I'm having trouble finding the Fourier transform of $g(t) = \cos^2{a x}$.

I know the answer has to be a sum of $3$ dirac delta functions, but I'm having trouble showing this. I'll show you where my work ran into a problem.

First, I used the identity $\cos^2 (x) = 1/2 \ (1 + \cos (2x))$. Then, I have:

$$\hat g(k) = 1/4 \pi \ \int_{-\infty}^{\infty} (1 + \cos \ (2ax)) e^{-ikx} dx $$

I can handle the $\cos \ (2ax)e^{-ikx}$ integral fine, and make it a dirac delta function, but I don't see how I can with a simple integration of $\int_{-\infty}^{\infty} e^{-ikx} dx$. To my knowledge, that's an indefinite integral equal to $ \frac{1}{-ik} e^{-ikx} + C$. That is not a dirac delta function. Just for context, the answer should be:

$1/4 \left( \delta (2a-k) + \delta (2a +k) + 2\delta (k)\right)$

But I don't see how this is possible. Can someone point out what I'm doing wrong?

Also, this is using a convention for the Fourier transform that I am forced to use in university. If I used a $e^{ikx}$ convention, this would be more straightforward.

  • $\begingroup$ you need to use two theorems: Duality and the bijective nature of the Fourier Transform $\endgroup$ Oct 17, 2018 at 20:28
  • $\begingroup$ @robertbristow-johnson What of the Duality nature of bijective nature of the Fourier transform allows me to understand how to solve this? I've attempted looking at them and it doesn't illuminate anything to me. $\endgroup$
    – sangstar
    Oct 17, 2018 at 20:52
  • $\begingroup$ i'm an electrical engineer and our notation regarding the Fourier Transform is a little different than the convention that mathematicians use. lemme try to put this in an answer. $\endgroup$ Oct 17, 2018 at 21:04
  • 1
    $\begingroup$ but it's easier to show what the F.T. of $\delta(x)$ is than show what the inverse F.T. of $e^{i k x}$ is. $\endgroup$ Oct 17, 2018 at 21:09

2 Answers 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, $$ \int_{-\infty}^{+\infty} e^{-ikx} dx = \lim_{\epsilon\to0^+}\int_{-\infty}^{+\infty} e^{-ikx -\epsilon|x|} dx= 2\pi \delta(k)\,. $$


The simple way of how engineers do it:

$$ \cos^2(t) = \left[ \frac{e^{jt} + e^{-jt}}{2} \right]^2 = \frac{1}{4} e^{j2t} + \frac{1}{2} + \frac{1}{4} e^{-j2t} $$ We also know that: $$ e^{2jt} \Longleftrightarrow 2 \pi \delta(\omega -2) \\ 1 \Longleftrightarrow \pi \delta(\omega) \\ e^{-j2t} \Longleftrightarrow 2 \pi \delta (\omega + 2) $$ After making use of the linearity of the Fourier transform: $$ \cos^2(t) \Longleftrightarrow \frac{1}{2} \pi \delta(\omega -2) + \pi \delta(\omega) + \frac{1}{2} \pi \delta (\omega + 2) $$

  • $\begingroup$ Tip: use \cos, not just cos. $\endgroup$
    – K.defaoite
    Sep 21, 2022 at 11:41

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