# Definite integral $\int_{0}^{+ \infty}e^{itx}e^{-x} \frac{x^n}{n!}dx$

Definite integral $$\int_{0}^{+ \infty}e^{itx}e^{-x} \frac{x^n}{n!}dx$$

I very much need this for probability, the answer is $$\frac{1}{(1-it)^{n+1}}.$$ I just don't know how to come to this myself. Can someone help out that has experience in integrating? I think the gamma function must be used.

• I tried using the gamma function, because i am convinced that it must be done using it, but to no avail. – Jerry West Jan 3 '16 at 20:53

One may observe that $$\frac1{\lambda}=\int_0^{+\infty} e^{-\lambda x}dx, \quad \Re \lambda>0.\tag1$$ Then differentiating $n$ times with respect to $\lambda$ gives $$\frac{(-1)^n\:n!}{\lambda^{n+1}}=(-1)^n\int_0^{+\infty} x^n e^{-\lambda x}dx, \quad \Re \lambda>0.\tag2$$ Apply $(2)$ with $\lambda=1-it$.
The following might help $$\int_0^\infty dx e^{x (\mathrm{i}t-1)}=\frac{\mathrm{i}}{\mathrm{i}+t},\qquad \mathrm{Im}(t)>-1$$ and $$\partial_t^{(n)} e^{\mathrm{i}tx} =(\mathrm{i}x)^n e^{\mathrm{i}tx}$$
• This answer is not helpful, since the integral has three factors as written. Just what do you mean for $u$ and $v$? Have you actually done the problem to completion so you are sure this hint works? – Rory Daulton Jan 3 '16 at 21:07
• @Rory Daulton let $u=\frac{x^n}{n!}$ and $dv=e^{-x(1-it)}$, is there something wrong with this method? – bob kelso Jan 3 '16 at 21:17