Integrate this!

We'll evaluate $\int_0^{\infty}\frac{\cos(3x)}{x^2+1}dx$ using complex integration.

First, we'll evaluate the related integral $\int_C\frac{e^{3iz}}{1+z^2}dz$ where $C$ is the loop in the complex pane consisting of the real interval $[-R,R]$ and the semicircle $\gamma$ lying above $[-R,R]$ , for some large real number $R$ .

We have $\int_C\frac{e^{3iz}}{1+z^2}dz=\int_{-R}^{R}\frac{e^{3ix}}{1+x^2}dx+\int_{\gamma}\frac{e^{3iz}}{1+z^2}dz. (*)$

Note $\frac{e^{3iz}}{1+z^2}=\frac{e^{3iz}}{(z-i)(z+i)}$ has two simple poles located at $i,-i$ . Only $i$ lies in the region enclosed by $C$ . By Cauchy's integration formula, $\int_C\frac{e^{3iz}}{1+z^2}dz=2\pi i\cdot \frac{e^{3i\cdot i}}{i+i}=\frac{\pi}{e^3}. (**)$

We find an upper bound on the integral over $\gamma$ by paramaterizing $z=Re^{it}$ , $t\in[0,\pi]$ :

$\displaystyle\bigg|\int_{\gamma}\frac{e^{3iz}}{1+z^2}dz\bigg|=\bigg|\int_{0}^{\pi}\frac{e^{3iRe^{it}}Rie^{it}}{1+R^2e^{2it}}dt\bigg|\le\int_{0}^{\pi}\bigg|\frac{e^{3iRe^{it}}Rie^{it}}{1+R^2e^{2it}}\bigg|dt$

$\displaystyle=\int_{0}^{\pi}\frac{|Re^{3iR(\cos(t)+i\sin(t))}|}{R^2-1}dt=\int_{0}^{\pi}\frac{Re^{-3R\sin(t)}}{R^2-1}$

$\le\int_{0}^{\pi}\frac{R}{R^2-1}dt=\frac{\pi R}{R^2-1}\to 0\text{ as }R\to\infty.$

Thus, as $R\to\infty$ , the integral over $\gamma$ goes to $0$ . $(***)$

Then letting $R\to\infty$ , we find from $(*),(**),$ and $(***)$ :

$\displaystyle\frac{\pi}{e^3}=\int_{-\infty}^{\infty}\frac{e^{3ix}}{1+x^2}dx$

$\displaystyle\Re\left(\frac{\pi}{e^3}\right)=\Re\left(\int_{-\infty}^{-{\infty}}\frac{e^{3ix}}{1+x^2}dx\right)$

$\displaystyle\frac{\pi}{e^3}=\int_{-\infty}^{\infty}\frac{\cos(3x)}{1+x^2}dx.$

Then since $\frac{\cos(3x)}{1+x^2}$ is an even function, we are left with

$\displaystyle\int_{0}^{\infty}\frac{\cos(3x)}{1+x^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(3x)}{1+x^2}dx=\frac{\pi}{2e^3}.$

We use a calculator to find $10^4\frac{\pi}{2e^3}\approx 782.1,$ so the answer is $\boxed{782}$ .

Instruction not followed.

Int(10000*7.8205344E-02) = 782