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I was playing around and wanted to consider the integral of $$I_n = \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x$$ using parts with $u = \sin^n x \implies \mathrm{d}u = n\cos x \sin^{n-1}x$ and $\mathrm{d}v = x^{-n} \implies v = \frac{x^{1-n}}{1-n}$

This gives $$I_n = \frac{n}{n-1}\int_0^{\infty} \left(\frac{\sin x}{x}\right)^{n-1} \cos x \, \mathrm{d}x$$

Which looks so close to $I_{n-1}$ but for that annoying $\cos x$ there, any help?

Note: I wanted to go down this route because I think I can note a pattern, as such: $$I_1 = \frac{\pi}{2} \\ I_2 = \frac{\pi}{2} \\ I_3 = \frac{3\pi}{8} \\ I_4 = \frac{\pi}{3} \\ I_5 = \frac{115\pi}{384} \\ I_6 = \frac{11\pi}{40}$$


marked as duplicate by David H, tired, Tom-Tom, Community Nov 4 '15 at 18:58

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    $\begingroup$ try to use the integration by parts once again $\endgroup$ – Michael Medvinsky Nov 4 '15 at 8:16
  • $\begingroup$ @MichaelMedvinsky With $u = \sin^{n-1}x \cos x$ and $\mathrm{d}v = x^{1-n}$? $\endgroup$ – Zain Patel Nov 4 '15 at 8:17
  • $\begingroup$ no, leave the cosinus for du to get $\cos^2x=1-\sin^2x$ $\endgroup$ – Michael Medvinsky Nov 4 '15 at 8:22
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    $\begingroup$ another integration by parts didn't work for me yet, also expected it would $\endgroup$ – Michael Medvinsky Nov 4 '15 at 9:32
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    $\begingroup$ It might be a good idea (I did not test it completely) to use Fourier transforms and convolutions. $\endgroup$ – mickep Nov 4 '15 at 9:43