Prove that: $e^{-\frac{|t|}{2}}= \int_{\mathbb R} \frac{e^{ixt}}{1+x^2} dx.$

Question

I need help for prove that $$e^{-\frac{|t|}{2}}= \int_{\mathbb R} \frac{e^{ixt}}{1+x^2} dx, \quad \forall t\in \mathbb R.$$ Thank you in advance

Can you prove that the value of the integral is real? (Hint: apply $e^{ix} = \cos x + i\sin x$.) — Jalex Stark, Mar 28 '18 at 18:22
@qbert Can you link to the previous iteration? That makes it easier to vote to close. — Jalex Stark, Mar 28 '18 at 18:24

score 1 · Answer 1 · edited Mar 28 '18 at 20:29

1

Hint:

If $t\geq 0$, examine $$ \int_{\Gamma_R} \frac{e^{izt}}{1+z^2}\mathrm dz $$ using the usual contour $\Gamma_R$ (semicircle union the real axis) in the upper half plane.

If $t<0$, examine $$ \int_{\Gamma_R'} \frac{e^{izt}}{1+z^2}\mathrm dz $$ where now $\Gamma_R'$ is the contour $\Gamma_R$ reflected across the real axis.

This will insure that the contribution of the arc is 0 in either case.

edited Mar 28 '18 at 20:29

amWhy

209,954

answered Mar 28 '18 at 18:27

operatorerror

29,103

"This has been asked before", you noted. Then you answer anyway. – amWhy Mar 28 '18 at 18:30
@amWhy I could not find the previous iteration and offered a hint on the chance that OP wants a hint, rather than a fully worked answer. If you feel strongly, I would be happy to delete it. – operatorerror Mar 28 '18 at 18:31

Prove that: $e^{-\frac{|t|}{2}}= \int_{\mathbb R} \frac{e^{ixt}}{1+x^2} dx.$

1 Answers1