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

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

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

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

• Theon is right : it should be $I=\int_{0}^{\infty}\dfrac{\ln^2{(1+t^4)}}{1+t^4}dt$ – Fabien Jul 21 '14 at 7:54

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 )

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

• +1. The substitution which leads to the Beta function should be $\displaystyle{\large\left(1 + x^{4}\right) \equiv {1 \over t}}$. I guess that's is a typo. Otherwise, it's a straightforward answer. – Felix Marin Aug 9 '14 at 20:54
• @FelixMarin: It is a typo. Thanks for the comment. – Mhenni Benghorbal Aug 10 '14 at 0:04

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.

• Personally, i believe this is how answers are ought to be written. – Lost1 Jul 21 '14 at 10:15
• Really nice answer : the maths for the result isn't necessary, the way of thinking is what matters. – Fabien Jul 21 '14 at 10:18

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}

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.