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A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.

Prove: $$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x dx=-\pi^2\left(4\,\zeta'(-1)+\frac23\right).$$

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    $\begingroup$ @VladimirReshetnikopv, just curious - from what math contest was this problem taken? $\endgroup$ Commented Oct 12, 2013 at 0:53
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    $\begingroup$ @JoseArnaldoDris Somewhere in Russia, I do not know exactly where. $\endgroup$ Commented Oct 12, 2013 at 15:32

6 Answers 6

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Here is a solution: Let $I$ denote the integral. Then

\begin{align*} I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\ &= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \frac{\mathrm{Li}_{2}(x)}{x} \, dx \\ &= -2 \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \mathrm{Li}_{2}(e^{-t}) \, dt \qquad (x = e^{-t}) \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t}{4\pi^{2} + t^{2}} \, e^{-nt} \, dt \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \left( \int_{0}^{\infty} \cos (2\pi u) e^{-tu} \, du \right) e^{-nt} \, dt \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi u)}{u + n} \, du \\ &= -2 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\infty} \frac{\cos (2\pi n u)}{u + 1} \, du \qquad (u \mapsto nu) \\ &= -2 \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{2}} \int_{0}^{1} \frac{\cos (2\pi n u)}{u + k} \, du \\ &= -2 \sum_{k=1}^{\infty} \int_{0}^{1} \frac{1}{u + k} \left( \sum_{n=1}^{\infty} \frac{\cos (2\pi n u)}{n^{2}} \right) \, du. \end{align*}

Now we invoke the Fourier series of the Bernoulli polynomial $B_{2}(x)$:

$$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^{2}} = \pi^{2} B_{2}(x) = \pi^{2} \left( x^{2} - x + \frac{1}{6} \right), \quad 0 \leq x \leq 1. $$

Then it follows that

\begin{align*} I &= -2\pi^{2} \sum_{k=1}^{\infty} \int_{0}^{1} \frac{u^{2} - u + \frac{1}{6}}{u + k} \, du \\ &= \pi^{2} \sum_{k=1}^{\infty} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\}. \end{align*}

Now we consider the exponential of the partial sum:

\begin{align*} &\exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] \\ &= e^{N^{2} + 2N} \prod_{k=1}^{N} \left( \frac{k}{k+1} \right)^{2k^{2} + 2k + \frac{1}{3}} \\ &= \frac{e^{N^{2} + 2N}}{(N + 1)^{2N^{2} + 2N + \frac{1}{3}}} \prod_{k=1}^{N} k^{4k} \\ &= \frac{e^{2N}}{\left(1 + \frac{1}{N}\right)^{2N^{2} + 2N + \frac{1}{3}}} \left\{ \frac{e^{N^{2}/4}}{N^{N^{2}/2 + N/2 + 1/12}} \prod_{k=1}^{N} k^{k} \right\}^{4}. \end{align*}

In view of the definition of Glaisher-Kinkelin constant $A$, we have

$$ \lim_{N\to\infty} \exp \left[ \sum_{k=1}^{N} \left\{ 2k + 1 + 2\left( k^{2} + k + \frac{1}{6} \right) \log \left( \frac{k}{k+1} \right) \right\} \right] = \frac{A^{4}}{e}. $$

This, together with the identity $ \log A = \frac{1}{12} - \zeta'(-1)$, yields

$$ I = \pi^{2} ( 4 \log A - 1 ) = -\pi^{2} \left(4\zeta'(-1) + \frac{2}{3} \right) $$

as desired.

