# Improper Integral Residue Theorem

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.
Also:
$$\int_{Cr} \frac{1}{(1+z^2)^3} dz = 0$$
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.
Thanks.

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, $$\dfrac{\varphi''(i)}2=-\dfrac{3i}{16}$$.

• 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? – jd94 Dec 13 '18 at 17:39
• 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. – José Carlos Santos Dec 13 '18 at 17:51
• 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? – jd94 Dec 13 '18 at 17:56
• I would not put it like that, but that's the right idea. – 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.

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

and

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

in this case let $$a = i$$

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