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How to use the Bromwich Integral

  1. Nov 8, 2008 #1
    I converted the function, [tex]{t^n}{e^{-t}}[/tex] 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) = [tex]\frac{n!}{(s+1)^{n+1}}[/tex]. The reason why I did this was because I wanted to find [tex]\int}{t^n}{e^{-t}}dt[/tex]. 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 [tex]\int}{\frac{n!}{(s+1)^{n+1}}}ds &=& \int}{{t^n}{e^{-t}}}dt[/tex], correct me if I am wrong. Thus, I got [tex]\int}{\frac{n!}{(s+1)^{n+1}}}ds &=& {\frac{n!}{-(n+2){{(s+1)^{n+2}}}}[/tex]. Thus, in order to find the [tex]\int}{t^n}{e^{-t}}dt[/tex], I just have to convert [tex]{\frac{n!}{-(n+2){{(s+1)^{n+2}}}}[/tex] 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. jcsd
  3. Nov 9, 2008 #2
    Mustn't it be integrated along a straight line parallel to the imaginary axis and intersecting the real axis in the point [tex]$ \gamma$[/tex] 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?
     
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