Is the following gamma function integral correct?

Question

$$\int_{0}^{\infty}\frac{x^{v-1}}{e^{-\beta\mu}{e^x}-1}=\Gamma(v)\left(e^{\beta\mu}+\frac{e^{2\beta\mu}}{2^v}+\frac{e^{3\beta\mu}}{3^v}+\dotsb\right)$$
This integral is given in my statistical physics assignment and I feel that the $\mu$ in the integrand should have the opposite sign (as my results would make more sense) but I don't know how to prove this and I can't find the formula anywhere. I would love to find a source for this formula.

Answer 1

On the RHS you have: $$\Gamma(v)\sum_{k=0}^\infty\frac{(e^{-\beta\mu})^k}{k^v}\tag{1}$$ we know that: $$\Gamma(v)=\int_0^\infty x^{v-1}e^{-x}dx$$ $$\sum_{k=0}^\infty\frac{(e^{-\beta\mu})^k}{k^v}=\Phi(e^{-\beta\mu},v,0)$$ I have changed it to a negative sign because it is the only way that this series will converge for $\beta,\mu>0$ and $\Phi$ represents the Lerch transcendent . On the same page you will find the identity: $$\Phi(z,s,a)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{t^{s-1}e^{-at}}{1-ze^{-t}}dt$$ which we can rearrange to give us the general formula: $$\int_0^\infty\frac{t^{s-1}e^{-at}}{1-ze^{-t}}dt=\Gamma(s)\Phi(z,s,a)$$

Try looking into these functions and seeing if there is any material on this derivation. Note: It is interesting that this summation (with $a=0$) is close in appearance to the Riemann zeta function

Answer 2

There's no sign error. You just need to identify a hidden convergent geometric series:$$\int_0^\infty\frac{x^{v-1}e^{\beta\mu}e^{-x}dx}{1-e^{\beta\mu}e^{-x}}=\sum_{n\ge1}e^{n\beta\mu}\int_0^\infty x^{v-1}e^{-nx}dx=\Gamma(v)\sum_n\frac{e^{n\beta\mu}}{n^v}.$$Incidentally, this can be rewritten with the polylogarithm as $\Gamma(v)\operatorname{Li}_v(e^{\beta\mu})$; convergence requires $\Re v>1,\,\Im\beta\mu<0$.