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    $\begingroup$ How did you see this??? I am amazed. $\endgroup$
    – Potato
    Commented Oct 12, 2013 at 2:25
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    $\begingroup$ I am awed by the range of mathematical knowledge displayed here. Bravo! $\endgroup$
    – Ron Gordon
    Commented Oct 12, 2013 at 13:26
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    $\begingroup$ @sos440 It is amazing! I can understand that steps with dilogarithm in the beginning and Glaisher-Kinkelin constant at the end are obvious, but I cannot imagine how you filled everything in between in less than an hour. I think I will never be able to do integrals like this. $\endgroup$ Commented Oct 12, 2013 at 15:39
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    $\begingroup$ Thank you! At first I had no idea, so I attacked this with my favorite techniques. Then I found that the intermediate series resembled this one. So I applied the same technique and hopefully this approach yielded the answer. $\endgroup$ Commented Oct 12, 2013 at 17:22
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    $\begingroup$ @BennettGardiner, The underlying idea is the same as in the summation by parts. The product is of the form $$P = \prod_{k=1}^{N} \frac{k^{f(k)}}{(k+1)^{f(k)}}. $$ Now splitting the product into both denominator and numerator, $$ P = \frac{\prod_{k=1}^{N} k^{f(k)}}{\prod_{k=1}^{N} (k+1)^{f(k)}} = \frac{\prod_{k=1}^{N} k^{f(k)}}{\prod_{k=2}^{N+1} k^{f(k-1)}}. $$ Merging the product again gives $$ P = \frac{1}{(N+1)^{f(N)}} \prod_{k=1}^{N} k^{f(k) - f(k-1)}. $$ $\endgroup$ Commented Oct 29, 2013 at 2:21
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In this answer, I will provide another approach to proving that $$I=\int_0^1 \log\left(1+\frac{\log^2x}{4\pi^2}\right)\frac{\log(1-x)}{x}dx=-4\pi^2\zeta'(-1)-\frac{2\pi^2}{3}.$$

First, by using the functional equation for the zeta function, we will prove the following lemma:

Lemma 1: Let $$L=\int_{0}^{1}\log\left(\frac{\log^{2}(x)}{4\pi^{2}}\right)\frac{\log(1-x)}{x}dx.$$ Then $L=-4\pi^{2}\zeta'(-1)+\frac{\pi^{2}}{3}.$

Proof: Expanding the series for $\log(1-x)/x$, we obtain $$\int_{0}^{1}\log\left(\frac{\log^{2}(x)}{4\pi^{2}}\right)\frac{\log(1-x)}{x}dx=-\sum_{n=0}^{\infty}\int_{0}^{1}\log\left(\frac{\log^{2}(x)}{4\pi^{2}}\right)\frac{x^{n}}{n}\frac{dx}{x}.$$ Letting $x=e^{-2\pi t}$, this is $$-2\pi\sum_{n=0}^{\infty}\frac{1}{n}\int_{0}^{\infty}\log\left(t^{2}\right)e^{-2\pi tn}dt=-4\pi\sum_{n=0}^{\infty}\frac{1}{n}\mathcal{L}\left(\log x\right)(2\pi n),$$ where $\mathcal{L}$ denotes the Laplace transform. Since $\mathcal{L}(\log x)(2\pi n)=\frac{-1}{2\pi n}(\log(2\pi n)+\gamma)$, we arrive at $$L=2\sum_{n=0}^{\infty}\frac{1}{n^{2}}\left(\log(2\pi n)+\frac{\gamma}{n^{2}}\right)=2\left(\frac{\pi^{2}}{6}\log2\pi+\frac{\pi^{2}\gamma}{6}-\zeta^{'}(2)\right).$$ Using the derivative of the functional equation for the zeta function, it follows that $$L=-4\pi^{2}\zeta'(-1)+\frac{\pi^{2}}{3}.$$

Lemma 2: We have that $I-L=-\pi^2$.

Proof: As before, by expanding the series for $\frac{\log(1-x)}{x}$ and substituting $x=e^{-2\pi t}$ we have that $$I=-2\pi\sum_{n=0}^{\infty}\frac{1}{n}\int_{0}^{\infty}\log\left(1+t^{2}\right)e^{-2\pi tn}dt.$$ Since $$\int_{0}^{1}\frac{2x}{x^{2}+t^{2}}=\log\left(1+t^{2}\right)-\log(t^{2}),$$ it follows that $$-4\pi\sum_{n=0}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\frac{x}{x^{2}+t^{2}}e^{-2\pi tn}dtdx=I-L.$$ Using the fact that $\frac{x}{x^{2}+t^{2}}$ is the Laplace transform of $\cos$, we have that $$I-L=-4\pi\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\int_{0}^{\infty}\cos(tv)e^{-xv}e^{-2\pi nt}dvdtdx$$

$$=-4\pi\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\left(\int_{0}^{\infty}\cos(tv)e^{-2\pi nt}dt\right)e^{-xv}dvdx $$

