Contour for Principal Value Integral

Question

I am trying to evaluate this principal value integral analytically $$ \mathcal{P}\int_0^\infty \mathrm{d}{x} \frac{1}{e^{x}-1}\frac{x}{x^2+\Omega^2}\frac{1}{x-\gamma} \,,$$ where $\gamma, \Omega \in \mathbb{R}$ and $\mathcal{P}$ denotes the principal value. So I need a contour that avoids the pole at $x=\gamma$ and goes across the positive real line. There's the added complication of the infinite number of simple poles on the y axis from the $1/(e^x -1)$ term. Any suggestions on a suitable contour, or an alternative method?

Answer 1

For contour integrals with integration limits $0$ and $\infty$, it is an elementary to trick to plug in a $\ln z$; i.e. we instead consider the contour integral: $$\oint_C \frac{z\ln z}{(z-\gamma)(z^2+\Omega^2)(e^z-1)}dz$$ where $C$ is the contour as shown in the image:

I will elaborate.

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Let $$f(z)= \frac{z}{(z-\gamma)(z^2+\Omega^2)(e^z-1)}dz$$ and let $J$ be your integral.

$$I_2+I_4=\int^\infty_0 f(t)\ln t dt$$ $$I_8+I_6=\int^0_\infty f(t)\ln(te^{2\pi i})dt=\int^0_\infty f(t)\ln t+f(t)\cdot 2\pi i dt$$

Then, $$I_2+I_4+I_6+I_8=-2\pi i\int^\infty_0 f(t)dt =-2\pi i J$$

Moreover, $I_1$ and $I_5$ vanish in the limit.

Also,

$$I_3=-\frac12\cdot 2\pi i\cdot\ln\gamma \lim_{z\to \gamma}f(z)(z-\gamma)$$ The negative sign is due to the clockwise direction of $I_3$ and because $I_3$ is a semicircle so there is a $\frac12$ factor.

Similarly, $$I_7=-\frac12\cdot2\pi i\cdot(\ln\gamma+2\pi i)\lim_{z\to\gamma}f(z)(z-\gamma)$$

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Just to sum things up a little bit, we have $$-2\pi iJ=2\pi i\sum\text{Res} -I_3-I_7$$

The calculation of the residue is quite straightforward.

At $+ i\Omega$, the residue is $$(\ln\Omega+\frac{i\pi}2)\lim_{z\to i\Omega}f(z)(z-i\Omega)$$

At $-i\Omega$, the residue is $$(\ln\Omega+\frac{3\pi i}2)\lim_{z\to -i\Omega}f(z)(z+i\Omega)$$

At $2\pi in$ where $n>0$, the residue is $$(\ln 2\pi n+\frac{i\pi}2)\lim_{z\to2\pi in}f(z)(z-2\pi i n)$$

At $2\pi in$ where $n<0$, the residue is $$(\ln 2\pi n+\frac{3i\pi}2)\lim_{z\to2\pi in}f(z)(z-2\pi i n)$$

Note that $J$ is real, so considering only the real part of the residue is sufficient.