Contour integration

The residue theorem, invented by Cauchy, is a method of evaluating definite integrals.

Consider the following contour in the complex plane :

where R is a limit going to infinity...or something

Let's find

$\int \limits _{-\infty }^{\infty }{\frac {1}{x^{2}+1}}\mathrm {d} x$

the first half of the contour is a straight line that goes from -R to R

hence, this can be represented as, (H is the entire contour, because i said so)

$\oint \limits _{H}{\frac {1}{x^{2}+1}}\mathrm {d} x=\int \limits _{\gamma _{R}}{\frac {1}{x^{2}+1}}\mathrm {d} x+\int \limits _{-R}^{R}{\frac {1}{x^{2}+1}}\mathrm {d} x$

the other half of the contour, $\gamma _{R}$, can be represented as the equation : $x=Re^{iv}$ where v is a variable that goes from $\pi$to 0

differentating, we have

$\mathrm {d} x=Rie^{iv}\mathrm {d} v$

plugging it back into our original integrals

$\oint \limits _{H}{\frac {1}{x^{2}+1}}\mathrm {d} x=\int \limits _{\pi }^{0}{\frac {Rie^{iv}\mathrm {d} v}{(Re^{iv})^{2}+1}}+\int \limits _{-R}^{R}{\frac {1}{x^{2}+1}}\mathrm {d} x$

if we set :

$P=\int \limits _{\pi }^{0}{\frac {Rie^{iv}\mathrm {d} v}{R^{2}e^{2iv}+1}}$

Now take the absolute value

$|P|=|\int \limits _{\pi }^{0}{\frac {Rie^{iv}}{R^{2}e^{2iv}+1}}\mathrm {d} v|$

from the triangle inequality for integrals :

$|P|\leq \int \limits _{\pi }^{0}|{\frac {Rie^{iv}}{R^{2}e^{2iv}+1}}|\mathrm {d} v$

$|P|\leq \int \limits _{\pi }^{0}{\frac {R|e^{iv}|}{|R^{2}e^{2iv}+1|}}\mathrm {d} v$

It's known that if z is real, then

$|e^{iz}|=1$

thus :

$|P|\leq \int \limits _{\pi }^{0}{\frac {R}{|R^{2}e^{2iv}+1|}}\mathrm {d} v$

Using the triangle inequality :

$|P|\leq \int \limits _{\pi }^{0}{\frac {R}{|R^{2}e^{2iv}|+|1|}}\mathrm {d} v$

$|P|\leq \int \limits _{\pi }^{0}{\frac {R}{|R^{2}|+|1|}}\mathrm {d} v$

Since there is no more v, we can move the thing out of the integral, or so i think

$|P|\leq {\frac {-R\pi }{R^{2}+1}}$

If we set a limit with R go to infinity :

$|P|\leq \lim _{R\to \infty }{\frac {-R\pi }{R^{2}+1}}$

$|P|\leq \lim _{R\to \infty }{\frac {-\pi }{R+1/R}}$

letting the limit go, we get

$|P|\leq 0$

Since the absolute value of something is always greater than 0 unless its 0, then P=0, this gives us

$\oint \limits _{H}{\frac {1}{x^{2}+1}}\mathrm {d} x=\int \limits _{-R}^{R}{\frac {1}{x^{2}+1}}\mathrm {d} x$

We can let R go to infinity

$\oint \limits _{H}{\frac {1}{x^{2}+1}}\mathrm {d} x=\int \limits _{-\infty }^{\infty }{\frac {1}{x^{2}+1}}\mathrm {d} x$

inside 1/(x^2+1), there is only 2 poles, where x=-i, and where x=i, inside our contour, there is only one pole, x=i, hence, using the residue theorem:

$2\pi i{\underset {x=i}{Res}}{\frac {1}{x^{2}+1}}=\int \limits _{-\infty }^{\infty }{\frac {1}{x^{2}+1}}\mathrm {d} x$

$2\pi i\lim _{x\to i}{\frac {(x-i)}{x^{2}+1}}=\int \limits _{-\infty }^{\infty }{\frac {1}{x^{2}+1}}\mathrm {d} x$

x^2+1=(x-i)(x+i), hence :

$2\pi i\lim _{x\to i}{\frac {1}{x+i}}=\int \limits _{-\infty }^{\infty }{\frac {1}{x^{2}+1}}\mathrm {d} x$

Setting the limit

$\pi =\int \limits _{-\infty }^{\infty }{\frac {1}{x^{2}+1}}\mathrm {d} x$