$$=-4\pi\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{1}\int_{0}^{\infty}\frac{2\pi n}{(2\pi n)^{2}+v^{2}}e^{-xv}dvdx$$

$$=-8\pi^{2}\int_{0}^{\infty}\left(\sum_{n=1}^{\infty}\frac{1}{(2\pi n)^{2}+v^{2}}\right)\left(\int_{0}^{1}e^{-xv}dx\right)dv.$$ Evaluating the integral in $x$, and using the cotangent identity $$\sum_{n=1}^{\infty}\frac{1}{\left(2\pi n\right)^{2}+v^{2}}=\frac{v\coth(v/2)-2}{4v^{2}},$$ we obtain $$I-L=-2\pi^{2}\int_{0}^{\infty}\left(\frac{v\coth(v/2)-2}{v^{2}}\right)\left(\frac{1-e^{-v}}{v}\right)dv.$$ Expanding out $\coth(v/2)$ in terms of exponentials, the integrand equals $\left(e^{-v}v+v+e^{-v}-2\right)/v^{3}$, which has a simple anti derivative. In particular $$\int_{0}^{\infty}\left(\frac{v\coth(v/2)-2}{v^{2}}\right)\left(\frac{1-e^{-v}}{v}\right)dv=-\frac{v+e^{-v}-1}{v^{2}}\biggr|_{v=0}^{v=\infty}=\frac{1}{2},$$ and so we have shown that $I-L=-\pi^{2}.$

Combining Lemma 1 and Lemma 2, it follows that $$I=-4\pi^2\zeta'(-1)-\frac{2\pi^2}{3},$$ as desired.

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I realize it's been a while, but I thought this one looked interesting, so I thought I would give it a go.

Make the obvious sub $\displaystyle t=\frac{-\ln(x)}{2\pi}, \;\ x=e^{-2\pi x}, \;\ dx=-2\pi e^{-2\pi x}dx$

$\displaystyle2\pi\int_{0}^{\infty}\ln(t^{2}+1)\ln(1-e^{-2\pi t})dt$.

Use parts $\displaystyle u=\ln(1-e^{-2\pi t}), \;\ dv=\ln(t^{2}+1)dt, \;\ v=x\ln(t^{2}+1)+2(\tan^{-1}(t)-t), \;\ du=\frac{2\pi}{e^{2\pi t}-1}dt$

$\displaystyle -2\pi \left(2\pi\int_{0}^{\infty}\left(t\ln(t^{2}+1)+2\tan^{-1}(t)-2t\right)\cdot \frac{1}{e^{2\pi t}-1}\right)dt$

$=\displaystyle -4\pi^{2} \left(\int_{0}^{\infty}\frac{t\ln(t^{2}+1)}{e^{2\pi t}-1}dt+2\int_{0}^{\infty}\frac{\tan^{-1}(t)}{e^{2\pi t}-1}dt-2\int_{0}^{\infty}\frac{t}{e^{2\pi t}-1}dt\right)$

Now, the two leftmost integrals form a rather well-known identity that can be derived from Hermite's integral (among other ways).

$\displaystyle\int_{0}^{\infty}\frac{t\ln(t^{2}+1)}{e^{2\pi t}-1}dt+2\int_{0}^{\infty}\frac{\tan^{-1}(t)}{e^{2\pi t}-1}dt=\zeta'(-1)+1/4$

The integral on the right end is rather elementary and evaluates to $\displaystyle 2\int_{0}^{\infty}\frac{t}{e^{2\pi t}-1}dt=1/12$

Putting the pieces together results in $\displaystyle-4\pi^{2}\left(\zeta'(-1)+1/4-1/12\right)=-4\pi^{2}\zeta'(-1)-\frac{2\pi^{2}}{3}$

Hermite's integral

$\displaystyle\zeta (a,t)=\sum_{n=0}^{\infty}\frac{1}{(n+t)^{a}}=\frac{t^{-a}}{2}+\frac{t^{1-a}}{a-1}+i\int_{0}^{\infty}\frac{(t+ix)^{-a}-(t-ix)^{-a}}{e^{2\pi x}-1}dx$

Diffing this w.r.t a leads to

$\displaystyle \zeta'(a,t)=\frac{-t^{-a}}{2}log(t)-\frac{t^{1-a}(1+(a-1)log(t))}{(a-1)^{2}}+i\int_{0}^{\infty}\frac{(t-ix)^{-a}log(t-ix)-(t+ix)^{-a}log(t+ix)}{e^{2\pi x}-1}dx$

By letting $a=-1, \;\ t=1$ leads to the two log/arctan integrals in question.

