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I'm doing some problems on integration of complex functions, and there's something here I don't understand. The integral is this: $$\int_{-\infty}^\infty \frac{x\sin(\omega x)}{x^4+1}\,dx$$ And in the first row it says: "The integrand is an even function, so the integral is equal to: $$\frac{1}{i}\int_{-\infty}^\infty \frac{xe^{\omega i x}}{x^4+1}\,dx$$ I don't understand what has been done there.

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Hint: $e^{i\omega x}=\cos(\omega x)+i\sin(\omega x)$. Then the real part of the integrand is an odd function, so its integral is... –  1015 Feb 6 '13 at 13:49

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You should remember that: $$e^{ix} = \cos(x)+i\sin(x)$$ But you also remeber that the integral of any odd function over $(-a,a)$ is equal to $0$, so we have $$\int_{-\infty}^\infty \frac{xe^{\omega i x}}{x^4+1}dx = \int_{-\infty}^\infty \frac{x(\cos(\omega x) + i\sin(\omega x))}{x^4+1}dx =\int_{-\infty}^\infty \frac{x\cos(\omega x)}{x^4+1}dx+\int_{-\infty}^\infty \frac{i\sin(\omega x)}{x^4+1}dx$$ But since cosine is an odd function, the integral involving just cosine scaled by an even function is equal to 0, so we know that $$\int_{-\infty}^\infty \frac{xe^{\omega i x}}{x^4+1}dx = i\int_{-\infty}^\infty \frac{\sin(\omega x)}{x^4+1}dx$$ And dividing both sides by $i$ gives the relation you were asking for.

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Thank you for your clear explanation –  MyUserIsThis Feb 6 '13 at 15:05

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