Involving Euler's constant and Gamma function

Question

$$\int_{0}^{\infty}{e^{-x}+x-1\over x(e^{x/a}-1)}\mathrm dx=a\gamma+\ln{\Gamma(1+a)}\tag1$$

$\gamma$ is Euler-Mascheroni constant,

How can we show that $(1)=a\gamma+\ln{\Gamma(1+a)}?$

Answer 1

Note that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{e^{-x}+x-1}{e^{x/a}-1}\frac{\mathrm{d}x}x &=\frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{e^{-ax}+ax-1}{e^x-1}\,\frac{\mathrm{d}x}x\\ &=\int_0^\infty\left(1-e^{-ax}\right)\frac{e^{-x}}{1-e^{-x}}\,\mathrm{d}x\\ &=\int_0^\infty\sum_{k=1}^\infty\left(e^{-kx}-e^{-(k+a)x}\right)\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+a}\right) \end{align} $$ Since $$ \left.\int_0^\infty\frac{e^{-ax}+ax-1}{e^x-1}\,\frac{\mathrm{d}x}x\right|_{a=0}=0 $$ and by Gautschi's Inequality $$ \frac{\Gamma(n+a+1)}{\Gamma(n+1)}=n^a\left(1+O\!\left(\frac1n\right)\right) $$ we get $$ \begin{align} \int_0^\infty\frac{e^{-ax}+ax-1}{e^x-1}\,\frac{\mathrm{d}x}x &=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac ak-\log\left(\frac{k+a}k\right)\right)\\ &=\lim_{n\to\infty}\left(aH_n-\log\left(\frac{\Gamma(n+a+1)}{\Gamma(a+1)\Gamma(n+1)}\right)\right)\\ &=\scriptsize\lim_{n\to\infty}\left(a\left(\log(n)+\gamma+O\!\left(\frac1n\right)\right)+\log(\Gamma(a+1))-a\log(n)+O\!\left(\frac1n\right)\right)\\[6pt] &=a\gamma+\log(\Gamma(a+1)) \end{align} $$

Answer 2

To evaluate this integral we will make use the following integral representation for the digamma function $\psi (x)$ $$\psi (x) = \int^\infty_0 \left (\frac{e^{-t}}{t} - \frac{e^{-xt}}{1 - e^{-t}} \right ) \, dt, \qquad x > 0. \tag1$$ Note that if we set $x = 1$ in the above integral representation for the digamma function, as $\psi (1) = -\gamma$ where $\gamma$ is the Euler–Mascheroni constant, we obtain the following integral representation for this constant $$\gamma = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{x e^x} \right ) \, dx, \tag2$$ and is the second result we intend to make use of.

Now, let $$I = \int^\infty_0 \frac{e^{-x} + x - 1}{x(e^{x/a} - 1)} \, dx.$$ Setting $x \mapsto a x$ the integral becomes $$I = \int^\infty_0 \frac{e^{-ax} + ax - 1}{x(e^x - 1)} \, dx.$$

Rearranging the numerator this can be rewritten \begin{align*} I &= \int^\infty_0 \frac{a e^{-x} + e^{-ax} - a e^{-x} + au + (a - 1) - a}{x(e^x - 1)} \, dx\\ &= a \int^\infty_0 \frac{e^{-x} + x - 1}{x(e^x - 1)} \, dx + \int^\infty_0 \frac{a e^{-ax} - a e^{-x} + a - 1}{x (e^x - 1)} \, dx. \tag3 \end{align*}

From (2), if we note that $$\gamma = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{xe^x} \right ) \, dx = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{e^{-x}}{x} \right ) \, dx = \int^\infty_0 \frac{x - 1 + e^{-x}}{x(e^x - 1)} \, dx,$$ the first integral appearing in (3) can be written in terms of the Euler-Mascheroni constant giving \begin{align*} I &= a \gamma + \int^\infty_0 \frac{e^{-ax} - a e^{-x} + a - 1}{x (e^x - 1)} \, dx, \end{align*} or \begin{align*} I &= a \gamma + \int^\infty_0 \frac{a (1 - e^{-x}) - (1 - e^{-ax})}{x (e^x - 1)} \, dx = a \gamma + \int^\infty_0 \left \{\frac{a}{x e^x} - \frac{1 - e^{-ax}}{x (e^x - 1)} \right \} \, dx, \tag4 \end{align*} after rearranging the integrand.