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If we let $x\mapsto e^{-x}$ and then use the series of $\log(1-e^{-x})$, we obtain that

$$\underbrace{\int_0^{\infty} \sum_{n=1}^{\infty}\log(4 \pi^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle \zeta(2) \log(4 \pi^2)} -\underbrace{\sum_{n=1}^{\infty}\int_0^{\infty} \log(4 \pi^2+y^2) \frac{e^{-y n}}{n} \ dy}_{\displaystyle -2\sum_{n=1}^{\infty}\frac{\operatorname{Ci}(2\pi n)}{n^2}+2\log(2)\zeta(2)+2\log(\pi)\zeta(2)}=\pi^{2} ( 4 \log A - 1 )$$

where in the last integral I used the exponential integral and $1.22a$ from this paper.

Q.E.D.

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Really late here, but please allow me to present a generalization: \begin{align*} &\int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{\ln ^2\left( x \right)}{\pi ^2\theta ^2} \right) \mathrm{d}x} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{\left(\theta -1 \right)^2}{2}-\frac{1}{6} \right)\pi ^2 \ln \left( \frac{\theta}{2} \right) -4\pi ^2\zeta '\left( -1,\frac{\theta}{2} \right) +\left(\frac{1}{3}-\frac{\theta ^2}{4}\right)\pi ^2 \end{align*} Derivation of this formula requires a few identities and lemmas:


Identity 0 $$ \int_0^1{\frac{\ln \left( 1-x \right)}{x}}\mathrm{d}x=-\frac{\pi ^2}{6} $$


Identity 1 $$ \coth \left( z \right) =\frac{1}{z}+\sum_{k=1}^{\infty}{\frac{2z}{z^2+n^2\pi ^2}} $$


Identity 2 $$ \int_0^{\infty}{\left( \frac{e^{-at}}{t}-\frac{be^{-ct}}{1-e^{-bt}} \right) \mathrm{d}t}=\psi \left( \frac{c}{b} \right) -\ln \left( \frac{a}{b} \right) $$


Identity 3 $$ \zeta '\left( -1,z \right) =\frac{1}{12}-\frac{z^2}{4}+\left( \frac{1}{12}-\frac{1}{2}z+\frac{1}{2}z^2 \right) \ln \left( z \right) +\mathcal{O} \left( \frac{1}{z^2} \right) $$


Identity 4 $$ \int_0^z{\ln \left( \Gamma \left( x \right) \right)}\mathrm{d}x=\frac{z}{2}\ln \left( 2\pi \right) +\frac{z\left( 1-z \right)}{2}-\zeta '\left( -1 \right) +\zeta '\left( -1,z \right) $$


Lemma 1 \begin{align*} &\int_1^t{\left( 2\ln \left( \Gamma \left( x \right) \right) -2x\ln \left( x \right) +2x+\ln \left( x \right) -\ln \left( 2\pi \right) -\frac{1}{6x} \right)}\mathrm{d}x \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-2\zeta '\left( -1 \right) +2\zeta '\left( -1,t \right) +\frac{1}{2}t^2-\left( t^2-t+\frac{1}{6} \right) \ln \left( t \right) -\frac{1}{2} \end{align*} This is basically done by using identity 4 and sheer calculation. Moreover, using identity 3, we have $$ \int_1^{\infty}{\left( 2\ln \left( \Gamma \left( x \right) \right) -2x\ln \left( x \right) +2x+\ln \left( x \right) -\ln \left( 2\pi \right) -\frac{1}{6x} \right) \mathrm{d}x}=-2\zeta '\left( -1 \right) -\frac{1}{3} $$


