Evaluating a real integral through a complex integral

Question

Given the integral

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

Evaluate it using the following complex intregral

$$\oint_{C(r)} \frac{1}{z^2+1} \,dz$$

where $C(r)$ is the closed semicircle in the upper half plane with endpoints $(-r,0)$ and $(r,0)$. The following hint is also provided:

$$ \frac{1}{z^2+1}=\frac{-1}{2i}\bigg(\frac{1}{z+i}-\frac{1}{z-i}\bigg)$$

and that you should show the integral along the open semicircle vanishes as $r \rightarrow \infty$

I have begun approaching the problem by breaking up $C(r)$ into $C_1$ and $C_2$ where $C_1$ is the open semicircle and $C_2$ is the line segment from $x=-r$ to $x=r$, giving us

$$ \frac{-1}{2i} \bigg(\int_{C_1}^{} \frac{1}{z+i}-\frac{1}{z-i} dz+\int_{C_2}^{}\frac{1}{z+i}-\frac{1}{z-i} dz\bigg)$$

For the $C_2$ integral, $z=x$, however, I'm unsure how to show the the $C_1$ integral vanishes as $r\rightarrow \infty$. How should I proceed?

Answer 1

For $r>1$, take the following parametrization for $C_1$

$$\gamma:[0,2\pi]\to \Bbb{C},\ \gamma(t)=re^{it}$$ Then

$$\int_{C_1}\frac{dz}{z^2+1}=\int_0^{\pi}\frac{re^{it}}{r^2e^{2it}+1}dt$$

And since $|re^{it}|=r$ and $|r^2e^{2it}+1|\ge |r^2e^{it}|-|1|=|r^2|-1=r^2-1$ $$\int_0^{\pi}\frac{re^{it}}{r^2e^{2it}+1}dt≤\frac{r\pi}{r^2-1}$$

which goes to $0$ as $r\to \infty$.

Answer 2

$$\int_{-\infty}^{+\infty} \frac{1}{x^2+2}\ \text{d}x$$

we stay in the upper half plane region.

we denote $$f(z) = \frac{1}{z^2+1}$$

and we find the singularities:

$$z_1 = -i ~~~~~~~~~~~ z_2 = +i$$

The singularity $$z_1 = -i$$

will be skipped because it's not in our region.

The singularity $$z_2 = +i$$

is instead the region (we are always speaking about the upper half complex plane, this is why) and so we will calculate the residue as

$$\text{Res}(i, f(z)) = -\frac{1}{2}i$$

Thence the final sum is

$$2\pi i \sum\text{Res}(z, f(z)) = \pi$$

And so

$$\int_{-\infty}^{+\infty} \frac{1}{x^2+1}\ \text{d}x = \pi$$