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I'm asked to solve the following improper integral:

$$\int_0^\infty \frac{\rm {Log}^2(t)}{1+t^2}dt. $$

Do I consider the function $f(z) = \frac{\rm{Log}^2(z)}{1+z^2}$ or some variant?

Is the correct contour the so called "keyhole" that skips the $(0,+\infty)$ interval?

Any hints and tips are appreciated. I don't want this problem to be completely solved, just need the starter and some ideas that I can use in this case and similar ones.

Thanks a lot!

Mhenni Benghorbal
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Bouvet Island
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2 Answers2

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To use a keyhole contour $C$, consider

$$\oint_C dz \frac{\log^3{z}}{1+z^2}$$

You can show that the integrals about the circular arcs about the origin, both large and small, vanish as theire respective radii go to $\infty$ and $0$. The contour integral is therefore equal to

$$\int_0^{\infty} dx \frac{\log^3{x} - (\log{x}+i 2 \pi)^3}{1+x^2} $$

which simplifies to

$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{1+x^2} + i 8 \pi^3 \int_0^{\infty} \frac{dx}{1+x^2} $$

The contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, namely $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$. (Note that it is crucial that $\arg{z} \in [0,2 \pi)$.) This sum is

$$\frac{-i \pi^3/8}{2 i} + \frac{-i 27 \pi^3/8}{-2 i} = \frac{13 \pi^3}{8} $$

Multiplying by $i 2 \pi$, we have

$$i \left [- 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + 8 \pi^3 \int_0^{\infty} \frac{dx}{1+x^2} \right ] + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{1+x^2} = i \frac{13 \pi^4}{4}$$

Equating imaginary parts and noting that the second integral in the brackets is simply $\pi/2$ (you should have permission to evaluate that without residues), we have

$$- 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} = \frac{13 \pi^4}{4} - 4 \pi^4 = -\frac{3 \pi^4}{4} $$

or

$$\int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} = \frac{\pi^3}{8}$$

Ron Gordon
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  • Thanks a lot Ron. Very useful. One more question: Do I have to always use $log^{n+1}$ when dealing with this type of integrals? Because that's my biggest problem with them: I just can't see what is the proper complex function that suits for each situation.

    Again, thank you very much!

     – Bouvet Island Dec 09 '13 at 03:00
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    @BouvetIsland: it depends on the contour. For example, see the contour I chose to solve this problem: http://math.stackexchange.com/questions/597638/calculate-displaystyle-int-0-infty-frac-ln-x1-x4-mathrmdx-using/597688#597688 If I had chosen to use a keyhole (which I could have done), then I would have used $\log^2{z}$. But there I did not. Study the solutions and see why I chose the integrands I did. – Ron Gordon Dec 09 '13 at 03:28
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Instead of contour integration, here is an easier way out. Note that $$\int_1^{\infty} \dfrac{\log^2(t)}{1+t^2}dt = \int_1^0 \dfrac{\log^2(1/t)}{1+1/t^2}\dfrac{-dt}{t^2} = \int_0^1\dfrac{\log^2(t)}{1+t^2}dt$$ Hence, the integral you are interested in is nothing but $$I = 2 \int_0^1 \dfrac{\log^2(t)}{1+t^2}dt = 2 \sum_{k=0}^{\infty}(-1)^k\int_0^1 t^{2k} \log^2(t) dt = 4 \sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^3} = \dfrac{\zeta(3,1/4)-\zeta(3,3/4)}{16} = \dfrac{\pi^3}{8}$$ where the last identity can be obtained from the properties of the PolyLogarithm function.

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    There's actually a much simpler way to evaluate that sum using...the residue theorem. – Ron Gordon Dec 09 '13 at 02:01
  • @RonGordon True. But I try to use techniques purely based on real variables, than invoking complex analysis. Sometimes, I think complex analysis is too powerful tool and there is a pleasure in solving a problem with slightly simpler tools. –  Dec 09 '13 at 02:03
  • Not to argue, as this is all subjective, but polylogs are hardly "simpler" than residues. – Ron Gordon Dec 09 '13 at 02:04
  • Thanks for the approach. The problem is that I'm on a complex analysis course and the idea is to use residues and stuff. Besides, I'm not familiar with polylogs... – Bouvet Island Dec 09 '13 at 02:23