showing $\frac{1}{(A+B)^{a}}=\frac{1}{\Gamma(a)} \frac{1}{2\pi i} \int_{-i\infty}^{+i \infty} dz \frac{B^z}{A^{a+z}} \Gamma(-z) \Gamma(a+z)$

Question

I want to prove

\begin{align} \frac{1}{(A+B)^{a}} = \frac{1}{\Gamma(a)} \frac{1}{2\pi i} \int_{-i\infty}^{+i \infty} dz \frac{B^z}{A^{a+z}} \Gamma(-z) \Gamma(a+z) \end{align} Starting from the integral representation of gamma function, still, the two products of gamma functions produce only $\frac{B}{A}$ form, not $A+B$.

I want to know the derivation of the above equation with an explicit process. (I found one, and understand the process by @Gary, If you know any other proofs, please give me another ways to showing this!)

---

After some searching this integral is called the "Mellin-Barnes identity" and its generalization is given by \begin{align} \frac{1}{(A_1+ \cdots + A_n)^{\lambda}} = \frac{1}{\Gamma(\lambda)} \frac{1}{(2\pi i)^{n-1}} \int_{-i \infty}^{+i \infty} dz_1 \cdots dz_{n-1} \prod_{i=1}^{n-1} A_{i}^{z_i} A_{n}^{-\lambda - z_1 - \cdots - z_{n-1}} \prod_{i=1}^{n-1} \Gamma(-z_i) \Gamma(\lambda + z_1 + \cdots + z_{n-1}) \end{align}

Answer 1

There might be many proofs: One short version is the usage of Mellin transformation.

From the inversion formula, \begin{align} \frac{1}{(1+x)^t} = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} x^{-s} B(s,t-s)ds = \frac{1}{2\pi i} \frac{1}{\Gamma(t)} \int_{c-i\infty}^{c+i\infty} x^{-s} \Gamma(s) \Gamma(t-s) ds, \quad 0<c<\operatorname{Re}[t]. \end{align}