I want to integrate $$\int_0^\infty \dfrac{\log t}{1+t^2}\,\mathrm dt$$ using the residue theorem. The poles are at $i,-i$. If the integral were from $-\infty$ to $\infty$, I would consider integrating along the semicircle with large radius $R$ and centered at the origin. But here it is from $0$ to $\infty$. What contour would be good to integrate along, if any?

  • $\begingroup$ maybe a wedge... $\endgroup$ – oldrinb Nov 3 '13 at 6:19

You can use a keyhole contour about the positive real axis, but you would consider the contour integral

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

where $C$ is the above described keyhole contour. The integrals over the large and small arcs about the origin vanish as the respective radii go to infinity and zero. Thus, we have that the above contour integral is equal to

$$\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{1+x^2} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{1+x^2}+4 \pi^2 \int_0^{\infty} dx \frac{1}{1+x^2}$$

This contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$:

$$\frac{i 2 \pi}{i 2} \left ( -\frac{\pi^2}{4} +\frac{9 \pi^2}{4}\right ) = 2 \pi^3$$

Setting the above two expressions equal to each other, you can show that the desired integral is zero.

  • $\begingroup$ Thanks, but I'm confused where the expression $\int_0^{\infty} dx \frac{\log^2{x}-(\log{x}+i 2 \pi)^2}{1+x^2}$ comes from. $\endgroup$ – Mika H. Nov 3 '13 at 6:50
  • $\begingroup$ From the multivaluedness of the log. Going back along the real axis, $z=x e^{i 2 \pi}$. $\endgroup$ – Ron Gordon Nov 3 '13 at 6:57
  • $\begingroup$ Actually, for the small arc over the origin, the $\log$ value is large in magnitude. Why does the integral vanish there? $\endgroup$ – Mika H. Nov 3 '13 at 16:19
  • $\begingroup$ Because $\lim_{x\to 0} x \log{x}=0$. $\endgroup$ – Ron Gordon Nov 3 '13 at 16:30

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