I've been trying to calculate the value of the integral $$ \int_{0}^\infty\frac{\log^2 x}{1+x^2}\,dx. $$

I am running into problems when forming even an appropriate contour. I've tried forming a specific contour and taking the principal branch of the logarithm, but I wasn't able to get good bounds and reduce the problem. I've calculated the residue to be $$\frac{\log^2(i)}{2i},$$but after struggling with what seems like an easy problem, I'm not sure I trust even that calculation.

Any help would be appreciated. Thanks.


We have $$I = \int_0^{\infty} \dfrac{\log^2(x)}{1+x^2} dx = \int_0^1 \dfrac{\log^2(x)}{1+x^2} dx + \int_1^{\infty} \dfrac{\log^2(x)}{1+x^2} dx$$ Now $$\overbrace{\int_1^{\infty} \dfrac{\log^2(x)}{1+x^2} dx = -\int_1^0 \dfrac{\log^2(x)}{1+x^2}dx}^{x \mapsto 1/x} = \int_0^1 \dfrac{\log^2(x)}{1+x^2}dx$$ Hence, we get that $$I = 2 \int_0^1 \dfrac{\log^2(x)}{1+x^2}dx = 2 \sum_{k=0}^{\infty}(-1)^k \int_0^1 x^{2k} \log^2(x) dx$$ $$\int_0^1 x^{2k} \log^2(x) dx = \dfrac2{(2k+1)^3} \,\,\,\, (\spadesuit)$$ Hence, $$I = 4 \sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^3} = \dfrac{\pi^3}8 \,\,\,\, (\clubsuit)$$

$(\spadesuit)$ is nothing but the $\Gamma$-function with some scaling factors. $$I_k = \underbrace{\int_0^1 x^{2k} \log^2(x) dx = \int_0^{\infty} t^2 e^{-(2k+1)t}dt}_{x \mapsto e^{-t}}$$ Now let $(2k+1)t = x$. We then get $$I_k = \int_0^{\infty} \dfrac{x^2}{(2k+1)^2} e^{-x} \dfrac{dx}{2k+1} = \dfrac{\Gamma(3)}{(2k+1)^3} = \dfrac2{(2k+1)^3}$$

$(\clubsuit)$ is evaluated as follows. We have $$\text{Li}_s(z) = \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$$ Now $$\text{Li}_3(i) = \sum_{k=1}^{\infty} \dfrac{i^k}{k^s}$$ $$\text{Li}_3(-i) = \sum_{k=1}^{\infty} \dfrac{(-1)^ki^k}{k^s}$$ Hence, $$\text{Li}_3(i) - \text{Li}_3(-i) = 2i \left(\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^3}\right) \,\,\,\, (\heartsuit)$$ Now the PolyLogarithmic function satisfies a very nice identity $$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ Taking $n=3$ and $x=\dfrac14$, gives us $$\text{Li}_3(i) - \text{Li}_n(-i) = - \dfrac{(2\pi i)^3}{3!}B_3(1/4) = i \dfrac{8 \pi^3}{6} \dfrac3{64} =i \dfrac{\pi^3}{16} \,\,\,\, (\diamondsuit)$$ Comparing $(\heartsuit)$ and $(\diamondsuit)$ gives us $(\clubsuit)$.

  • $\begingroup$ Marvis, are the last two identities something I should recognize? I'm guessing it is some sort of a reduction formula to get $$\int_0^1x^{2k}\log^2(x)\,dx=\frac{2}{(2k+1)^3}$$ and then the sum is somehow calculated to be $\frac{\pi^3}{32}$ (the multiplication by $4$ then changing the denominator appropriately)?? $\endgroup$ – Clayton Mar 27 '13 at 4:30
  • $\begingroup$ @Clayton For the integral, replace $x$ by $e^{-t}$, we then get $$\int_{0}^{\infty} e^{-(2k+1)t} t^2 dt$$ which gives us $\dfrac2{(2k+1)^3}$ from the definition of $\Gamma$-function. For the second one, I derived the infinite sum sometime back on stackexchange. Let me search for it and get back to you. $\endgroup$ – user17762 Mar 27 '13 at 4:34
  • $\begingroup$ @Clayton For the summation, you need to essentially make use of a very nice identity satisfied by the PolyLogarithmic function namely $$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ For instance, look here at one of my old answers. I am still searching for this exact summation, I have in my answer. $\endgroup$ – user17762 Mar 27 '13 at 4:42
  • $\begingroup$ Thank you for the help! I really like your method of solution! $\endgroup$ – Clayton Mar 27 '13 at 4:43
  • $\begingroup$ @Clayton Unfortunately, I couldn't find my previous version on stackexchange. Hence, I have written the solution in my answer. $\endgroup$ – user17762 Mar 27 '13 at 5:06

Hint: In this answer, it is shown that $$ \frac\pi2\sec\left(\frac\pi2\alpha\right) =\int_0^\infty\frac{z^\alpha}{1+z^2}\,\mathrm{d}z $$ Differentiate twice with respect to $\alpha$ and set $\alpha=0$.

Two other methods are given in this answer.

  • $\begingroup$ Thank you very much. The contour you give in the second link is really the sort of solution I was looking for, although it seems arbitrary that, for that problem, you integrate with $\log^3(z)$ instead of $\log^2(z)$, yet the imaginary part break up in such a way you can calculate the integral. Is there a natural way to see such real functions so that their imaginary part turns into the sought after function? $\endgroup$ – Clayton Mar 27 '13 at 4:42
  • $\begingroup$ @Clayton: That we used $\log(z)^3$ instead of $\log(z)^2$ is not so arbitrary if you try to use $\log(z)^2$ in its place. Since we integrate $$\frac{(\log(z)+2\pi i)^2-\log(z)^2}{1+z^2}=\frac{4\pi i\log(z)-4\pi^2}{1+z^2}$$ we get a relation between the integrals of $\frac{\log(z)}{1+z^2}$ and $\frac1{1+z^2}$. The natural thing to do is to use $\log(z)^3$ to get a relation among the integrals of $\frac{\log(z)^2}{1+z^2}$, $\frac{\log(z)}{1+z^2}$, and $\frac1{1+z^2}$ $\endgroup$ – robjohn Mar 27 '13 at 10:20

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