 n + 1} b} x} } {\paren {2 n + 1} b + i a} } 0 \infty}\) Primitive of $e^{a x}$

We have, as $b, n > 0$:

\(\ds \size {\lim_{x \mathop \to \infty} e^{\paren {i a - \paren {2 n + 1} b} x} }\) \(=\) \(\ds \lim_{x \mathop \to \infty} \size {e^{\paren {i a - \paren {2 n + 1} b} x} }\) Modulus of Limit
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to \infty} \size {e^{i a x} } \size {e^{-\paren {2 n + 1} b x} }\) Exponential of Sum
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to \infty} e^{-\paren {2 n + 1} b x}\)
\(\ds \) \(=\) \(\ds 0\) Exponential Tends to Zero and Infinity

We similarly have:

\(\ds \size {\lim_{x \mathop \to \infty} e^{-\paren {i a + \paren {2 n + 1} b} x} }\) \(=\) \(\ds \lim_{x \mathop \to \infty} \size {e^{-\paren {i a + \paren {2 n + 1} b} x} }\) Modulus of Limit
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to \infty} \size {e^{-i a x} } \size {e^{-\paren {2 n + 1} b x} }\) Exponential of Sum
\(\ds \) \(=\) \(\ds \lim_{x \mathop \to \infty} e^{-\paren {2 n + 1} b x}\)
\(\ds \) \(=\) \(\ds 0\) Exponential Tends to Zero and Infinity

So:

\(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \paren {\intlimits {-\frac {e^{\paren {i a - \paren {2 n + 1} b} x} } {\paren {2 n + 1} b - i a} } 0 \infty - \intlimits {-\frac {e^{-\paren {i a + \paren {2 n + 1} b} x} } {\paren {2 n + 1} b + i a} } 0 \infty}\) \(=\) \(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \paren {\frac 1 {\paren {2 n + 1} b - i a} - \frac 1 {\paren {2 n + 1} b + i a} }\) Exponential of Zero
\(\ds \) \(=\) \(\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {2 i a} {a^2 + \paren {2 n + 1}^2 b^2}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \frac 2 b \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 + \paren {2 n + 1}^2}\)
\(\ds \) \(=\) \(\ds \frac 4 b \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a {2 b} } } {4 \paren {\frac a {2 b} }^2 + \paren {2 n + 1}^2}\)

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:

\(\ds \int_0^\infty \frac {\sin a x} {\sinh b x} \rd x\) \(=\) \(\ds \frac 4 b \sum_{n \mathop = 0}^\infty \frac {\paren {\frac a {2 b} } } {4 \paren {\frac a {2 b} }^2 + \paren {2 n + 1}^2}\)
\(\ds \) \(=\) \(\ds \frac \pi {2 b} \map {\tanh} {\frac {a \pi} {2 b} }\)

$\blacksquare$


Sources

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