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I'm trying to follow a proof (with many steps omitted) and got to the left-hand side of the equation below. The proof, however, continues with the right-hand side.

$$\int_{-\infty}^{\infty}\frac{e^{imx}-xime^{imx}}{(x^2+a^2)^{3/2}}dx = 2\int_{0}^{\infty} \frac{\cos (mx)+mx \sin (mx)}{(x^2+a^2)^{3/2}}dx$$

Is there a good reason for the complex component of the integral on the left to be zero? Sorry, my mathematics isn't great...

Thank you in advance!

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  • $\begingroup$ @J.Yu I'm not sure how this helps... doesn't the complex component also scale with $o(x^{-2})$? Botond Thank you! $\endgroup$ – Tom Waits Jun 25 '18 at 9:28
  • $\begingroup$ sorry I misread, this question is about even and odd functions, use f(x) = -f(-x) to solve it $\endgroup$ – J. Yu Jun 25 '18 at 9:31
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If $f$ is an odd function then, $\forall a \in \mathbb{R}$ $$\int_{-a}^{a} f=0$$ And if $\exists \int_{0}^{\infty} f$ and $\exists\int_{-\infty}^{0} f$ (it's not infinity), then $$\int_{-\infty}^{\infty} f=0$$ And in the case of even function, $\forall a \in \mathbb{R}$ $$\int_{-a}^{a}f=2\int_{0}^{a}f$$ And if $\exists \int_{0}^{\infty} f$ and $\exists\int_{-\infty}^{0} f$ (it's not infinity), then $$\int_{-\infty}^{\infty} f=2\int_{0}^{\infty}f$$ In your case, $\frac{i\sin(mx)}{(x^2+a^2)^{3/2}}$ and $\frac{-mix\cos(x)}{(x^2+a^2)^{3/2}}$ are odd, and $\frac{\cos(mx)}{(x^2+a^2)^{3/2}}$ and $\frac{mx\sin(mx)}{(x^2+a^2)^{3/2}}$ are even.

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Note that by Euler's formula $(1-i m x) \exp (i m x) = i \sin (m x)+m x \sin (m x)-i m x \cos (m x)+\cos (m x)$. Hence the odd part of the integrand is equal to its imaginary part. However, the symmetric integral over an odd integrand vanishes trivially.

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