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Question:-Prove that $$\int_0^1( \ln x)^2 \frac{1+x^2}{1+x^4} \, dx=\frac {3\sqrt{2}\pi^3}{64}$$

My attempt:

$$\int_0^1 \left( \ln\frac{1}{x}\right)^2 \frac{1+x^{-2}}{x^{-2}+x^2} \, dx$$

Put $\left(x-\frac{1}{x}\right)=t$, we get

$$\int_{-\infty}^0\left( \ln\frac{(t+\sqrt{t^2+4})}{2}\right)^2 \frac{1}{t^2+2} \, dt$$

After that I got stuck.

Can anybody help me!

Michael Hardy
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Paras
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  • This is a duplicate of this question, as with the substitution $x\to \frac{1}{x}$ we have: $$\int_1^\infty \frac{1+x^2}{1+x^4}\ln^2 x,dx=\int_0^1 \frac{1+x^2}{1+x^4}\ln^2 x,dx$$ $$\Rightarrow 2\int_0^1 \frac{1+x^2}{1+x^4}\ln^2 x,dx=\int_0^\infty \frac{1+x^2}{1+x^4}\ln^2 x,dx$$ – Zacky May 12 '20 at 07:14

1 Answers1

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Let $f(x)=\frac{1+x^2}{1+x^4}\ln^2 x$. By the substitution $x\mapsto\frac1x$, it can proved that $I=\int^1_0f(x)dx=\int^\infty_1f(x)dx$. Hence $2I=\int^\infty_0 f(x)dx$.

Employing Feynman's technique, $$2I=\frac{\partial^2}{\partial a^2}\underbrace{\int^\infty_0\frac{1+x^2}{1+x^4}x^adx}_{J}\bigg\vert_{a=0}$$

We rewrite it as $$J=\frac12\int^\infty_0\frac{1+z}{1+z^2}z^b dz\qquad b=\frac{a-1}{2}$$ by enforcing $z=x^2$, so as to facilitate the evaluation of $J$ by residue calculus.

We take the branch cut of $z^b$ along the positive real axis and consider a keyhole contour ($C$) integral.

Then, $$\oint_C \frac{1+z}{1+z^2}z^b dz=2\pi i(\operatorname{Res}_{z=i}+\operatorname{Res}_{z=-i})$$

Note that as $b<0$, the integrand is $o(z^{-1})$ and thus the large circular contour integral vanishes as its radius tends to infinity.

The integral above positive real axis is $2J$ while that below is $-e^{-2\pi ib}\cdot2J$.

Hence, $$2(1-e^{2\pi ib})J=2\pi i\left[\frac{1+i}{2i}i^b+\frac{1-i}{-2i}(-i)^b\right]$$ $$2(1-e^{2\pi ib})J=\pi \left[(1+i)e^{\pi ib/2}-(1-i)e^{3\pi ib/2}\right]$$ $$2(e^{-\pi ib}-e^{\pi ib})J=\pi \left[(1+i)e^{-\pi ib/2}-(1-i)e^{\pi ib/2}\right]$$ $$-4i\sin(\pi b)J=\pi \left[e^{-\pi ib/2}-e^{\pi ib/2}+i(e^{-\pi ib/2}+e^{\pi ib/2})\right]$$ $$-4i\sin(\pi b)J=\pi \left(-2i\sin(\pi b/2)+2i(\cos(\pi b/2)\right)$$ $$J=\frac{\pi}{4}(-\csc(\pi b/2)+\sec(\pi b/2))$$

Note that in the last step we used double-angle formula $\sin\theta=2\sin\frac{\theta}2\cos\frac{\theta}2$.

$$I=\frac12\frac{\partial^2 J}{\partial a^2}\bigg\vert_{a=0}=\frac12\frac{\partial^2 J}{(2\partial b)^2}\bigg\vert_{b=-1/2}=\frac12\cdot\frac14\cdot\frac{\pi}{4}\cdot\frac{\pi^2}{4}(-\left(-3\sqrt2)+3\sqrt2\right)=\frac{3\sqrt 2}{64}\pi^3$$

Szeto
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