Is this : ${\Gamma(1-\alpha)}\int_{0}^\infty \frac{e^{-t}}{{(1+xt)}^{(1-\alpha)}} dt = \int_0^\infty \frac{t^{-\alpha}e^{-t}}{{(1+xt)}} dt$ true?

Question

Some of my friends sent me the following identity regarding irrationality of Gamma function topic :

$${\Gamma(1-\alpha)}\int_{0}^\infty \frac{e^{-t}}{{(1+xt)}^{(1-\alpha)}}\, \mathrm{d}t = \int_0^\infty \frac{t^{-\alpha}e^{-t}}{{(1+xt)}}\, \mathrm{d}t \tag{01}$$

such that : $\alpha$ be a complex number such that : $Re(\alpha) <1$ and $x$ is a positive real number , I have used the usual definition of Riemann zeta function which it has similar form for the RHS of (01) , but integral representation of Riemann zeta function diverges for Re(s) <1 , and i took $x=\frac{1}{z}$ i got a complicated form for starting , pleas anyway ?

Answer 1

This is just switching the order of integration: \begin{align}\mathrm{LHS}&=\int_0^\infty e^{-t}\frac{\Gamma(1-\alpha)}{(1+xt)^{1-\alpha}}\,dt\\&=\int_0^\infty e^{-t}\int_0^\infty u^{-\alpha}e^{-(1+xt)u}\,du\,dt\\&=\int_0^\infty u^{-\alpha}e^{-u}\int_0^\infty e^{-(1+xu)t}\,dt\,du\\&=\int_0^\infty\frac{u^{-\alpha}e^{-u}}{1+xu}\,du=\mathrm{RHS}.\end{align}