Prove $\mathrm{e}^{(\Delta -1)t} = \sum_{n \geq 0} \mathrm{e}^{-t}\,(1-\mathrm{e}^{-t})^n\,\frac{\Gamma(\Delta+n)}{\Gamma(\Delta)\,\Gamma(n+1)}$

Question

I encountered the following series representation (but without proof) stating that for any $t > 0$ and $\Delta > 0$, we have $$\mathrm{e}^{(\Delta -1)t} = \sum_{n \geq 0} \mathrm{e}^{-t}\,(1-\mathrm{e}^{-t})^n\,\frac{\Gamma(\Delta+n)}{\Gamma(\Delta)\,\Gamma(n+1)},$$ where $\Gamma(\cdot)$ represents the Gamma function. Can anyone provide a (rigorous) proof of this identity? Thank you very much for your help!

Answer 1

It suffices to show $$e^{\Delta t} = \sum_{n\geq 0}(1-e^{-t})^n\frac{\Gamma(\Delta+n)}{\Gamma(\Delta)\Gamma(n+1)}.$$ Note that $$e^{\Delta t} = (-(1-e^{-t})+1)^{-\Delta}.$$ Since $t > 0$ it holds $|1-e^{-t}| < 1$, so by Newton's binomial theorem we have $$(-(1-e^{-t})+1)^{-\Delta} = \sum_{n\geq 0}(-1)^n(1-e^{-t})^n\frac{(-\Delta)(-\Delta-1)\cdots(-\Delta - n +1)}{n!} = \sum_{n\geq 0}(1-e^{-t})^n\frac{\Delta(\Delta+1)\cdots(\Delta+n-1)}{n!} = \sum_{n\geq 0}(1-e^{-t})^n\frac{\Gamma(\Delta + n)}{\Gamma(n+1)\Gamma(\Delta)}.$$

Answer 2

Probably, not the simpliest way :) $$(1-e^{-t})^n\frac{\Gamma(\Delta+n)}{\Gamma(\Delta)\Gamma(n+1)}=(1-e^{-t})^n\frac{\Gamma(\Delta+n)\Gamma(1-\Delta)}{\Gamma(1-\Delta)\Gamma(\Delta)\Gamma(n+1)}$$ $$=(1-e^{-t})^n\frac{\sin\pi \Delta}{\pi}\frac{\Gamma(\Delta+n)\Gamma(1-\Delta)}{\Gamma(n+1)}=(1-e^{-t})^n\frac{\sin\pi \Delta}{\pi}\int_0^1s^{\Delta+n-1}(1-s)^{-\Delta}ds$$ Performing summation with respect to $n$ $$\sum_{n=0}^\infty(1-e^{-t})^n\frac{\Gamma(\Delta+n)}{\Gamma(\Delta)\Gamma(n+1)}=\frac{\sin\pi \Delta}{\pi}\int_0^1\left(\frac{1-s}s\right)^{-\Delta}\frac{ds}{s^2\big(\frac {1-s}s+e^{-t}\big)}$$ Making the substitution $x=\frac{1-s}s$ $$=\frac{\sin\pi \Delta}{\pi}\int_0^\infty\frac{x^{-\Delta}}{x+e^{-t}}dx\overset{x=se^{-t}}{=}\frac{\sin\pi \Delta}{\pi}e^{\Delta t}\int_0^\infty\frac{s^{-\Delta}}{1+s}ds=\frac{\sin\pi \Delta}{\pi}e^{\Delta t}B(1-\Delta;\Delta)=e^{\Delta t}$$