1
$\begingroup$

I couldn't calculate $\int \limits_{-\infty}^{+\infty} \frac{e^{-itx}}{2\pi} \frac{1}{a^2+x^2} dx. $

I can either turn this into something along the lines of $\large \int \limits_0^{\pi/2} \cos( t \cdot \tan x ) dx$ or $ \large \int \limits_0^{+\infty} \frac{\cos tx}{a + ix} dx$ neither of which I can solve.

I've been told, that some tools of complex calculus could simplify this, but my book hasn't covered any before giving the exercise, so I wonder if there is a way without it.

Thanks.

3 Answers 3

0

Hint: you may find the initial Fourier transform in any table.

2

Its better we use duality property of Fourier transform. It says that if $f(t)\Longleftrightarrow F(ix)$ then $F(it)\Longleftrightarrow 2\pi f(-x)$Thus by knowing that $\dfrac{1}{2a}e^{-a|t|}\Longleftrightarrow \dfrac{1}{a^2+x^2}$, by invoking duality, we get $\dfrac{1}{a^2+t^2}\Longleftrightarrow \dfrac{\pi}{a}e^{-a|x|}$Hope this helps.

Thanks...

0

Integrals containing $a^2 + x^2$

http://en.wikipedia.org/wiki/Trigonometric_substitution#Integrals_containing_a2_.E2.88.92_x2