1
$\begingroup$

Evaluate for $t\in \mathbb{R}$ $$\int_{-\infty}^\infty{e^{itx} \over (1+x^2)^2}dx.$$

Here is what I have done: Let $f(z)={e^{itz}\over (1+z^2)^2}$. This has two poles $z=i$ $z=-i$ and an essential singularity at $z=- \infty$. When $t\leq 1$ I think the integral becomes ${1\over 2}\pi (1-t)e^t$ and when $t\leq 0$ I think the integral becomes ${1\over 2}\pi(1+|t|)e^{-|t|}$. I'm not sure how to show this. Any solutions or help is greatly appreciated.

$\endgroup$
1
$\begingroup$

Hint: Use contour $\Gamma$ consisting of the upper semicircle $C_\rho^+ : y = \sqrt{\rho^2 - x^2}$ and the line $\gamma$ joining $(-\rho, 0)$ to $(\rho, 0)$. Apply Jordan's lemma to see that the integral vanishes over $C_\rho^+$ (or do this yourself using the M-L estimate).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.