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Calculating in closed form $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^4(m^2+n^2)}$

Find the value:

$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$

I use computer have this reslut

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$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{2}\dfrac{\pi\ln^2{2}}{\sqrt{2}}-\dfrac{3}{2}\dfrac{\pi^2\ln{2}}{\sqrt{2}}$$

where $C$ is Catalan constant.

My idea: let $$\tan{x}=t,\sin{2x}=\dfrac{2t}{1+t^2},dx=\dfrac{1}{1+x^2}$$

then

$$I=\int_{0}^{\infty}\ln^2{(1+t^4)}\dfrac{\dfrac{2}{1+t^2}}{2-\left(\dfrac{2t}{1+t^2}\right)^2}\cdot\dfrac{dt}{1+t^2}=\int_{0}^{\infty}\dfrac{\ln^2{(1+t^4)}}{(t^2-1)^2}dt$$

Then I can’t.Thank you

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Here is an approach. I’ll continue from the step you have already reached with the correction suggested in Fabien comment which is considering the integral

$$ I = \int_{0}^{\infty} \frac{\ln^2(1+x^4)}{1+x^4} dx. $$

To evaluate the above integral we consider the integral

$$ F = \int_{0}^{\infty} (1+x^4)^{\alpha} dx = \frac{\pi}{2\sqrt{2}\Gamma( 3/4 )}{\frac {\Gamma ( -1/4-\alpha ) }{

\Gamma \left( -\alpha \right) }},$$

which can be evaluated using the beta function techniques (use the substitution $(1+x^4)=\frac{1}{t}$). Now $I$ follows from $F$ as (see related techniques )

$$ I = \lim_{\alpha \to -1}\frac{d^2F(\alpha)}{d\alpha^2}=\frac{\pi \,\sqrt {2}}{48}\left({\pi}^{2}-36\,\pi \,\ln \left( 2 \right) +108 \ln^2( 2 )+12\,\psi’\left( 3/4 \right) \right) .$$

**Hint:** $\quad~I(n)~=~\displaystyle\int_0^\infty\Big(1+x^k\Big)^n~dx\quad~=>\quad~I”(-1)~=~\displaystyle\int_0^\infty\frac{\ln^2\big(1+x^k\big)}{1+x^k}~dx.~$ But, at

the same time, $I(n)$ can be shown to equal $~\dfrac1k\cdot B\bigg(\dfrac1k~,-\dfrac1k-n\bigg),~$ with the help of the simple

substitution $t=\dfrac1{1+x^k}$ . Then, expressing it in terms of the $\Gamma$ function, differentiating, and using

the relationship between the polygamma function $\psi_0$ and harmonic numbers, along with Euler’s

formula for their generalization to non-natural arguments, we are thus finally able to arrive at an

expression whose only “mysterious” quantity is $\psi_{_1}\Big(\frac34\Big).~$ Since this function has been studied for

several centuries, I’m sure you will be able to find its closed form expression somewhere, along with

the proof behind it. Also, Euler’s reflection formula for the $\Gamma$ function might be of some assistance.

You have made a mistake with the denominator, which cannot be zero. See, $2-sin(2x)^2$ is always bigger or equal than $1$.

I think that the mistake was in expanding $2-(\frac{2t}{1+t^2})^2$: you have changed that initial $2$ for a $1$ and then completed the square. Repeat that expansion slowly.

After performing the initial arctangent substitution, the resultant integral may be rewritten as the second derivative of a beta function and evaluated accordingly:

$$\begin{align}

I

&=\int_{0}^{\frac{\pi}{2}}\frac{2\cos^2{\theta}}{2-\sin^2{(2\theta)}}\log^2{\left(1+\tan^4{\theta}\right)}\,\mathrm{d}\theta\\

&=\int_{0}^{\infty}\frac{\log^2{(1+x^4)}}{1+x^4}\mathrm{d}x\\

&=\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\int_{0}^{\infty}\frac{\mathrm{d}x}{(1+x^4)^{\alpha}}\\

&=\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\int_{0}^{\infty}\frac{\frac14u^{-\frac34}}{(1+u)^{\alpha}}\mathrm{d}u\\

&=\frac14\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\operatorname{B}{\left(\frac14,\alpha-\frac14\right)}\\

&=\frac14\Gamma{\left(\frac14\right)}\Gamma{\left(\frac34\right)}\left[-\zeta{(2)}+\left(\gamma+\Psi{\left(\frac34\right)}\right)^2+\Psi^\prime{\left(\frac34\right)}\right]\\

&=\frac{\sqrt{2}\,\pi}{4}\left[-\frac{\pi^2}{6}+\left(\frac{\pi}{2}-\log{8}\right)^2+\pi^2-8C\right]\\

&=-2\sqrt{2}\,\pi\,C+\frac{13\pi^3}{24\sqrt{2}}+\frac{\pi\log^2{8}}{2\sqrt{2}}-\frac{\pi^2\log{8}}{2\sqrt{2}}.~~~\blacksquare

\end{align}$$

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