Definite Integral to Infinity of Reciprocal of Exponential of x minus One minus Exponential of -x over x

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Theorem

$\ds \int_0^\infty \paren {\frac 1 {e^x - 1} - \frac {e^{-x} } x} \rd x = \gamma$

where $\gamma$ denotes the Euler-Mascheroni constant.


Proof

\(\ds \gamma\) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\sum_{k \mathop = 1}^n \frac 1 k - \ln n}\) Definition of Euler-Mascheroni Constant
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\sum_{k \mathop = 1}^n \paren {\int_0^1 x^{k - 1} \rd x } + \int_0^1 \frac {1 - x^{n - 1} } {\ln x} \rd x}\) Primitive of Power, Definite Integral from $0$ to $1$ of $\dfrac {x^m - x^n} {\ln x}$
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\int_0^1 \paren {\sum_{k = 1}^n x^{k - 1} } \rd x + \int_0^1 \frac {1 - x^{n - 1} } {\ln x} \rd x}\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\int_0^1 \paren {\frac {1 - x^n} {1 - x} + \frac {1 - x^{n - 1} } {\ln x} } \rd x}\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \int_0^1 \lim_{n \mathop \to \infty} \paren {\frac {1 - x^n} {1 - x} + \frac {1 - x^{n - 1} } {\ln x} } \rd x\) Lebesgue's Dominated Convergence Theorem
\(\ds \) \(=\) \(\ds \int_0^1 \paren {\frac 1 {1 - x} + \frac 1 {\ln x} } \rd x\) Sequence of Powers of Number less than One

Let:

$x = e^{-t}$

By Derivative of $e^{a x}$:

$\dfrac {\d x} {\d t} = -e^{-t}$

By Exponential of Zero, we have:

as $x \to 1$, $t \to 0$.

By Exponential Tends to Zero and Infinity, we have:

as $x \to 0$, $t \to \infty$.

We therefore have:

\(\ds \int_0^1 \paren {\frac 1 {1 - x} + \frac 1 {\ln x} } \rd x\) \(=\) \(\ds -\int_\infty^0 e^{-t} \paren {\frac 1 {1 - e^{-t} } + \frac 1 {\map \ln {e^{-t} } } } \rd t\) substituting $x = e^{-t}$
\(\ds \) \(=\) \(\ds \int_0^\infty e^{-t} \paren {\frac {e^t} {e^t - 1} - \frac 1 t} \rd t\) Reversal of Limits of Definite Integral, Definition of Natural Logarithm
\(\ds \) \(=\) \(\ds \int_0^\infty \paren {\frac 1 {e^t - 1} - \frac {e^{-t} } t} \rd t\)

$\blacksquare$


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