Definite Integral to Infinity of Logarithm of Exponential of x plus One over Exponential of x minus One

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Theorem

$\ds \int_0^\infty \map \ln {\frac {e^x + 1} {e^x - 1} } \rd x = \frac {\pi^2} 4$


Proof

We can write:

\(\ds \int_0^\infty \map \ln {\frac {e^x + 1} {e^x - 1} } \rd x\) \(=\) \(\ds \int_0^\infty \map \ln {\frac {e^{x/2} \paren {e^{x/2} + e^{-x/2} } } {e^{x/2} \paren {e^{x/2} - e^{-x/2} } } } \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty \map \ln {\coth \frac x 2} \rd x\) Definition of Hyperbolic Cotangent

Let:

$u = \coth \dfrac x 2$

We have, by Derivative of Hyperbolic Cotangent Function:

$\dfrac {\d u} {\d x} = -\dfrac 1 2 \csch^2 \dfrac x 2$

From Difference of Squares of Hyperbolic Cotangent and Cosecant, this can be written:

$\dfrac {\d u} {\d x} = \dfrac 1 2 \paren {1 - \coth^2 \dfrac x 2} = \dfrac 1 2 \paren {1 - u^2}$

From Limit to Infinity of Hyperbolic Cotangent Function, we have:

as $x \to \infty$, $u \to 1$.

We also have:

as $x \to 0^+$, $u \to \infty$.

With this, we have:

\(\ds \int_0^\infty \map \ln {\coth \frac x 2} \rd x\) \(=\) \(\ds 2 \int_\infty^1 \frac {\ln u} {1 - u^2} \rd u\) substituting $u = \coth \dfrac x 2$
\(\ds \) \(=\) \(\ds -2 \int_1^\infty \frac {\ln u} {1 - u^2} \rd u\) Reversal of Limits of Definite Integral
\(\ds \) \(=\) \(\ds -2 \int_1^0 \paren {-\frac 1 {v^2} } \frac {\map \ln {\frac 1 v} } {1 - \paren {\frac 1 {v^2} }^2} \rd v\) substituting $v = \dfrac 1 v$
\(\ds \) \(=\) \(\ds -2 \int_0^1 \frac {\ln v} {1 - v^2} \rd v\) Reversal of Limits of Definite Integral
\(\ds \) \(=\) \(\ds -2 \int_0^1 \ln v \paren {\sum_{n \mathop = 0}^\infty \paren {v^2}^n} \rd v\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds -2 \sum_{n \mathop = 0}^\infty \int_0^1 v^{2 n} \ln v \rd v\) Fubini's Theorem
\(\ds \) \(=\) \(\ds 2 \sum_{n \mathop = 0}^\infty \frac {\map \Gamma 2} {\paren {2 n + 1}^2}\) Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$
\(\ds \) \(=\) \(\ds 2 \times 1! \times \frac {\pi^2} 8\) Gamma Function Extends Factorial, Sum of Reciprocals of Squares of Odd Integers
\(\ds \) \(=\) \(\ds \frac {\pi^2} 4\)

$\blacksquare$


Sources

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