I think this integral has no exact solution, but I've found that you can approximate it for large $\mu$ as:
$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1} dx \approx \sqrt{\frac{\pi \mu}{2}} e^{-\mu}$$
This is used a couple of times in a book I'm reading. I've checked it numerically and it works, but I wanted to get the symbolic expression.
I've tried expanding the square root:
$$\sqrt{x^2-\mu^2}=\mu\sqrt{-(1-x^2/\mu^2)}\approx i\mu \left(1-\frac{1}{2}\frac{x^2}{\mu^2}-\frac{1}{8}\frac{x^4}{\mu^4}-…\right)$$
That means that I get an imaginary solution as far as I know, because the first term is a simple logarithmic integral:
$$\int_{\mu}^{\infty} \frac{i\mu}{e^x-1}dx=-i\mu\ln{|e^{-\mu}-1|}$$
And the following terms are integrals that I don't know how to solve. So I guess that this is not the way to proceed.
Any ideas on how to do this approximation, or a more exact solution? Appreciate your help.
Best Answer
With the substitution $z=x-\mu$, the integral becomes
$$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1}\,dx=\int_0^\infty \frac{\sqrt{z^2+2\mu z}}{e^{\mu+z}-1}\,dz=\frac{\sqrt{2\mu}}{e^\mu }\int_0^\infty\frac{\sqrt{z+z^2/(2\mu)}}{e^z-e^{-\mu}}\,dz$$
Taking the limit $\mu\to\infty$ on the integrand and substituting $w=\sqrt{z}$, the integral becomes $$\int_0^\infty \sqrt{z}\,e^{-z}\,dz=\int_0^\infty 2w^2 e^{-w^2}\,dx=\int_{-\infty}^\infty w^2 e^{-w^2}\,dw.$$ This is a standard variation on the Gaussian integral and evaluates to $\sqrt{\pi}/2$. Hence $$\int_{\mu}^{\infty} \frac{\sqrt{x^2-\mu^2}}{e^x-1}\,dx\sim \frac{\sqrt{2\mu}}{e^\mu}\frac{\sqrt{\pi}}{2}=\sqrt{\frac{\pi u}{2}}e^{-\mu}$$ in the $\mu\to\infty$ limit as claimed.