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$$ \mbox{Consider}\quad \int_{0}^{\infty}{\log\left(x\right) \over \left(\, x + a\,\right)^3} \,\mathrm{d}x\quad \mbox{where}\ a\ \mbox{is}\ positive. $$

  • First, we make a branch cut along the negative imaginary axis, and then consider the standard contour with a small semicircle around 0 and a small semicircle centered at $-a$.
  • Letting $\,\mathrm{f}\left(z\right)$ be the integrand above with $z$ in place of $x$, the residue at $-a$ is $-1/\left(2a^{2}\right)$.
  • I'm able to show that the integral around the big semicircle goes to 0 as the radius goes to infinity and similarly with the arc around 0 as the radius goes to 0, but there's a little trouble with the one around $-a$.
  • If $-a$ were a simple pole, then it would just evaluate to $-\pi\mathrm{i}$ times the residue, but I have an order $3$ pole.

Any suggestions ?. I was also thinking of rotating the standard contour by $\pi/2$ so that the pole is inside the contour rather than outside.

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  • $\begingroup$ @Vlad I understand that your edits to the title make it look better and improves readability, but your edits actually take up too much front page space. $\endgroup$ – Irregular User Jun 30 '16 at 2:02
  • $\begingroup$ Why not integrate by part? Aside from dealing with the two limits carefully, the integral is elementary. $\endgroup$ – achille hui Jun 30 '16 at 2:04
  • $\begingroup$ @IrregularUser Sorry, I assumed we accidentally edited it simultaneously. $\endgroup$ – Vlad Jun 30 '16 at 2:05
  • $\begingroup$ @Vlad No problem, nothing to worry about :) $\endgroup$ – Irregular User Jun 30 '16 at 2:06
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    $\begingroup$ @Hoang-NganNguyen. I suspect that the result is $\frac{ \log(a)-1}{2a^2}$ provided that $a>0$. $\endgroup$ – Claude Leibovici Jun 30 '16 at 3:01
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Suppose we seek to evaluate

$$K = \int_0^\infty \frac{\log x}{(x+a)^3} \; dx$$

with $a$ positive.

We use $$f(z) = \frac{(\log z)^2}{(z+a)^3}$$

and integrate around a keyhole contour with the branch cut of the logarithm on the positive real axis and the argument between zero and $2\pi.$

We obtain

$$2\pi i \mathrm{Res}_{z=-a} f(z) \\ = \int_0^\infty \frac{(\log z)^2}{(z+a)^3} dz - \int_0^\infty \frac{(\log z +2\pi i)^2}{(z+a)^3} dz \\ = -4\pi i \int_0^\infty \frac{\log z}{(z+a)^3} dz + 4\pi^2 \int_0^\infty \frac{1}{(z+a)^3} dz.$$

The simple one yields

$$\left[-\frac{1}{2} \frac{1}{(x+a)^2}\right]_0^\infty = \frac{1}{2a^2}.$$

Now for the residue we get

$$\frac{1}{2}\left.\left((\log z)^2\right)''\right|_{z=-a} = \left.\left(\frac{1}{z} \log z\right)'\right|_{z=-a} = \left.\left(\frac{1}{z^2} - \frac{1}{z^2} \log z\right)\right|_{z=-a} \\ = \frac{1}{a^2} (1-\log a - i\pi).$$

We thus have

$$ \frac{2\pi i}{a^2} (1-\log a - i\pi) = -4\pi i K + \frac{2\pi^2}{a^2}$$

or

$$ \frac{2\pi i}{a^2} (1-\log a) = -4\pi i K $$

and finally

$$K = \frac{1}{2a^2} (\log a - 1).$$

This can also be obtained comparing real and imaginary parts of the initial integral of $f(z).$

Remark. To be rigorous we need to apply ML to the circular components. We get for the large circle that with $z=R\exp(i\theta)$ $$|\log^2 z| = |\log z|^2 = \sqrt{\log^2 R + \theta^2}^2$$

and hence

$$\lim_{R\rightarrow\infty} 2\pi R \times \frac{\log^2 R + 4\pi^2}{(R-a)^3} = 0$$

and

$$\lim_{\epsilon\rightarrow 0} 2\pi \epsilon \times \frac{\log^2 \epsilon + 4\pi^2}{(a-\epsilon)^3} = 0.$$

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  • $\begingroup$ Thanks, this method seemed to work much nicer than some of the other ones. Would you say moving up a power of $\log$ only helps when you have a real pole? If the pole were not real, then we could take an arc from $R$ to $-R$ and an arc around 0, then look at a the integral on the positive/negative axis and solve for the positive one. Since the pole isn't real, no extra arc around it would be necessary. $\endgroup$ – Curious Jul 1 '16 at 21:32
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    $\begingroup$ There is no arc around the pole, just a plain keyhole contour aligned with the positive real axis and the pole inside the keyhole in the left half plane. $\endgroup$ – Marko Riedel Jul 1 '16 at 21:50
  • $\begingroup$ I know, I think I just phrased my question/explanation poorly. I think it just goes back to missing the intuition behind you knew to do the keyhole contour versus, say, a semi circle in the half plane with a semicircle on the negative real axis around $-a$ and a semicircle around 0. $\endgroup$ – Curious Jul 2 '16 at 9:21

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