# How to use the Bromwich Integral

1. Nov 8, 2008

### flouran

I converted the function, $${t^n}{e^{-t}}$$ where t is the variable (time domain) and n is any real whole integer greater than or equal to 0, to the s-domain using a unilateral Laplace transform. In the end, I got f(s) = $$\frac{n!}{(s+1)^{n+1}}$$. The reason why I did this was because I wanted to find $$\int}{t^n}{e^{-t}}dt$$. I know how to do this with integration by parts, but I wanted to "spice things up a bit". Once I had converted the function to the s-domain, I integrated it with respect to s, in other words, I thought $$\int}{\frac{n!}{(s+1)^{n+1}}}ds &=& \int}{{t^n}{e^{-t}}}dt$$, correct me if I am wrong. Thus, I got $$\int}{\frac{n!}{(s+1)^{n+1}}}ds &=& {\frac{n!}{-(n+2){{(s+1)^{n+2}}}}$$. Thus, in order to find the $$\int}{t^n}{e^{-t}}dt$$, I just have to convert $${\frac{n!}{-(n+2){{(s+1)^{n+2}}}}$$ back to the time domain using the Bromwich Integral, but I don't know how to do this (again, correct me if I am wrong). Could somebody help me?

Last edited: Nov 8, 2008
2. Nov 9, 2008

### flouran

Mustn't it be integrated along a straight line parallel to the imaginary axis and intersecting the real axis in the point $$\gamma$$ which must be chosen so that it is greater than the real parts of all singularities of F(s)? I just don't know how do this. I've only done math until BC Calculus. Also, in practice, computing the complex integral can be done by using the Cauchy residue theorem, right?