Tricky integral

  • Thread starter capicue
  • Start date
  • #1
4
0
I need a hint. Wolfram alpha says it equals pi/2, but I don't know how to get that.

\int_{-\infty}^{\infty} \frac{\sin^4(x)}{x^2}\, dx
 

Answers and Replies

  • #2
761
13
Did you make a typo?
 
  • #3
4
0
I don't think so. Here's the Wolfram alpha solution.
 
  • #4
761
13
Are you familiar with contour integration?
 
  • #5
4
0
Yes. I thought about trying something like that, but the usual substitution z = e^{ix} to write the sin part as an exponentials doesn't work with the x^2 on the bottom.

What were you thinking about with the contour integration?

My other thought was to try to relate it back to \int sin(x)/x which is something I can do.
 
  • #6
761
13
I haven't worked it out but if you write out the sin^4 and then use contour integration it will probably work:
[tex] \sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}\ [/tex]
 
Last edited:
  • #7
4
0
You're right, that would probably work. I'll give it a go. Thanks for your help!

Alternatively, I figured out how to reduce it:

[tex]
\int_{-\infty}^{\infty} \frac{\sin^4{x}}{x^2}\, dx = \int_{-\infty}^{\infty} \frac{4 \sin^3{x} \cos{x}}{x}\, dx
[/tex]
(IBP)
[tex]
= \int_{-\infty}^{\infty} \frac{\sin{2x}(1 - \cos{2x})}{x}\, dx
[/tex]
(trig identities)
[tex]
= \int_{-\infty}^{\infty} \frac{\sin{2x}}{x}\, dx - \int_{-\infty}^{\infty} \frac{\sin{4x}}{2x}\, dx
[/tex]
[tex]
= \int_{-\infty}^{\infty} \frac{\sin{x}}{x}\, dx - \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin{x}}{x}\, dx
[/tex]
(change of variables)
[tex]
= \pi - \frac{\pi}{2}
[/tex]
(sinc integral)
 
  • #8
761
13
Yes, well found!
 

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