Lemma 2 $$ \int_0^1{\frac{-\theta \ln \left( 1-x \right)}{x\left[ \theta ^2+r^2\ln ^2\left( x \right) \right]}\mathrm{d}x}=\frac{\pi}{r}\left[ \ln \left( \Gamma \left( \frac{\theta}{2\pi r} \right) \right) -\frac{\theta}{2\pi r}\ln \left( \frac{\theta}{2\pi r} \right) +\frac{\theta}{2\pi r}+\frac{1}{2}\ln \left( \frac{\theta}{4\pi ^2r} \right) \right] $$ I shall prove this one: \begin{align*} \int_0^1{\frac{\ln \left( 1-x \right)}{x}\frac{-\theta}{\theta ^2+r^2\ln ^2\left( x \right)}\mathrm{d}x}=&\int_0^1{\left( \sum_{n=1}^{\infty}{\frac{x^{n-1}}{n}} \right) \left( \int_0^{\infty}{\cos \left( r\ln \left( x \right) t \right) e^{-\theta t}\mathrm{d}t} \right) \mathrm{d}x} \\ =&\int_0^1{e^{-\theta t}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^1{\cos \left( r\ln \left( x \right) t \right) x^{n-1}\mathrm{d}x}}\mathrm{d}t} \\ =&\int_0^{\infty}{e^{-\theta t}\sum_{n=1}^{\infty}{\frac{1}{n}\Re \left\{ \int_0^{\infty}{x^{irt+n-1}\mathrm{d}x} \right\}}\mathrm{d}t} \\ =&\int_0^{\infty}{e^{-\theta t}\sum_{n=1}^{\infty}{\frac{1}{r^2t^2+n^2}}\mathrm{d}t} \\ =&\int_0^{\infty}{\frac{e^{-\theta t}}{2rt}\left( \pi \coth \left( \pi rt \right) -\frac{1}{rt} \right) \mathrm{d}t} \\ =&\frac{1}{4r^2}\int_0^{\infty}{\frac{e^{-\theta t}}{t}\left( 2\pi r\frac{1+e^{-2\pi rt}}{1-e^{-2\pi rt}}-\frac{2}{t} \right) \mathrm{d}t} \\ =&\frac{1}{4r^2}\int_{\infty}^{\theta}{\left( \frac{\mathrm{d}}{\mathrm{d}a}\int_0^{\infty}{\frac{e^{-at}}{t}\left( 2\pi r\frac{1+e^{-2\pi rt}}{1-e^{-2\pi rt}}-\frac{2}{t} \right) \mathrm{d}t} \right) \mathrm{d}a} \\ =&\frac{1}{4r^2}\int_{\infty}^{\theta}{\int_0^{\infty}{e^{-a}\left( \frac{2}{t}-2\pi r\frac{1+e^{-2\pi rt}}{1-e^{-2\pi rt}} \right) \mathrm{d}t}\mathrm{d}a} \\ =&\frac{1}{4r^2}\int_{\infty}^{\theta}{\left( \int_0^{\infty}{\left( \frac{e^{-at}}{t}-\frac{2\pi re^{-at}}{1-e^{-2\pi rt}} \right) \mathrm{d}t}+\int_0^{\infty}{\left( \frac{e^{-at}}{t}-\frac{2\pi re^{-\left( 2\pi r+a \right) t}}{1-e^{-2\pi rt}} \right) \mathrm{d}t} \right) \mathrm{d}a} \\ =&\frac{1}{4r^2}\int_{\infty}^{\theta}{\left( 2\psi \left( \frac{a}{2\pi r} \right) -2\ln \left( \frac{a}{2\pi r} \right) +\frac{2\pi r}{a} \right) \mathrm{d}a} \\ =&\frac{\pi}{r}\left[ \ln \left( \Gamma \left( x \right) \right) -x\ln \left( x \right) +x+\frac{1}{2}\ln \left( x \right) \right] _{R\rightarrow \infty}^{\frac{\theta}{2\pi r}} \\ =&\frac{\pi}{r}\left[ \ln \left( \Gamma \left( \frac{\theta}{2\pi r} \right) \right) -\frac{\theta}{2\pi r}\ln \left( \frac{\theta}{2\pi r} \right) +\frac{\theta}{2\pi r}+\frac{1}{2}\ln \left( \frac{\theta}{4\pi ^2r} \right) \right] \end{align*}


