# How can we prove that $8\int_{0}^{\infty}{\ln x\over x}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx=-32C+4\gamma^2-5\pi^2?$

An integral exhibits $3$ interesting constants.

$$8\int_{0}^{\infty}{\ln x\over x}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx=-32\color{red}C+4\color{blue}\gamma^2-5\color{green}\pi^2\tag1$$

I am only interested in $(1)$, because rarely Catalan's constant and Euler-Masheroni's constant they appear together!

Making an attempt:

It is too difficult here to make an attempt, apart from differentiating under the integral

$$I(a)=8\int_{0}^{\infty}{\ln x\over x^{a}}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx\tag2$$

$$I{'}(a)=8\int_{0}^{\infty}{1\over x^{a}}\left(e^{-x}-{1\over \sqrt[4]{1+8x}}\right)\mathrm dx\tag3$$

I guess $(3)$ diverges

How may we prove $(1)?$

• Study the integrals $$I_1=\int_{0}^{+\infty}\frac{\log x}{x}\left(e^{-x}-\frac{1}{x+1}\right)\,dx,\qquad I_2=\int_{0}^{+\infty}\frac{\log x}{x}\left(\frac{1}{1+x}-\frac{1}{(1+8x)^{1/4}}\right)\,dx$$ separately, through differentiation under the integral sign and the Laplace transform. – Jack D'Aurizio Apr 21 '17 at 8:46
• Hint: $$\mathcal{L}(\log x) = -\frac{\gamma+\log s}{s},\qquad \mathcal{L}^{-1}\left(\frac{1}{x e^x}\right)=\mathbb{1}_{s\geq 1}(s).$$ The remaining part is straightforward. – Jack D'Aurizio Apr 21 '17 at 8:49
• General solution: $\int_0^{\infty } \frac{\log (x) \left(\exp (-x)-\frac{1}{(1+a x)^{1/b}}\right)}{x} \,dx=\frac{1}{2} \left(-\left(\log (a)+\psi ^{(0)}\left(\frac{1}{b}\right)\right) \left(2\gamma +\log (a)+\psi ^{(0)}\left(\frac{1}{b}\right)\right)-\psi^{(1)}\left(\frac{1}{b}\right)\right)$ – Mariusz Iwaniuk Apr 21 '17 at 9:20

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &8\int_{0}^{\infty}{\ln\pars{x} \over x} \pars{\expo{-x} - {1 \over \root[4]{1 + 8x}}}\,\dd x \,\,\,\stackrel{\mrm{IBP}}{=}\,\,\, -4\int_{0}^{\infty}\ln^{2}\pars{x} \bracks{-\expo{-x} + {2 \over \pars{1 + 8x}^{5/4}}}\,\dd x \\[5mm] & = 4\ \overbrace{\pars{\gamma^{2} + {\pi^{2} \over 6}}}^{\ds{\Gamma''\pars{1}}} - 8\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{1 + 8x}^{5/4}}\,\dd x = 4\gamma^{2} + {2 \over 3}\,\pi^{2} - \left.8\,\partiald[2]{}{\mu}\int_{0}^{\infty}{x^{\mu} \over \pars{1 + 8x}^{5/4}}\,\dd x\,\right\vert_{\ \mu\ =\ 0} \end{align}
$$\mbox{and}\ \left.\partiald[2]{}{\mu}\bracks{{1 \over \Gamma\pars{5/4}}\, 8^{-\mu - 1}\,\Gamma\pars{\mu + 1}\Gamma\pars{{1 \over 4} - \mu}} \right\vert_{\ \mu\ =\ 0} = 4C + {17 \over 24}\,\pi^{2}$$
For any $a\in(0,1)$ we have $$\int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{dx}{x^a} = -\frac{\pi}{\sin(\pi a)}+\Gamma(1-a)\tag{1}$$ due to the Laplace transform and the integral definition of the $\Gamma$ function.
By differentiating with respect to $a$ we get: $$\int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{\log x}{x^a}\,dx = \psi(1-a)\Gamma(1-a)-\frac{\pi^2\cos(\pi a)}{\sin^2(\pi a)}\tag{2}$$ and by considering the limit as $a\to 1^-$ we get $$\int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{\log x}{x}\,dx =\frac{6\gamma^2-\pi^2}{12}.\tag{3}$$ On the other hand $$\begin{eqnarray*} \int \frac{1}{x}\left(\frac{1}{x+1}-\frac{1}{(1+8x)^{1/4}}\right)\,dx &=& \log(x)-\log(x+1)+2\arctan((1+8x)^{1/4})\\&+&\log(1-(1+8x)^{1/4})-\log(1+(1+8x)^{1/4})\end{eqnarray*}\tag{4}$$ hence the remaining part can be tackled by integration by parts, or by exploiting Mariusz Iwaniuk's comment.