I need a hand with the following integral, I'm not sure how to go about it. I'm trying to verify the following relation, found in Pathria and Beale's Statistical Mechanics, appendix E.
$$\dfrac{\partial}{\partial z} f_v (z) = z^{-1} f_{v-1} (z)$$
Where $$ f_v (z) = \dfrac{1}{\Gamma (v)} \int_0^\infty \dfrac{x^{v-1}}{z^{-1}e^x + 1} dx $$
I get stuck after the $z$-derivative. \begin{align} \dfrac{\partial}{\partial z} f_v (z) &= \dfrac{\partial}{\partial z} \dfrac{1}{\Gamma (v)} \int_0^\infty \dfrac{x^{v-1}}{z^{-1}e^x + 1} dx \\[.5em] &=\dfrac{1}{\Gamma (v)} \int_0^\infty \dfrac{\partial}{\partial z}\dfrac{x^{v-1}}{z^{-1}e^x + 1} dx \\[.5em] &= \dfrac{1}{\Gamma (v)} \int_0^\infty \dfrac{x^{v-1} z^{-2} e^x}{(z^{-1}e^x + 1)^2} dx \end{align}
I'm not sure where to go from here. Any help would be appreciated. Thank you!
Use integration by parts: $$f'_v(z)=\frac{z^{-1}}{\Gamma(v)}\,\left(-\left.\frac{x^{v-1}}{z^{-1}\,\exp(x)+1}\right|_{x=0}^{x=\infty}+(v-1)\,\int_0^\infty\,\frac{x^{v-2}}{z^{-1}\,\exp(x)+1}\,\text{d}x\right)\,.$$ Assuming that $\text{Re}(v)>1$, we get $$f'_v(z)=\frac{z^{-1}\,(v-1)}{\Gamma(v)}\,\int_0^\infty\,\frac{x^{v-2}}{z^{-1}\,\exp(x)+1}\,\text{d}x=\frac{z^{-1}}{\Gamma(v-1)}\,\int_0^\infty\,\frac{x^{v-2}}{z^{-1}\,\exp(x)+1}\,\text{d}x\,.$$ Hence, $f'_v(z)=z^{-1}\,f_{v-1}(z)$ for all $v\in\mathbb{C}$ with $\text{Re}(v)>1$ and $z\in\mathbb{C}$ with $z\notin\mathbb{R}_{\leq -1}$. Note that $f_1(z)=\ln(1+z)$ for all $z\in\mathbb{C}\setminus\mathbb{R}_{\leq -1}$ and $$f_v(z)=-\text{Li}_v(-z)\text{ for all }v\text{ with }\text{Re}(v)>1\text{ and }z\in\mathbb{C}\setminus\mathbb{R}_{\leq -1}\,,$$ where $\text{Li}_v$ is the polylogarithmic function with parameter $v$.