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While studying Fourier Integral topic, I encountered this problem. I have to prove that

$$\int_{-\infty}^{\infty}{\frac{\sin\omega \cos{2\omega}}{\omega}} d\omega =0$$

I've no idea how to do this. Could someone guide me, please?

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  • $\begingroup$ Hint : convert the product to sum. $\endgroup$ – Zaid Alyafeai Mar 18 '17 at 9:38
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HINT $$\sin(x)\cos(y) = \frac{\sin(x+y) + \sin(x-y)}{2}$$

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    $\begingroup$ so, $\int_{-\infty}^{\infty}{\frac{\sin\omega \cos{2\omega}}{\omega}}=\frac{1}{2}\left(\int_{-\infty}^{\infty}{\sin(3\omega)} + \int_{-\infty}^{\infty}{\sin(-\omega)}\right)$, but these integrals does not converge. $\endgroup$ – 01000110 Mar 18 '17 at 10:06
  • $\begingroup$ You was left $\frac{1}{w}$ $\endgroup$ – kotomord Mar 18 '17 at 10:12
  • $\begingroup$ You're right! >_< $\endgroup$ – 01000110 Mar 18 '17 at 10:15
  • $\begingroup$ @01000110, note the two integrals have equal values with opposite signs. $\endgroup$ – Zaid Alyafeai Mar 18 '17 at 10:15

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