I'm stuck on a question involving evaluating improper integrals using the residue theorem.
Here's what I'm trying to evaluate:
$\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^3} dx$

To begin with, we can define a contour integral over the the semi-circle disc above the Real axis called $C$ which can be broken up in two parts involving the Real number line, and the curve of the semi-circular disc
$\int_{C} \frac{1}{(1+z^2)^3} dz = \int_{-R}^{R} \frac{1}{(1+x^2)^3} dx + \int_{Cr} \frac{1}{(1+z^2)^3} dz$

We can see that the left-hand side of the equation has poles at $i$ and $-i$ with multiplicity $3$ respectively. Using the Residue Theorem
$\int_{C} \frac{1}{(1+z^2)^3} dz = 2 \pi i (3 \frac{1}{((i)+i)^3})$

Because there are $3$ instances of the $i$ being the pole inside the region defined.
$\int_{Cr} \frac{1}{(1+z^2)^3} dz = 0$
Because parameterizing $z = Re^{it}$ and setting the limit as R goes to infinity will show the integrand to be zero.

That leaves us with the final answer of
$-\frac{6 \pi}{8}$

But this is not correct.
The correct answer is $\frac{3 \pi}{8}$ and I'm not quite sure how that is formed.

Any help would be appreciated.


The residue of $\dfrac1{(1+z^2)^3}$ at $i$ is $-\dfrac{3i}{16}$ and therefore your integral is equal to$$2\pi i\times\frac{-3i}{16}=\frac{3\pi}8$$indeed.

In order to compute that residue, you can use the fact that$$\dfrac1{(1+z^2)^3}=\frac{\frac1{(z+i)^3}}{(z-i)^3}$$and that therefore the residue that we are interested in is $\dfrac{\varphi''(i)}2$, where $\varphi(z)=\dfrac1{(z+i)^3}$. And, as I wrote, $\dfrac{\varphi''(i)}2=-\dfrac{3i}{16}$.

  • $\begingroup$ Why can't we find the residue by multiplying the original function by $(z-1)^3$, isn't that how you find a residue usually? $\endgroup$ – jd94 Dec 13 '18 at 17:39
  • $\begingroup$ Why $(z-1)^3$? The number $1$ has nothing to do with this problem. It should be $(z-i)^3$, and that is exactly what I did. $\endgroup$ – José Carlos Santos Dec 13 '18 at 17:51
  • $\begingroup$ Typo on my end. So the idea is, if we have a higher order pole, we differentiate it until the order of it is 1, and divide by how many times we differentiated factorial? $\endgroup$ – jd94 Dec 13 '18 at 17:56
  • $\begingroup$ I would not put it like that, but that's the right idea. $\endgroup$ – José Carlos Santos Dec 13 '18 at 18:08

By the Cauchy integral formula

if $a$ is inside the contour, and $f(z)$ is holomorphic inside the contour.

$\int_\gamma \frac {f(z)}{z-a} dz = 2\pi i f(a)$


$\int_\gamma \frac {f(z)}{(z-a)^n} dz = 2\pi i \frac {1}{(n-1)!} f^{(n-1)}(a)$

in this case let $a = i$

and $f(z) = \frac {1}{(z+i)^3}$


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