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I Would like to know the value of this integral. $$\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$$ I think $$I=\frac{a}{b}(\pi^3\ln2)+\frac{c}{d}(\pi\ln^32)+\frac{e}{f}(\pi\ln^22)+\frac{g}{h}(\pi\ln2)+\frac{i}{j}(\pi^3)+\frac{k}{m}\zeta({3})??$$ Where a,b,c,d...are integers Thanks.

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    $\begingroup$ Where did this horrible, integral-creature come from? $\endgroup$
    – graydad
    Feb 17, 2015 at 22:42
  • $\begingroup$ W|A does not give me a value. But I think I've seen something similar in the past. I would start to tackle it using the series expansion of $\ln (1+x)$. I also think that complex analysis will come in handy here. Especially contour integration. $\endgroup$
    – Tolaso
    Feb 17, 2015 at 22:49

2 Answers 2

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Let $\displaystyle\small\gamma=\lim_{R\to\infty}[-R,R]\cup Re^{i[0,\pi]}$. Observe that $$\small\oint_\gamma\frac{\ln^4(1-iz)}{z^2}dz=\frac{1}{8}\int^\infty_0\frac{\ln^4(1+x^2)}{x^2}dx-3\int^\infty_0\frac{\ln^2(1+x^2)\arctan^2{x}}{x^2}dx+2\int^\infty_0\frac{\arctan^4{x}}{x^2}dx=0$$ since

  • The integral over the arc vanishes as $\small\mathcal{O}\left(\dfrac{\ln^4{R}}{R}\right)$.
  • The imaginary part is odd and vanishes over a symmetric interval.
  • The singularity at $\small 0$ is removable.

Therefore, \begin{align} \small\int^\infty_0\frac{\ln^2(1+x^2)\arctan^2{x}}{x^2}dx &\small=\frac{1}{24}\int^\infty_0\frac{\ln^4(1+x^2)}{x^2}dx+\frac{2}{3}\int^\infty_0\frac{\arctan^4{x}}{x^2}dx\\ &\small=\frac{1}{3}\int^\infty_0\frac{\ln^3(1+x^2)}{1+x^2}dx+\frac{2}{3}\int^\frac{\pi}{2}_0x^4\csc^2{x}\ dx\\ &\small=-\frac{8}{3}\int^\frac{\pi}{2}_0\ln^3(\cos{x})\ dx+\frac{8}{3}\int^\frac{\pi}{2}_0x^3\cot{x}\ dx\\ &\small=-\frac{1}{6}\frac{\partial^3\operatorname{B}}{\partial b^3}\left(\frac{1}{2},\frac{1}{2}\right)+\frac{16}{3}\sum^\infty_{n=1}\int^\frac{\pi}{2}_0x^3\sin(2nx)dx\\ &\small=2\pi\zeta(3)+\frac{\pi^3}{3}\ln{2}+\frac{4\pi}{3}\ln^3{2}+2\pi\sum^\infty_{n=1}\frac{(-1)^n}{n^3}+\frac{\pi^3}{3}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\\ &\small=2\pi\zeta(3)+\frac{\pi^3}{3}\ln{2}+\frac{4\pi}{3}\ln^3{2}+2\pi\left(-\frac{3}{4}\zeta(3)\right)+\frac{\pi^3}{3}\ln{2}\\ &\small=\frac{\pi}{2}\zeta(3)+\frac{2\pi^3}{3}\ln{2}+\frac{4\pi}{3}\ln^3{2} \end{align}

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    $\begingroup$ Very Well,great +1.Thanks $\endgroup$
    – user178256
    Feb 18, 2015 at 7:42
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It has a closed form a bit different from what you conjectured: $$I=\frac{4\pi}3\ln^32+\frac{2\pi^3}3\ln2+\frac\pi2\zeta(3)$$

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