To find the last integral that has appears, as $$\psi (x) = \frac{d}{dx} \ln \Gamma (x),$$ then $$\ln \Gamma (x) = \int^x_1 \psi (u) \, du.$$ From the integral representation for the digamma function, namely (1), we can write the above expression for $\ln \Gamma (x)$ as a double integral, namely $$\ln \Gamma (x) = \int^\infty_0 \int^x_1 \left (\frac{e^{-t}}{t} - \frac{e^{-ut}}{1 - e^{-t}} \right ) \, du dt,$$ after the order of integration has been changed. The $u$-integration can be readily performed. Thus \begin{align*} \ln \Gamma (x) &= \int^\infty_0 \left [\frac{e^{-t}}{t} u + \frac{e^{-ut}}{t (1 - e^{-t})} \right ]^x_1 \, dt\\ &= \int^\infty_0 \left [(x - 1) - \frac{1 - e^{-(x - 1)t}}{1 - e^{-t}} \right ] \frac{e^{-t}}{t} \, dt. \end{align*} Now if $x \mapsto x + 1$, then $$\ln \Gamma (x + 1) = \int^\infty_0 \left [\frac{x}{t e^t} - \frac{1 - e^{-xt}}{t(e^t - 1)} \right ] \, dt.$$ Thus the integral appearing in (4) is equal to $\Gamma (a + 1)$ and yields $$\int^\infty_0 \frac{e^{-x} + x - 1}{x(e^{x/a} - 1)} \, dx = a \gamma + \ln \Gamma (a + 1),$$ as required.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}{\expo{-x} + x - 1 \over x\pars{\expo{x/a} - 1}}\,\dd x}} = \int_{0}^{\infty}{\sum_{m = 2}^{\infty}\pars{-x}^{m}/m! \over x\pars{1 - \expo{-x/a}}}\expo{-x/a}\,\dd x \\ = &\ \sum_{m = 2}^{\infty}{\pars{-1}^{m} \over m!} \int_{0}^{\infty}x^{m - 1}\sum_{n = 0}^{\infty}\expo{-\pars{n + 1}x/a}\,\dd x = \sum_{m = 2}^{\infty}{\pars{-1}^{m} \over m!}\sum_{n = 0}^{\infty}\ \overbrace{\int_{0}^{\infty}x^{m - 1}\expo{-\pars{n + 1}x/a}\dd x} ^{\ds{\pars{m - 1}!\pars{a \over n + 1}^{m}}} \\[5mm] = &\ \sum_{n = 0}^{\infty}\ \underbrace{\sum_{m = 2}^{\infty}{1 \over m}\pars{-\,{a \over n + 1}}^{m}} _{\ds{{a \over n + 1} - \ln\pars{1 + {a \over n + 1}}}}\ = \lim_{N \to \infty}\sum_{n = 1}^{N} \bracks{{a \over n} - \ln\pars{1 + {a \over n}}} \\[5mm] = &\ a\ \underbrace{\lim_{N \to \infty}\pars{\sum_{n = 1}^{N}{1 \over n} - \ln\pars{N}}}_{\ds{=\ \gamma}}\ +\ \lim_{N \to \infty}\bracks{a\ln\pars{N} + \sum_{n = 1}^{N}\ln\pars{n \over n + a}} \\[5mm] = &\ a\gamma + \lim_{N \to \infty}\sum_{n = 1}^{N}\ln\pars{N^{a/N}n \over n + a} = a\gamma + \lim_{N \to \infty}\ln\pars{\prod_{n = 1}^{N}{N^{a/N}n \over n + a}} = a\gamma + \lim_{N \to \infty}\ln\pars{N^{a}\, N! \over \bracks{1 + a}^{\,\overline{N}}} \\[5mm] = &\ a\gamma + \lim_{N \to \infty}\ln\pars{N^{a}\, N! \over \Gamma\pars{1 + a + N}/\Gamma\pars{1 + a}} = a\gamma + \lim_{N \to \infty}\ln\pars{\Gamma\pars{1 + a} \,{N^{a}\, N! \over \bracks{N + a}!}} \\[5mm] = &\ a\gamma + \lim_{N \to \infty}\ln\pars{\Gamma\pars{1 + a} \,{N^{a}\, \root{2\pi}N^{N + 1/2}\expo{-N} \over \root{2\pi}\pars{N + a}^{N + a + 1/2}\expo{-\pars{N + a}}}} \\[5mm] = &\ a\gamma + \lim_{N \to \infty}\ln\pars{\Gamma\pars{1 + a} \,{1 \over \pars{1 + a/N}^{N}\expo{-a}}} = \bbx{a\gamma + \ln\pars{\Gamma\pars{1 + a}}} \end{align}

because $\ds{\lim_{N \to \infty}\pars{1 + a/N}^{N} = \expo{a}}$ and $\ds{\lim_{N \to \infty}\pars{1 + a/N}^{a + 1/2} = 1}$.