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Evaluate $$I =\int_0^{\infty}\frac{(1+x^2) \ln^2x}{1+x^4}dx$$

I didn't think much and applied integration by parts. After applying ibp repeatedly, I got stuck at this integral $$\int \frac{\ln^2(x)}{1+\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)}dx$$

I wasn't able to find an elementary anti-derivative of this function. I can apply ibp at this point too, but the things will go very lengthy and complicated from here.

I was wondering if there is some another method (obviously simpler) for this problem.

Any help is greatly appreciated.

Quanto
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Blue Cat Blues
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    Perhaps not simpler, but you could iterate the power of the log and apply the residue theorem. example – user170231 Jan 22 '25 at 19:06
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    Another example using Taylor series – user170231 Jan 22 '25 at 19:22
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    $$I=\int_0^\infty\ln^2(x)\cdot\frac{1+x^2}{1+x^4}dx\overset{x=e^{-\frac t2}}{=}\frac18\int_{-\infty}^\infty t^2\frac{\cosh \frac t2}{\cosh t}dt=\frac18\frac{\partial^2}{\partial a^2},\bigg|{a=\frac12}\int{-\infty}^\infty \frac{\cosh at}{\cosh t}dt$$ Using the rectangular contour in the complex plane $$\int_{-\infty}^\infty\frac{e^{at}}{\cosh t}dt=\frac{\pi}{\cos\frac{\pi a}2},,\Rightarrow,,\int_{-\infty}^\infty \frac{\cosh at}{\cosh t}dt=\frac{\pi}{\cos\frac{\pi a}2}$$ – Svyatoslav Jan 23 '25 at 02:09
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    Corrected a typo $$I=\frac18\frac{\partial^2}{\partial a^2},\bigg|{a=\frac12}\frac{\pi}{\cos\frac{\pi a}2}=\frac\pi8\left(\frac\pi2\right)^2\frac{\partial^2}{\partial x^2},\bigg|{x=\frac\pi4}\frac1{\cos x}=\frac{3\sqrt2,\pi^3}{32}$$ – Svyatoslav Jan 23 '25 at 06:52

3 Answers3

13

\begin{align} \int_0^{\infty}&\ln^2x\cdot\frac{1+x^2}{1+x^4}dx = \frac{d^2}{da^2} \bigg( \int_0^{\infty}\frac{x^a+x^{a+2}}{1+x^4}dx\bigg)_{a=0}\\ = &\ \frac{d^2}{da^2} \bigg[\frac\pi4\csc\frac{(a+1)\pi}4 + \frac\pi4\csc\frac{(a+3)\pi}4\bigg]_{a=0}=\frac{3\pi^3}{16\sqrt2}\\ \end{align}

Quanto
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$$I=\int\log^2(x)\,\frac{1+x^2}{1+x^4}\,dx$$

Write $$\frac{1+x^2}{1+x^4}=\frac {(x+i)(x-i)} {\prod _{i=1}^4 (x-r_i)}$$ and use partial fraction decomposition to face $$\frac{1+x^2}{1+x^4}=\sum _{i=1}^4 \frac {A_i}{x-r_i}$$ and use $$\int \frac {\log^2(x)}{x-r_i}\,dx=\log ^2(x) \log \left(1-\frac{x}{r_i}\right)+2 \log (x) \,\text{Li}_2\left(\frac{x}{r_i}\right)-2\, \text{Li}_3\left(\frac{x}{r_i}\right)$$

Recombine everything and compute $$J(p)=\int_0^p\log^2(x)\,\frac{1+x^2}{1+x^4}\,dx$$ Expanding as a series for large $p$ $$J(p)=\frac{3 \pi ^3}{16 \sqrt{2}}-\frac{\log ^2(p)+2 \log (p)+2}{p}-\frac{9 \log ^2(p)+6 \log (p)+2}{27 p^3}+O\left(\frac{1}{p^5}\right)$$

Edit

You could use this approach for $$K_{m,n}=\int_0^\infty \log^2(x)\,\frac{1+x^m}{1+x^n }\,dx$$ For $n=m+2$, this would generate (starting from $m=0$) $$\left\{\frac{\pi ^3}{4},\frac{20 \pi ^3}{81 \sqrt{3}},\frac{3 \pi ^3}{16 \sqrt{2}},\frac{2\sqrt{850+362 \sqrt{5}}}{625} \pi ^3,\frac{7 \pi ^3}{54}\right\}$$ which is close to $$K_{m,m+2}\sim 4+\frac{\pi ^3}{85 m}$$

Claude Leibovici
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\begin{align} &\int_0^{\infty}\ln^2x\cdot\frac{1+x^2}{1+x^4}dx\\ &= \int_0^{1}\ln^2x\cdot\frac{1+x^2}{1+x^4}dx+\int_1^{\infty}\ln^2x\cdot\frac{1+x^2}{1+x^4}\overset{x\to1/x}{dx}\\ &= 2\int_0^{1}\ln^2x\cdot\frac{1+x^2}{1+x^4}dx\\ &=2\int_0^{1}\sum_{n=0}^\infty(-1)^n(1+x^2)x^{4n}\ln^2xdx\\ &=2\sum_{n=0}^\infty(-1)^n\int_0^{1}(1+x^2)x^{4n}\ln^2xdx\\ &=2\sum_{n=0}^\infty(-1)^n\int_0^{1}(x^{4n}+x^{4n+2})\ln^2xdx\\ &=4\sum_{n=0}^\infty(-1)^n\bigg[\frac1{(4n+1)^3}+\frac1{(4n+3)^3}\bigg]\\ &=4\sum_{n=\infty}^\infty(-1)^n\frac1{(4n+1)^3}\\ &=-4\pi\text{Res}\left(\frac{1}{(4z+1)^3\sin(\pi z)},z=-\frac14\right)\\ &=\frac{3\pi^3}{16\sqrt2}. \end{align}

xpaul
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