what is the Fourier cosine transform of $e^{-ax}$

Question

What is the Fourier cosine transform of $e^{-ax}$

I got

$$ \int_{0}^{\infty}\cos(kx)e^{-ax}dx = \frac{e^{-ax}(k\sin(kx) -\cos(kx))} {a^{2}+k^{2}}\Bigr|_{0}^{\infty} $$

But how do you continue from here?

Answer 1

Observe that if $a>0$ then $$ \lim_{x\to+\infty}e^{-ax}(k\sin(kx) -\cos(kx))=0. $$

Answer 2

I think it makes sense to chose $f(x) = e^{-a|x|}$ and $a>0$ otherwise the integral does not make any sense.

$$\mathcal{F}_{\cos}(f)(k) =\int_{\Bbb R}\cos(kx)e^{-a|x|}dx \\= 2\int_{\Bbb R}\cos(kx)e^{-ax}dx =2\frac{e^{-ax}(k\sin(kx) -\cos(kx))} {a^{2}+k^{2}}\Bigr|_{0}^{\infty} =\frac{1}{a^2+k^2}$$ Given that $$\lim_{x\to+\infty}e^{-ax}(k\sin(kx) -\cos(kx))=0.$$