Prove of the generalization \begin{align*} \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{\ln ^2\left( x \right)}{\theta ^2\pi ^2} \right) \mathrm{d}x}=&\int_0^1{\frac{\ln \left( 1-x \right)}{x}\int_0^1{\frac{2t\ln ^2\left( x \right)}{\theta ^2\pi ^2+t^2\ln ^2\left( x \right)}\mathrm{d}t}\mathrm{d}x} \\ =&2\int_0^1{\int_0^1{\frac{\ln \left( 1-x \right)}{xt}\left( \frac{t^2\ln ^2\left( x \right)}{\theta ^2\pi ^2+t^2\ln ^2\left( x \right)} \right)}\mathrm{d}x\mathrm{d}t} \\ =&2\int_0^1{\frac{1}{t}\int_0^1{\frac{\ln \left( 1-x \right)}{x}\left( 1-\frac{\theta ^2\pi ^2}{\theta ^2\pi ^2+t^2\ln ^2\left( x \right)} \right)}\mathrm{d}x\mathrm{d}t} \\ =&2\int_0^1{\frac{1}{t}\left( \int_0^1{\frac{\ln \left( 1-x \right)}{x}}\mathrm{d}x+\theta \pi \int_0^1{\frac{\ln \left( 1-x \right)}{x}\frac{-\theta \pi}{\theta ^2\pi ^2+t^2\ln ^2\left( x \right)}\mathrm{d}x} \right) \mathrm{d}t} \\ =&2\pi ^2\int_0^1{\frac{1}{t}\left( \frac{\theta}{t}\left[ \ln \left( \Gamma \left( \frac{\theta}{2t} \right) \right) -\frac{\theta}{2t}\ln \left( \frac{\theta}{2t} \right) +\frac{\theta}{2t}+\frac{1}{2}\ln \left( \frac{\theta}{4\pi t} \right) \right] -\frac{1}{6} \right) \mathrm{d}t} \\ =&2\pi ^2\int_{\frac{\theta}{2}}^{\infty}{\left( 2\ln \left( \Gamma \left( t \right) \right) -2t\ln \left( t \right) +2t+\ln \left( t \right) -\ln \left( 2\pi \right) -\frac{1}{6t} \right) \mathrm{d}t} \\ =&2\pi ^2\left( \int_1^{\infty}{-\int_1^{\frac{\theta}{2}}{\,\,}} \right) \left( 2\ln \left( \Gamma \left( t \right) \right) -2t\ln \left( t \right) +2t+\ln \left( t \right) -\ln \left( 2\pi \right) -\frac{1}{6t} \right) \mathrm{d}t \\ =&2\pi ^2\left( -2\zeta '\left( -1 \right) -\frac{1}{3}-\left( -2\zeta '\left( -1 \right) +2\zeta '\left( -1,\frac{\theta}{2} \right) +\frac{1}{2}\left( \frac{\theta}{2} \right) ^2-\left( \left( \frac{\theta}{2} \right) ^2-\frac{\theta}{2}+\frac{1}{6} \right) \ln \left( \frac{\theta}{2} \right) \right) +\frac{1}{2} \right) \\ =&\frac{\pi ^2}{2}\left( \left( \theta -1 \right) ^2-\frac{1}{3} \right) \ln \left( \frac{\theta}{2} \right) -4\pi ^2\zeta '\left( -1,\frac{\theta}{2} \right) +\frac{\pi ^2}{3}-\frac{\theta ^2\pi ^2}{4} \end{align*}


Results
The following is the formula evaluated at $\theta=\frac{1}{3}$, $\frac{1}{2}$, $\frac{2}{3}$, $1$, $\frac{4}{3}$, $\frac{3}{2}$, $\frac{5}{3}$, and $2$ respectively.

