# Using residues to evaluate an improper integral

Use residues to evaluate the improper integral \begin{align} \int_{0}^{\infty} \frac{1}{(x^2+1)^2} \, dx \end{align}

First, I said $f(z) = \frac{1}{(z^2+1)^2}$. My only question so far is how do I establish the region $C$ (from the given real limits of $0$ to $\infty$) so I can do countour integration and find residues in $C$?

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## 3 Answers

First recognize that since your integrand is even, you have

$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^2}dx = \int_0^{\infty}\frac{1}{(1+x^2)^2}dx.$$

Then use the residue theorem with a semicircular contour in the upper (or lower) half plane. Of course you will need to argue that the integral along the semicircular arc goes to zero.

If you look at the image below, you'll see that as $a\rightarrow\infty$, you'll be integrating over the whole real line plus a semicircular arc at infinity:

This is the true reason we do the semicircular contour. Let's set up our semicircular contour: $\gamma(t) = [-R,R]\cup\{Re^{it}:0\le t\le\pi\}$. This parameterizes the above contour. Let's integrate over this. Define $I_R$ by

$$I_R = \int_{\gamma} \frac{1}{(1+z^2)^2}dz = \int_{-R}^R\frac{1}{(1+x^2)^2}dx + \int_0^{\pi} \frac{1}{(1+(Re^{it})^2)^2}iRe^{it}dt.$$

As we take $R\rightarrow\infty$, notice that we would get the integral we were interested in to begin with.

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My professor said the same thing about the upper (or lower) half plane. I don't understand why do we know to use the half circle for $C$. (My initial assumption was to use the whole circle for $C$.) –  les guimauves Mar 9 '14 at 23:15
The problem is that you ultimately want your contour to contain your integral in some way (via a limiting process or otherwise). We can't do that with the whole circle. Additionally, the integral around the whole circle would go to zero either because the denominator decays very rapidly or because you include both poles which cancel each other when employing the residue theorem. –  Cameron Williams Mar 9 '14 at 23:20
To recapitulate your response (just so I really get this), using residue theorem yields $\int_C f(z) dz = 2\pi i (B_1 + B_2 + \cdots + B_n)$, where each $B_i$ (for $0 \le i \le n$) is the residue of each singularity point inside $C$. I take it that if we tried a whole circle, with two residues inside it, then $B_1$ in top half of $C$ is positive and $B_2$ in bottom half of $C$ is negative. Hence the integral would be $0$. So this is why we use the half-circle instead. –  les guimauves Mar 9 '14 at 23:28
There's a lot more to it than that. I'll edit my post. –  Cameron Williams Mar 9 '14 at 23:35

In response to @Cameron Williams' hint and comments, I am going to attempt the solution.

We have $f(z) = \frac{1}{(z^2+1)^2}$. Let $C$ be the half circle as described by @Cameron Williams. Now, we have $z = i$ to be the singularity point inside $C$.

In finding the residue, \begin{align} \text{Res}_{z = i} f(z) &= \text{Res}_{z = i} \frac{1}{(z^2+1)^2} = \text{Res}_{z = i} \frac{1}{(z+i)^2(z-i)^2} \\ &= \frac{d}{dz} \frac{1}{(z+i)^2} \Bigg\vert_{z=i} = -\frac{2}{(z+i)^3} \Bigg\vert_{z=i} = \frac{1}{4i} \end{align}

For the horizontal line and half-circle arc, we have $z = x$ and $z=Re^{i \theta}$ respectively. Employing the residue theorem for integrals, we have \begin{align} \int_C f(z) \, dz = \int_{-R}^{R} \frac{1}{(x^2+1)^2} \, dx + \int_{0}^{\pi} \frac{1}{(Re^{i \theta}+1)^2} (iRe^{i \theta} \, d\theta) = 2\pi i \, \text{Res}_{z = i} f(z) = \frac{\pi}{2} \end{align}

Taking the limit as $R \rightarrow \infty$, we get \begin{align} \int_{-\infty}^{\infty} \frac{1}{(x^2+1)^2} \, dx + 0 = \frac{\pi}{2} \end{align} Therefore, \begin{align} \int_{0}^{\infty} \frac{1}{(x^2+1)^2} \, dx = \frac{\pi}{2} \end{align}

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Looks good to me. Though it seems like you had some typos in your LaTeX formatting. Do you understand now why we pick the semicircular contour? –  Cameron Williams Mar 10 '14 at 0:26
Yes, now I understand. The whole objective is to find $\int_{-\infty}^{\infty} \frac{1}{(x^2+1)^2} \, dx$. The half-circle around one singularity point will help us with that; the horizontal portion of the half-circle is what we needed. That horizontal portion is not present in the whole circle to begin with, so the whole circle won't help us at all in the first place. –  les guimauves Mar 10 '14 at 0:30
Great! You got it :) You should have $Res_{z=i}f(z) = \frac{1}{4i}$ I think. –  Cameron Williams Mar 10 '14 at 0:37

Hint. Try $\gamma=\gamma_1\cup\gamma_2$, where $$\gamma_1(t)=t, \,\,\,t\in[-R,R],$$ and $$\gamma_2(t)=R\,\mathrm{e}^{it},\,\,\,t\in[0,\pi],$$ and let $R\to\infty$.

Answer. $\int_0^\infty \frac{dx}{(1+x^2)^2}=\frac{\pi}{4}$.

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