- #1

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

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- Thread starter capicue
- Start date

- #1

- 4

- 0

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

- #2

- 761

- 13

Did you make a typo?

- #3

- 4

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I don't think so. Here's the Wolfram alpha solution.

- #4

- 761

- 13

Are you familiar with contour integration?

- #5

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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]

[tex] \sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}\ [/tex]

Last edited:

- #7

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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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