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

Question

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 $(1)?$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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\begin{align} &\int_{0}^{\infty}{x^{\mu} \over \pars{1 + 8x}^{5/4}}\,\dd x = \int_{0}^{\infty}x^{\mu}\bracks{{1 \over \Gamma\pars{5/4}}\int_{0}^{\infty}t^{1/4}\expo{-\pars{1 + 8x}t}\,\dd t}\,\dd x \\[5mm] = &\ {1 \over \Gamma\pars{5/4}} \int_{0}^{\infty}t^{1/4}\expo{-t}\int_{0}^{\infty}x^{\mu}\expo{-8tx} \,\dd x\,\dd t = {1 \over \Gamma\pars{5/4}}\int_{0}^{\infty}t^{1/4}\expo{-t} \,{\Gamma\pars{\mu + 1} \over \pars{8t}^{\mu + 1}}\,\dd t \\[5mm] = &\ {8^{-\mu - 1}\,\Gamma\pars{\mu + 1} \over \Gamma\pars{5/4}} \int_{0}^{\infty}t^{-3/4 - \mu}\expo{-t}\,\dd t = {1 \over \Gamma\pars{5/4}}\,8^{-\mu - 1}\,\Gamma\pars{\mu + 1} \Gamma\pars{{1 \over 4} - \mu} \end{align}

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$$ \mbox{and}\ \left.\partiald[2]{}{\mu}\bracks{{1 \over \Gamma\pars{5/4}}\, 8^{-\mu - 1}\,\Gamma\pars{\mu + 1}\Gamma\pars{{1 \over 4} - \mu}} \right\vert_{\ \mu\ =\ 0} = 4C + {17 \over 24}\,\pi^{2} $$

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\begin{align} &8\int_{0}^{\infty}{\ln\pars{x} \over x} \pars{\expo{-x} - {1 \over \root[4]{1 + 8x}}}\,\dd x = 4\gamma^{2} + {2 \over 3}\,\pi^{2} - 8\pars{4C + {17 \over 24}\,\pi^{2}} \\[5mm] = & \bbx{\ds{-32C + 4\gamma^{2} - 5\pi^{2}}} \end{align}

Answer 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.