Consider this integral $(1)$
$$\int_{0}^{\infty}\color{red}{{\gamma+\ln x\over e^x}}\cdot{1-\cos x\over x}\,\mathrm dx={1\over 2}\cdot{\pi-\ln 4\over 4}\cdot{\pi+\ln 4\over 4}\tag1$$
Recall a well-known integral for $\gamma$:
$$\int_{0}^{\infty}e^{-x}\ln x\,\mathrm dx=-\gamma.$$
Making an attempt:
I am not sure, what to do...
Recall: $\cos x={e^{ix}+e^{-ix}\over 2}$, then $(1)$ becomes
$$\int_{0}^{\infty}{\gamma+\ln x\over e^x}\cdot{2-e^{ix}-e^{-ix}\over 2x}\,\mathrm dx\tag2$$
Or using $e^{-x}$ series, then $(1)$ becomes
$$\sum_{n=0}^{\infty}{(-1)^n\over n!}\color{blue}{\int_{0}^{\infty}(\gamma+\ln x)(1-\cos x)x^{n-1}\,\mathrm dx}\tag3$$
$$\color{blue}{blue}=\int_{0}^{\infty}(\gamma+\ln x)x^{n-1}\,\mathrm dx-\int_{0}^{\infty}(\gamma+\ln x)\cos(x) x^{n-1}\,\mathrm dx=I_1-I_2\tag4$$
Indefinite integral of $$I_1={x^n\over n^2}(n\ln x+n\gamma-1)+C$$
Put in the limit and $I_1$ is diverges and $I_2$ it is a lengthy IBP and it also diverges.
How to prove (1)?
Answer
Assuming that $a>1$,
$$ \begin{align} I(a) &= \int_{0}^{\infty} \frac{\gamma+\ln x}{e^{ax}} \frac{1-\cos x}{x} \, dx \\ &= -\int_{0}^{\infty} \frac{\gamma+\ln x}{e^{ax}} \sum_{n=1}^{\infty} \frac{(-1)^{n}x^{2n-1}}{(2n)!} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{0}^{\infty} (\gamma + \ln x) e^{-ax} x^{2n-1} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \left(\gamma \, \frac{\Gamma(2n)}{a^{2n}} + \frac{\Gamma'(2n) -\ln(a) \Gamma(2n)}{a^{2n}}\right) \\ &=-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n} \left(\frac{\gamma + \psi_{0}(2n)- \ln(a)}{a^{2n}} \right) \tag{1} \\ &= -\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n} \frac{H_{2n}- \frac{1}{2n} -\ln (a) }{a^{2n}} \tag{2} \\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n}\frac{H_{2n}}{a^{2n}} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{4n^{2}} \frac{1}{a^{2n}} + \ln (a) \sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n} \frac{1}{a^{2n}} \\\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n}\frac{H_{2n}}{a^{2n}} + \, \frac{1}{4} \, \operatorname{Li}_{2} \left(- \frac{1}{a^{2}} \right) - \frac{\ln (a)}{2} \, \ln\left(1+ \frac{1}{a^{2}} \right) \tag{3}. \end{align}$$
From the ordinary generating function for the harmonic numbers, we see that $$f(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n} H_{n}}{n} \, z^{n} = -\int_{0}^{z} \frac{\log(1+t)}{t(1+t)} \, dt = \operatorname{Li}_{2}(-z) + \frac{1}{2} \, \ln^{2}(1+z), \quad |z| <1. $$
Therefore,$$\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n}\frac{H_{2n}}{a^{2n}}&= \frac{1}{2} \left[f\left(\frac{i}{a}\right)+f\left(-\frac{i}{a} \right) \right] \\ &= \frac{1}{2} \left[\operatorname{Li}_{2} \left(-\frac{i}{a}\right)+ \frac{1}{2} \, \ln^{2} \left(1+ \frac{i}{a} \right)+ \operatorname{Li}_{2} \left(\frac{i}{a}\right) + \frac{1}{2} \, \ln^{2} \left(1- \frac{i}{a} \right) \right] \\ &= \frac{1}{4} \left[\operatorname{Li}_{2} \left(-\frac{1}{a^{2}} \right) +\ln^{2} \left(1+ \frac{i}{a} \right) + \ln^{2} \left(1- \frac{i}{a} \right)\right], \tag{5} \end{align}$$
and $$\begin{align} I(a) &= -\frac{1}{4} \left[ \ln^{2} \left(1+ \frac{i}{a}\right)+ \ln^{2} \left(1- \frac{i}{a} \right)+2 \ln (a) \ln \left(1+ \frac{1}{a^{2}} \right)\right] \\ &= - \frac{1}{8} \left[ \ln^{2} \left(1+ \frac{1}{a^{2}} \right)-4\arctan^{2} \left(\frac{1}{a} \right) + 4 \ln (a) \ln \left(1+ \frac{1}{a^{2}} \right)\right]. \end{align}$$
Letting $a \downarrow 1$, we get $$I(1) = - \frac{1}{8} \, \ln^{2}(2)+\frac{1}{2} \left(\frac{\pi^{2}}{16} \right) = \frac{\pi^{2}-4 \ln^{2}(2)}{32} = \frac{\pi^{2}-\ln^{2}(4)}{32}.$$
For evaluation purposes, I assumed that $a >1$. But the result should hold for $a>0$.
$(1)$ https://en.wikipedia.org/wiki/Digamma_function
$(2)$ https://en.wikipedia.org/wiki/Digamma_function#Relation_to_harmonic_numbers
$(3)$ https://en.wikipedia.org/wiki/Polylogarithm
$(4)$ https://en.wikipedia.org/wiki/Polylogarithm#Properties
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