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The question below is really bugging me ...Given that the following function,

$$f(x) = \Theta(x + 1) - \Theta(x - 1)\,,$$

(where $\Theta(x)$ is the Heaviside step function) has a Fourier transform:

$$\widetilde{f}(k) = 2\frac{\sin k}{k}$$

where in this case we define the Fourier transform as

$$\widetilde{f}(k) = \int_{-\infty}^{\infty} f(x) e^{-ikx}\,dx\,,$$

evaluate the following integral:

$$\int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx\,.$$


So far, I've considered that the Fourier transform of $c(x) = f(x) \ast g(x)$ (with '$\ast$' denoting convolution) is equal to $\widetilde{c}(k) = \widetilde{f}(k) \widetilde{g}(k)$ by the convolution theorem, hence if $g(x) = f(x)$, this gives a starting point. I have so far failed to manipulate any integrals into the correct form, though!

I feel like the solution is much simpler than I am imagining it to be ...

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You have the inverse Fourier transform $$\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} \tilde{f}(k)^2 e^{ikx} dk = \mathcal{F}^{-1}[\tilde{f}(k)^2] = f(x) * f(x) = \int\limits_{-\infty}^{+\infty} f(y)f(x-y) dy$$ where the right-hand side is the convolution product. Setting $x=0$ you get $$\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty} \tilde{f}(k)^2 dk = \int\limits_{-\infty}^{+\infty} f(y)f(-y) dy = 2$$ where the evaluation is obtained by explicitly computing the integral of products of theta functions. Then, simpligying gives $$4 \pi = \int\limits_{-\infty}^{+\infty} \tilde{f}(k)^2 dk = 4 \int\limits_{-\infty}^{+\infty}\frac{\sin^2 k}{k^2} dk$$ and therefore $$\int\limits_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2} dx = \pi \, . $$

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  • $\begingroup$ This answer saved my physics degree $\endgroup$ – Archimaredes May 15 '18 at 11:06
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You can solve it using the convolution theorem but it's a little more involved and it seems to me you would have to use Plancharel's theorem at some point anyways so might as well use it directly.

Plancharel's theorem gives,

$$2\pi\|f(x)\|_{L^2}^2 = \|\tilde f(k)\|_{L^2}^2$$

The left hand side is $2\pi\int\limits_{-\infty}^\infty (\Theta(x+1)-\Theta(x-1))^2dx$

The right hand side is $\int\limits_{-\infty}^\infty (2\frac{\sin(k)}{k})^2dk$

So we just calculate the the left hand side and we can determine the integral we want. The left hand side is just 2 (rectangle) and so we have

$$2\pi\int\limits_{-\infty}^\infty \frac{\sin^2(k)}{k^2}dk=2\pi\frac{2}{4}=\pi$$

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  • $\begingroup$ Thank you! However apparently the final value is $\pi$ according to Wolfram Alpha and the other answer $\endgroup$ – Archimaredes May 14 '18 at 21:23
  • $\begingroup$ Yeah I think there is a factor of $2\pi$ that is missing $\endgroup$ – TSF May 14 '18 at 21:24
  • $\begingroup$ There is a problem in the normalization in the definition of the Fourier transform you are using, there should be an additional factor of $2 \pi$. $\endgroup$ – Antoine May 14 '18 at 21:24
  • $\begingroup$ Yeah I use the symmetric definition and I just noticed that the OP used $e^{-ikx}$ rather than $e^{-2\pi i \xi x}$. $\endgroup$ – TSF May 14 '18 at 21:25

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