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

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Theorem

$\displaystyle \int_0^\infty \frac {\sinh a x} {e^{b x} + 1} \rd x = \frac \pi {2 b} \csc \frac {a \pi} b - \frac 1 {2 a}$

where:

$a$ and $b$ are positive real numbers with $b > a$
$\csc$ is the cosecant function.


Proof

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

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:

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

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:

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

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:

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

$\blacksquare$


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