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The problem here is to evaluate $$ \int_0^\infty \frac{x}{(x^2 + a^2) \, \sin(\mu x)} dx $$ for $a,\mu >0.$ Clearly this integral doesn't converge in the usual sense, but we can calculate its Cauchy Principal Value.

My attempt was to integrate the function in the complex domain along a quartercircle-contour in the first quadrant, with little bumps at $z=ia$ and $z = \frac{n \pi}{\mu}, n \in \mathbb{N}_{>0}.$ The infinitely many poles along the positive real axis worry me. The residue at $\frac{n \pi}{\mu}$ is $ \frac{(-1)^n}{ a^2\mu^2 + \pi^2 n^2}$, so we should evaluate $\sum_{n=1}^{\infty} \frac{(-1)^n}{ a^2\mu^2 + \pi^2 n^2}. $

How can we do that? How can we evaluate this integral? I'd really appreciate an approach with contour integration.

Note: The solution is $$ PV \int_0^\infty \frac{x}{(x^2 + a^2) \, \sin(\mu x)} dx = \frac{\pi}{2 \, \sinh(\mu a)},$$ but I want to prove it.

Breaking M
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  • make $ \mu$ complex to move away the poles away from the real axis and analtically continue back to real $\mu$ in the end of the calculation – tired Feb 26 '16 at 16:03
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    btw. ur residues are wrong. they are given by $\frac{(-1)^n}{\mu^2 a^2+(n\pi)^2}$ and the coressponding sum converges according to leibnitz – tired Feb 26 '16 at 16:06
  • Furthermore because we are dealing with an even function, u can use the standard semicircle as integration contour (maybe minus small bumps on the real axis depending on how exactly u want to tackle the problem). – tired Feb 26 '16 at 16:09

1 Answers1

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The integrand is even, hence

$$\operatorname{v.p.} \int_0^{\infty} \frac{x}{(x^2+a^2)\sin (\mu x)}\,dx = \frac{1}{2} \operatorname{v.p.} \int_{-\infty}^{\infty} \frac{x}{(x^2+a^2)\sin (\mu x)}\,dx.\tag{$\ast$}$$

For this we can use a standard semicircular contour, modified by small semicircles around the poles on the real axis. Let's choose the small semicircles around the real poles and the large semicircle in the upper half-plane, then, if $C_R$ denotes the contour where the large semicircle has radius $R > a$, we have

$$\int_{C_R} \frac{z}{(z^2+a^2)\sin (\mu z)}\,dz = 2\pi i \operatorname{Res} \biggl(\frac{z}{(z^2+a^2)\sin (\mu z)}; ia\biggr) = 2\pi i \frac{ia}{2ia\sin (\mu ia)} = \frac{\pi}{\sinh (\mu a)},$$

independent of $R$. Since $\lvert \sin (\mu z)\rvert$ grows exponentially as $\operatorname{Im} z \to \infty$, the integral over the large semicircle tends to $0$ as $R \to \infty$ in such a way that the poles on the real axis are avoided.

If we let the radius $\varepsilon$ of the small semicircles around the poles shrink to $0$, the integral over the small semicircle around $\frac{k\pi}{\mu}$ converges to $r_k := -\pi i \operatorname{Res}\bigl(\frac{z}{(z^2+a^2)\sin (\mu z)}; \frac{k\pi}{\mu}\bigr)$. But since the integrand is an even function, we have $r_{-k} = -r_k$, and hence

$$\sum_{0 < \lvert k\rvert < \mu R/\pi} r_k = 0$$

for all $R$ (subject to the restrictions above).

Therefore,

$$\operatorname{v.p.} \int_{-\infty}^{\infty} \frac{x}{(x^2+a^2)\sin (\mu x)}\,dx = \frac{\pi}{\sinh (\mu a)}.$$

By $(\ast)$,

$$\operatorname{v.p.} \int_{0}^{\infty} \frac{x}{(x^2+a^2)\sin (\mu x)}\,dx = \frac{\pi}{2\sinh (\mu a)}.$$

Daniel Fischer
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