\begin{align*} \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{\pi ^2} \right) \mathrm{d}x}=&\frac{1}{36}\pi ^2\ln \left( \frac{1}{432} \right) +\frac{1}{4}\pi ^2+\frac{1}{3\sqrt{3}}\pi ^3+\frac{2}{3}\pi ^2\ln \left( \mathbf{A} \right) -\frac{\pi}{2\sqrt{3}}\psi ^{\left( 1 \right)}\left( \frac{1}{3} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{4\ln ^2\left( x \right)}{\pi ^2} \right) \mathrm{d}x}=&\frac{1}{12}\pi ^2\ln \left( 2 \right) +\frac{5}{16}\pi ^2-\pi \mathbf{G}-\frac{1}{2}\pi ^2\ln \left( \mathbf{A} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{4\pi ^2} \right) \mathrm{d}x}=&\frac{1}{18}\pi ^2\ln \left( 27 \right) +\frac{1}{3}\pi ^2+\frac{2}{9\sqrt{3}}\pi ^3-\frac{4}{3}\pi ^2\ln \left( \mathbf{A} \right) -\frac{\pi}{3\sqrt{3}}\psi ^{\left( 1 \right)}\left( \frac{1}{3} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{\ln ^2\left( x \right)}{\pi ^2} \right) \mathrm{d}x}\,\,\,\,=&\frac{1}{3}\pi ^2\ln \left( 2 \right) +\frac{1}{4}\pi ^2-2\pi ^2\ln \left( \mathbf{A} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{16\pi ^2} \right) \mathrm{d}x}=&\frac{1}{18}\pi ^2\ln \left( \frac{27}{4} \right) -\frac{2}{9\sqrt{3}}\pi ^3-\frac{4}{3}\pi ^2\ln \left( \mathbf{A} \right) +\frac{\pi}{3\sqrt{3}}\psi ^{\left( 1 \right)}\left( \frac{1}{3} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{4\ln ^2\left( x \right)}{9\pi ^2} \right) \mathrm{d}x}=&\frac{1}{24}\pi ^2\ln \left( \frac{4}{3} \right) -\frac{3}{16}\pi ^2+\pi \mathbf{G}-\frac{1}{2}\pi ^2\ln \left( \mathbf{A} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{25\pi ^2} \right) \mathrm{d}x}=&\frac{1}{36}\pi ^2\ln \left( \frac{25}{432} \right) -\frac{5}{12}\pi ^2-\frac{1}{3\sqrt{3}}\pi ^3+\frac{2}{3}\pi ^2\ln \left( \mathbf{A} \right) +\frac{\pi}{2\sqrt{3}}\psi ^{\left( 1 \right)}\left( \frac{1}{3} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{\ln ^2\left( x \right)}{4\pi ^2} \right) \mathrm{d}x}\,\,\,\,=&4\pi ^2\ln \left( \mathbf{A} \right) -\pi ^2 \end{align*} It seems like the integral have this conjugacy behavior (sort of) when thaking the value of $\theta$ and $2-\theta$, and if we add them together, we have \begin{align*} \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{\pi ^2} \right) \mathrm{d}x}+\int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{25\pi ^2} \right) \mathrm{d}x}=&\frac{1}{18}\pi ^2\ln \left( \frac{5}{432} \right) -\frac{1}{6}\pi ^2+\frac{4}{3}\pi ^2\ln \left( \mathbf{A} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{4\ln ^2\left( x \right)}{\pi ^2} \right) \mathrm{d}x}+\int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{4\ln ^2\left( x \right)}{9\pi ^2} \right) \mathrm{d}x}=&\frac{1}{24}\pi ^2\ln \left( \frac{16}{3} \right) +\frac{1}{8}\pi ^2-\pi ^2\ln \left( \mathbf{A} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{4\pi ^2} \right) \mathrm{d}x}+\int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{9\ln ^2\left( x \right)}{16\pi ^2} \right) \mathrm{d}x}=&\frac{1}{9}\pi ^2\ln \left( \frac{27}{2} \right) +\frac{1}{3}\pi ^2-\frac{8}{3}\pi ^2\ln \left( \mathbf{A} \right) \\ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{\ln ^2\left( x \right)}{\pi ^2} \right) \mathrm{d}x}\,\,\,\,+\,\,\int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\frac{\ln ^2\left( x \right)}{4\pi ^2} \right) \mathrm{d}x}\,\,=&\frac{1}{3}\pi ^2\ln \left( 2 \right) -\frac{3}{4}\pi ^2+2\pi ^2\ln \left( \mathbf{A} \right) \end{align*} Could it be for $2\pi \mid \theta _1+\theta _2$ $$ \int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\left( \frac{\ln \left( x \right)}{\theta _1} \right) ^2 \right) \mathrm{d}x}+\int_0^1{\frac{\ln \left( 1-x \right)}{x}\ln \left( 1+\left( \frac{\ln \left( x \right)}{\theta _2} \right) ^2 \right) \mathrm{d}x}\\ \in \mathbb{Q} \pi ^2\ln \left( \mathbb{Q} \right) +\mathbb{Q} \pi ^2+\mathbb{Q} \pi ^2\ln \left( \mathbf{A} \right) $$ Just some thoughts, though.

