# 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.

Now, there are things called "tempered distributions", which do have Fourier transforms "in the sense of distributions", even thought they're not necessarily integrable functions. Exactly what that means is a story too long for a MSE post, but the Fourier trnasform in that sense is not obtained by evaluating an integral. (See for example the chapter on distributions and Fourier transforms in Folland Real Analysis. Note the theory of tempered distributions gives the mathematical justification (orcorrect "interpretation") for various statements in the Wikipedia article referenced in the other answer...)

But your $f$ is not a tempered distribution either. It does not have a Fourier transform in any standard sense that I know of.

Possibly the person who asked you to find this Fourier transform was confused. Or possibly there was a typo or an omission in your statement of the problem.

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...