In the language of distributions, the Dirac delta distribution is the map $\delta$ from the space of test functions (smooth compactly supported functions) to, say $\mathbb{R}$ with the "operation" $(\delta, f) = f(0)$ for every test function $f$.

To figure out the Fourier transform of a distribution, you need to determine the Fourier transform of a test function $f$.

$$\widehat{f}(\xi) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-ix \xi}f(x) \, dx$$

By definition, the Fourier transform of a distribution $\varphi$ is defined by $(\widehat{\varphi},f)=(\varphi,\widehat{f})$ for every test function $f$.

EDIT: As commenters below pointed out, I should say Schwartz function instead of test function and tempered distribution instead of distribution..

Therefore

$$(\widehat{\delta},f) = (\delta,\widehat{f}) = \widehat{f}(0) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} e^{-i x \cdot 0} f(x) \, dx =\frac{1}{\sqrt{2\pi}}\int_\mathbb{R} f(x) \, dx = (\frac{1}{\sqrt{2\pi}},f)$$

where the last equality is because the "constant" distribution 1 is regular, i.e., can be represented in integral form. Therefore, as a distribution, $\widehat{\delta} = (2\pi)^{-1/2}$.

The problem you encounter essentially boils down to proving
$$
\int_{-\infty}^{+\infty}e^{-ikx}dx = 2\pi \delta(k)\,.
$$
There are many ways to prove this fact. For instance, one can first prove that the Fourier transform extends in an invertible way to tempered distribution (to which $\delta(x)$ belongs), then note that
$$
\int_{-\infty}^{+\infty} e^{ikx}\delta(k)\,dk = 1\,,
$$
and finally apply the inverse Fourier transform to obtain the desired identity. Another way I like is the following (non-formal) approach based on a regularization of the integral: for $\epsilon>0$,
$$
\int_{0}^{+\infty} e^{-(\epsilon+i k)x} dx = \frac{1}{\epsilon+ik}\xrightarrow[\epsilon\to 0^+]{} -i \,\mathrm{PV}\frac{1}{x}+\pi \delta(k)\,,
$$
where PV denotes the principal value, while
$$
\int_{-\infty}^0 e^{(\epsilon-ik)x} dx = \frac{1}{\epsilon-ik}\xrightarrow[\epsilon\to0^+]{}+i\,\mathrm{PV}\frac{1}{x}+\pi \delta(k)\,.
$$
Hence,
$$
\int_{-\infty}^{+\infty} e^{-ikx} dx = \lim_{\epsilon\to0^+}\int_{-\infty}^{+\infty} e^{-ikx -\epsilon|x|} dx= 2\pi \delta(k)\,.
$$

## Best Answer

EditWhat 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

notobtained by evaluating an integral. (See for example the chapter on distributions and Fourier transforms in FollandReal 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

havea 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 itisa 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...