What is wrong in this calculation $\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx$?

Question

$\def\Res{\operatorname{Res}}\def\Re{\operatorname{Re}}$ First solution: Complex function $f(z)=\frac{\cos z}{1+z^2}$ has a pole $z=i$ on the upper complex plane. $\Res (f,i)=\frac{e+1/e}{4i}$, so $$ \int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx=2\pi i\Res (f,i)=\frac{\pi}{2}(e+1/e) $$

However, this integration is in fact $\frac{\pi}{e}$. Here is another solution.

Consider complex function $g(z)=\frac{e^{iz}}{1+z^2}$. $\Res (g,i)=\frac{1}{2ie}$ and $\frac{\cos x}{1+x^2}=\Re (g(x))$ when $x$ is real. $$\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx=\Re (2\pi i\Res (g,i))= \frac{\pi}{e} $$

Why is the first solution wrong?

Answer 1

When you use $\operatorname{Res}$, you need to choose a closed contour. In the second solution the closed contour is (I assume) the real axis and the upper semicirle with radius of $\infty$. Why can we get rid of the integral over the semicircle? Because $\left|\frac{e^{iR}}{1+R^2}\right|\rightarrow\frac{1}{R^2}$ when $|R|\rightarrow\infty$, and the absolute value of the integral over the semicircle can be evaluated to be less than $\frac{1}{R^2}\times 2\pi R$, which go to zero when $R$ go to $\infty$. This happen since $e^{iz}$ is a bounded function over $\mathbb{C}$.

In the first solution, $\cos(z)$ over $\mathbb{C}$ is not a bounded function. you can see that $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ so $\cos(iR)=\frac{e^{R}+e^{-R}}{2}$, and the integral over the infinite semicircle is not zero anymore.