# Consider the integral $I=\int_{-\infty}^{\infty} \frac{1}{x^{2}+1}$

Consider the integral $I=\int_{-\infty}^{\infty} \frac{1}{x^{2}+1}\, dx$. Show how to evaluate this integral by considering $\oint_{C_{(R)}} \frac{1}{z^{2}+1}, dz$ where $C_{R}$ is the closed semicircle in the upper half plane with endpoints at $(-R, 0)$ and $(R, 0)$ plus the $x$ axis.

Could someone help me through this problem?

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While it's certainly an improvement in politeness, simply adding "Could someone help me through this problem?" to your assignment (as you've done for your last several questions) is still not enough. Don't you try these questions yourself? If you have any thoughts whatsoever - even if they are just "I have no idea what the question is asking" - please say what they are. Currently, all I get is the impression that you have no interest in doing your own work, and are now simply being more polite about asking others to do it for you. – Zev Chonoles Apr 26 '12 at 22:58
I'm rusty on these problems, but you want to show, for any fixed $R$, the integral is nonzero only on $[-R,R]$. Taking the limit of this expression as $R \rightarrow \infty$ gives the value of the integral on $[-\infty, \infty]$. – Austin Mohr Apr 26 '12 at 22:58

Let $f(z)=\frac{1}{z^2+1}$ and the contour $C$ be the closed half circle of radius R as described above. Let $C_R$ be the arc of the half circle $C$ with radius $R$. Using residue theorem, we have

$$\oint_{C}f(z) dz=\int_{-\infty}^{\infty}f(x)dx+\int_{C_R}f(z) dz=2\pi i\sum\text{Residues} \Rightarrow \int_{-\infty}^{\infty}f(x)dx = 2\pi i\sum\text{Residues}-\int_{C_R}f(z) dz$$

We start by proving

$$\lim_{R \to \infty}\int_{C_R}f(z) dz=0$$

We let M be the maximum value of the function on the contour, and L be the length of the contour. $$\left|\int_{C_R}\frac{dz}{z^2+1}\right| \leq ML$$

By inspection, $L = \pi R$. We proceed to find M:

$$M=\left|\frac{1}{z^2+1}\right|=\frac{1}{|z^2+1|}=\frac{1}{|z|^2+1}=\frac{1}{R^2+1}$$

Thus

$$\lim_{R \to \infty}\left|\int_{C_R}\frac{dz}{z^2+1}\right| \leq \lim_{R \to \infty} ML=\lim_{R \to \infty} \frac{\pi R}{R^2+1}=0$$

So this integral becomes zero. Now we have

$$\int_{-\infty}^{\infty}\frac{dx}{x^2+1} = 2\pi i\sum\text{Residues}$$

so we must find the required residues. To compute the residues, we expand $f(z)$ and find that there are singularities when $z= \pm i$

$$f(z)=\frac{1}{(z+i)(z-i)}$$

The singularity $z=i$ is the only one within the contour $C$, so this is the only residue we need to compute. To compute this residue (let it be called $b$), we do the following:

$$b=\lim_{z \to i} (z-i)f(z) =\lim_{z \to i} \frac{(z-i)}{(z+i)(z-i)}=\frac{1}{(i+i)}=\frac{1}{2i}$$

This is the only residue that needs to be computed, thus we can find the value of the integral:

$$\int_{-\infty}^{\infty}\frac{dx}{x^2+1} = 2\pi i b = \frac{2 \pi i}{2 i}=\pi$$

So the integral is found to equal $\pi$.

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Denote the semicircle part $A$ and the $(-R,R)$ part $B$ (these are contours that make up $C$).

• Using the Residue theorem, what is $\oint_{C(R)}\frac{dz}{z^2+1}$?
• Can you find an upper bound for $\int_{A(R)}\frac{dz}{z^2+1}$ in terms of $R$? First bound the integrand.
• Notice the fact that $$\oint_{C(R)}=\int_{A(R)}+\int_{B(R)}.$$ You're interested in $\int_{B(R)}$ as $R\to\infty$. Can you use the previous two facts and take a limit?
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