Q:

$\int_{0}^{\infty}(n+1)(1-e^{-x})^{n-1}e^{x(it-1)}e^{-x}dx$

I have to compute the following integral 
$\int_{0}^{\infty}(n+1)(1-e^{-x})^{n-1}e^{x(it-1)}e^{-x}dx$
which should be equal to $\frac{1-it}{n}\int_{0}^{\infty}(n+1)(1-e^{-x})^{n}e^{itx}e^{-x}dx$.
Any idea how to compute it ??

A:

Let $I=\int_{0}^{\infty}(n+1)(1-e^{-x})^{n-1}e^{x(it-1)}e^{-x}dx$
you can see thet $ \frac{d}{dx}( \frac{1}{n}(1-e^{-x})^{n})=(1-e^{-x})^{n-1}e^{-x}$
$$I=\left. \frac{n+1}{n}(1-e^{-x})^{n}e^{x(it-1)} \right\rvert_{0}^{\infty} - \int_{0}^{\infty}(n+1)\frac{1}{n}(1-e^{-x})^{n}(it-1)e^{x(it-1)}dx$$
as we have $\left. \frac{n+1}{n}(1-e^{-x})^{n}e^{x(it-1)} \right\rvert_{0}^{\infty}=0 \quad because \quad   e^{ixt} \quad bounded \quad and \quad  e^{-x} \rightarrow 0 \quad when \quad x \rightarrow \infty$ 
so 
$$I= - \frac{(it-1)}{n}\int_{0}^{\infty}(n+1)\frac{1}{n}(1-e^{-x})^{n}(it-1)e^{x(it-1)}dx$$
$$I= \frac{(1-it)}{n}\int_{0}^{\infty}(n+1)\frac{1}{n}(1-e^{-x})^{n}(it-1)e^{x(it-1)}dx$$

