Definite Integral to Infinity of Hyperbolic Sine of a x over Exponential of b x minus One

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Theorem

$\ds \int_0^\infty \frac {\sinh a x} {e^{b x} - 1} \rd x = \frac 1 {2 a} - \frac \pi {2 b} \cot \frac {a \pi} b$

where:

$a$ and $b$ are positive real numbers with $b > a$
$\cot$ denotes the cotangent function.


Proof

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

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:

$\ds \lim_{x \mathop \to \infty} \frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} = 0$

and:

$\ds \lim_{x \mathop \to \infty} \frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} = 0$

We therefore have:

\(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\intlimits {\frac {e^{\paren {a - \paren {n + 1} b} x} } {a - \paren {n + 1} b} } 0 \infty - \intlimits {-\frac {e^{-\paren {a + \paren {n + 1} b} x} } {a + \paren {n + 1} b} } 0 \infty}\) \(=\) \(\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac 1 {\paren {n + 1} b - a} - \frac 1 {\paren {n + 1} b + a} }\) Exponential of Zero
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac a {\paren {n + 1}^2 b^2 - a^2}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds -\frac 1 b \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}\) shifting the index, extracting factors

By Mittag-Leffler Expansion for Cotangent Function, we have:

$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$

where $z$ is not an integer.

Setting $z = \dfrac a b$, we have:

$\ds \pi \cot \frac {\pi a} b = \frac b a + 2 \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$

Rearrangement gives:

$\ds \frac \pi 2 \cot \frac {\pi a} b - \frac b {2 a} = \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}$

So:

\(\ds \int_0^\infty \frac {\sinh a x} {e^{b x} - 1} \rd x\) \(=\) \(\ds -\frac 1 b \sum_{n \mathop = 1}^\infty \frac {\paren {\frac a b} } {\paren {\frac a b}^2 - n^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} - \frac \pi {2 b} \cot \frac {a \pi} b\)

$\blacksquare$


Sources

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