# Fourier transform of $e^{-t}\cos(t)$

I'm trying to find the Fourier transform of $f(t) = e^{-t}\cos(t)$ using the following definition:

$$\mathcal{F}(x) = \mathcal{F}[f(t)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ixt}dt$$

however after substituting in $f(t)$ into the integral and attempting to integrate by parts, I get stuck.

Any tips on how to proceed from here? Moreover, I was told the final answer would involve terms containing the Dirac Delta function, and I don't quite understand how this would result from evaluating the integral.

• The integral does not converge for any value of $x$. Nov 24, 2017 at 16:00

Edit What I said in the first version of this answer is not true. (I said $f$ is a tempered diistribution. It's not.)
That integral is the definition of the Fourier transform of a (Lebesgue) integrable function $f$. Your function is not integrable, so it simply does not have a Fourier transform in this sense.
But your $f$ is not a tempered distribution either. It does not have a Fourier transform in any standard sense that I know of.
Details: Of course the function $e^t\cos(t^{10})$ does define a tempered distribution, because of cancellation. A clean way to see there's not enough cancellation in $f(t)=e^t\cos(t)$: Suppose it is a tempered distribution. Since multiplication by $\cos(t)$ maps the Schwarz space to itself it follows that $e^t\cos^2(t)$ is a tempered distribution. And now so is $e^t\sin^2(t)$, being a multiple of a translate of $e^t\cos^2(t)$. Hence $e^t$ is a tempered distribution...