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Initially we have: $$\displaystyle \int_{0}^{1}{\ln\left(1+\frac{\ln^2 x}{4\pi^2} \right)\frac{\ln\left(1-x \right)}{x}}dx=2\pi\int_{0}^{\infty}{\ln\left(1+x^2 \right)\ln\left(1-e^{-2\pi x} \right)}dx$$

$$\displaystyle =-2\pi\sum_{k=1}^{\infty}{\frac{1}{k}\int_{0}^{\infty}{\ln\left(1+x^2 \right)e^{-2\pi kx}dx}}$$ $$ \displaystyle =-2\sum_{k=1}^{\infty}{\frac{1}{k^2}\int_{0}^{\infty}{\frac{xe^{-2\pi k x}}{1+x^2}}}dx.$$

Let $$\displaystyle I(a)=\int_{0}^{\infty}{\frac{e^{-ax}x}{x^2 +1}}dx,$$ then it follows that $$\displaystyle I'(a)=\int_{0}^{\infty}{\frac{e^{-ax}(-x^2)}{x^2 +1}}dx=\int_{0}^{\infty}{\frac{e^{-ax}}{1+x^2}}dx-\frac{1}{a}.$$ The last integral has been calculated here(*).

So it arises $$\displaystyle \int_{0}^{\infty}{\frac{x}{x^2 +1}}e^{-2\pi k x}dx=-Ci\left(2\pi k \right)=-\gamma -\ln\left(2\pi k \right)-\int_{0}^{2\pi k}{\frac{\cos t-1}{t}dt}.$$

Finally we have:

$$\displaystyle \pi\int_{0}^{\infty}{\ln\left(1+x^2 \right)\ln\left(1-e^{-2\pi x} \right)}dx=\frac{\gamma \pi^2}{6}$$ $$\displaystyle +\frac{\ln\left(2\pi \right)\pi^2}{6}+\sum_{k=1}^{\infty}{\int_{0}^{1}{\frac{\cos\left(2\pi k x \right)-1}{k^2x}}}dx+\sum_{k=1}^{\infty}{\frac{\ln k}{k^2}}.$$

Also using elementary Fourier analysis it is proved that: $$\displaystyle \sum_{k=1}^{\infty}{\frac{\cos\left(2\pi k x \right)-1}{k^2}}=\pi^2 \left(x^2 -x \right),$$ $0<x<1$, and even uniformly from Weierstass' M-test.

Consequently \begin{align*} % Your equations here with ampersands (&) for alignment points \sum_{k=1}^{\infty} {\int_{0}^{1} {\frac{\cos\left(2\pi k x \right)-1}{k^2 x}}} dx &= \int_{0}^{1} {\pi^2} \left(x-1 \right) dx \\ &= -\frac{\pi^2}{2} \end{align*}

Finally, because $$\displaystyle \zeta'\left(2 \right)=\sum_{k=2}^{\infty}{\frac{\ln k}{k^2}}=\frac{\pi^2}{6}\left(12\ln A-\gamma -\ln 2\pi \right),$$ it follows that $$\displaystyle \int_{0}^{1}{\ln\left(1+\left(\frac{\ln x}{2\pi} \right)^2 \right)\frac{\ln\left(1-x \right)}{x}}dx=\pi^2\left(4\ln A-1 \right)=-\pi^2\left(4\zeta'\left(-1 \right)+\frac{2}{3} \right).$$

Addendum: The value of $\displaystyle \zeta'(2)$ the function of $\displaystyle \zeta'(1)$, is derived from a derivation of the functional relation $$\displaystyle \Gamma\left(\frac{s}{2} \right)\zeta(s)\pi^{-s/2}=\Gamma\left(\frac{1-s}{2} \right)\zeta\left(1-s \right)\pi^{-\frac{1-s}{2}}.$$

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