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?(adsbygoogle = window.adsbygoogle || []).push({});

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# How to use the Bromwich Integral

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