Value of $\int_{-\infty}^\infty\frac{\cos x}{1+x^2} \, dx$

Question

I am a bit puzzled by the expression $\displaystyle I=\int_{-\infty}^\infty\frac{\cos x}{1+x^2}\,dx$.

If I try solving it using Cauchy's formula, I arrive to $I=2\pi i \frac{\cos i}{2i} = \pi\cos i$.

EDIT: In detail, I am trying the common trick of calculating the integral over a semi-circumference of radii $r$ and the line from $-r$ to $r$. Since the integral over the semi-circunference goes to zero as r goes to infinity, I can use that to calculate the integral over the real line.

But the result I expected is $\frac{\pi}{e}$, since that is what i get if I first substitute $\cos x$ by $e^{ix}$ in the expression, which should be a legal move since $e^{ix}= \cos x + i \sin x$ and $\sin x$ is an odd function, so its integral from $-\infty$ to $\infty$ is $0$.

What is going on?

You'd like someone to find the error in what you did but you don't bother showing us what you did? — David C. Ullrich, Jul 16 '16 at 15:06
You need to explain what contour (or family of contours) you used and why the integral along the part of the contour not along the real axis vanished in some limit. Because $e^{iz}$ goes to zero as $\Im z \rightarrow\infty$, but $\cos z$ does not. — mjqxxxx, Jul 16 '16 at 15:10
@mjqxxxx totally right. I forgot that $\cos(x)$ is not $\le 1$ in $\mathbb{C}$. — Jsevillamol, Jul 16 '16 at 15:14
You can solve it also using Laplace transform and elementary way — Aman Rajput, Jul 16 '16 at 16:29
Does this answer your question? What is wrong in this calculation $\int_{-\infty}^{\infty}\frac{\cos x}{1+x^2}dx$? — Anne Bauval, Jun 14 '23 at 20:09

score 3 · Accepted Answer · edited Jul 16 '16 at 15:32

3

Yes what you used should be a bit modified as ,

$$ I = \int_{-\infty}^\infty \frac{\cos x}{1+x^2} \, dx = \operatorname{Re} \left(\int_{-\infty}^\infty \frac{e^{ix}}{1+x^2} \, dx\right) $$

Now if you compute the residue you would get the value as $\displaystyle 2\pi i \left(\frac{1}{2ie}\right) = \frac{\pi}{e}$ which is the desired answer using complex analysis.

It is well know example to convert trigo problems in exponential forms so as it vanishes under the semicircular arc.

edited Jul 16 '16 at 15:32

Michael Hardy

1

answered Jul 16 '16 at 15:22

Aditya Narayan Sharma

2,896

It was a duplicate. – Anne Bauval Jun 14 '23 at 20:10

Value of $\int_{-\infty}^\infty\frac{\cos x}{1+x^2} \, dx$

1 Answers1

Linked