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so I'm trying to evaluate the integral in the title,

$$\int_{-\infty}^{\infty}\frac{1}{x^2}dx$$

by using complex plane integration. I've chosen my contour to be a infinte half circle with it's diameter on the real axis. (integration is preformed ccw).

when R tends to infinity, the arch part of the contour yields zero, and so we are left with the part along the real axis, which is the one I'm trying to evalute.

there are no other poles in my contour, only a second order pole at $z=0$ lying on it. the residue of this pole is $0$ so the integral sums up to be zero (by using Cauchy principal value.)

However my function is always positive and greater than $0$, so this doesn't make sense.

Any help would be appreciated

Somos
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user635635
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    Residue Theorem is not applicable when there is a pole on the contour. – Kavi Rama Murthy Jan 17 '19 at 11:49
  • I took an infinitely small half cirlce around that point, so it only contributes half of what is did if it was inside my contour so it contributes i\piRes(z=0). but since the Res is 0 it contributes nothing. – user635635 Jan 17 '19 at 11:58
  • When there are serious questions about the very existence of the integral you are assuming that it exists and the integral over a line slightly below the real axis tend to integral over the real line. This is way off any rigorous argument. – Kavi Rama Murthy Jan 17 '19 at 12:07
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    Note this is a reply to a comment below - I can't put the reply in its proper place because that answer is deleted: Yes, the integral over the small circle is $0$. But the integral over half of the small circle is not half the integral over the small circle. – David C. Ullrich Jan 17 '19 at 15:16

2 Answers2

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The integral diverges even if you chose the principal value. To show this by using residue theorem, first chose a contour C composed by the following: $$C_R: \text{ big semicircle centered at origin, with radius $R$, counterclockwise}$$ $$C_1: \text{ straight line from $-R$ to $-r$, towards $+\infty$}$$ $$C_r: \text{ small semicircle centered at origin, with radius $r$, clockwise}$$ $$C_2: \text{ straight line from $R$ to $r$, towards $+\infty$}$$ $$\int_C\frac{dz}{z^2}=0$$ by residue theorem. $$\int_{-\infty}^\infty\frac{dx}{x^2}=0-\lim_{R\to\infty}\int_{C_R}\frac{dz}{z^2}-\lim_{r\to0}\int_{C_r}\frac{dz}{z^2}\\ =0-0-(-\infty)=\infty$$ The calculation of the limits is left to readers.

Kemono Chen
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  • perhaps i should clarify that I'm doing this in order to find the integral over real line of (\sinc{x})^2 which is \pi. Works fine if I take the integral in the title to be zero, but as you said, I believe it should converge. – user635635 Jan 17 '19 at 12:06
  • @user635635, the integral in your title must diverge I promise. I think, your calculation may be wrong. – Kemono Chen Jan 17 '19 at 12:08
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Usually you'd need this kind of contour if you happen to have poles on the real axis. We take the small radius $r\to 0$ and the large radius $R\to\infty$. The value on the bigger half circle should decay fast enough so the value of the integral would depend on the limit of the smaller half circle.

You can see this example of a similar calculation.

fig 1

BigbearZzz
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  • You should probably mention explicitly that if $f==1/z^2$ then the integral over the smaller half-circle does not tend to $0$ as $r\to0$. – David C. Ullrich Jan 17 '19 at 15:13