I'm pretty new to Fourier Transforms and I stumbled upon this exercise, which asks me to calculate this integral $$\int_{-\infty}^{+\infty}\left(\frac{\sin(ax)}x\right)^4\,dx$$ The exercise suggests to use fourier transforms, but I really don't get how! I recognized that the term inside the parentheses is a Dirichlet integral, so I know its transform, which is $$\frac{1}2\sqrt\frac{\pi}2 \left(\text{sgn}(a-\lambda) +\text{sgn}(a+\lambda\right))$$ but I can't just multiply everything by $e^{-ia\lambda}$ as if were nobody's business! And then inside that integral I have the 4th power the transform, how do I deal with it?

Thanks a lot for your patience

  • $\begingroup$ Fourier transforms convert multiplication to convolution and vice versa. $\endgroup$ – Lord Shark the Unknown Jul 16 '17 at 14:59
  • $\begingroup$ you caould apply Paresval to the function $\sin^2( x)/x^2$ which is just the triangle function $\endgroup$ – tired Jul 16 '17 at 14:59
  • $\begingroup$ @MarkViola fools seldom differ :p $\endgroup$ – tired Jul 16 '17 at 15:00
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    $\begingroup$ @MarkViola haha... personally i am more on the foolish side ^^ $\endgroup$ – tired Jul 16 '17 at 15:02
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    $\begingroup$ @tired Unfortunately, I seem to be with you. $\endgroup$ – Mark Viola Jul 16 '17 at 15:07

First note that enforcing the substitution $x\to x/a$, reveals

$$\int_{-\infty}^\infty \frac{\sin^4(ax)}{x^4}\,dx=a^3\int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx$$

To find the integral $\int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx=\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}\frac{\sin^2(x)}{x^2}\,dx$, we simply need to convolve the Fourier Transform of $\frac{\sin^2(x)}{x^2}$ with itself, and evaluate this at $\omega=0$.


We have the Fourier Transform pairs

$$\begin{align} f(x) &\leftrightarrow F(\omega)\\\\ f^2(x) &\leftrightarrow \frac{1}{\sqrt {2\pi}}F(\omega)*F(\omega)\\\\ \frac{\sin(x)}{x}&\leftrightarrow \sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\\\\ \frac{\sin^2(x)}{x^2}&\leftrightarrow \frac{1}{\sqrt{2\pi}}\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)*\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)=\sqrt{\frac\pi8}2 \text{tri}(\omega/2) \end{align}$$

where $\text{rect}(t)$ and $\text{tri}(t)$ are the Rectangle Function and Triangle Function, respectively.

Therefore, we have

$$\begin{align} \int_{-\infty}^\infty \frac{\sin^4(x)}{x^4}\,dx &=\left.\left(\left(\sqrt{\frac\pi8}2 \text{tri}(\omega/2)\right)*\left(\sqrt{\frac\pi8}2 \text{tri}(\omega/2)\right)\right)\right|_{\omega =0}\\\\ &=\int_{-2}^2 \left(\sqrt{\frac{\pi}{8}}\frac{}{}(2-|\omega'|)\right)^2\,d\omega'\\\\ &=\left(\frac{\pi}{8}\right)\,2\int_0^2 (\omega'-2)^2\,d\omega'\\\\ &=\left(\frac{\pi}{8}\right)\,2\left(\frac83\right)\\\\ &=\frac{2\pi}{3} \end{align}$$

Hence, we can assert that

$$\bbox[5px,border:2px solid #C0A000]{\int_{\infty}^\infty \frac{\sin^4(ax)}{x^4}\,dx=\frac{2\pi a^3}{3}}$$

  • $\begingroup$ You are a godsent, thanks you a ton! I always get lost in silly calculations like this $\endgroup$ – AnxiousObject Jul 16 '17 at 15:47
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jul 16 '17 at 15:49
  • $\begingroup$ This is brilliant @MarkViola $\endgroup$ – Zeno San Jul 16 '17 at 19:36
  • $\begingroup$ @lelouch.d.light Wow. Thank you. $\endgroup$ – Mark Viola Jul 16 '17 at 20:11

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