Solve Laplace Transform: t^m = m!/s^m+1

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SUMMARY

The Laplace Transform of the function t^m is defined as L[t^m] = m!/s^(m+1). The integral from 0 to infinity of (t^m)(e^(-st))dt can be evaluated using integration by parts and induction. By differentiating under the integral sign, the expression can be simplified to show that taking the derivative with respect to s m times yields the factor t^m. This method is valid for positive integer values of m.

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catcherintherye
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I am trying to show that L[t^m] = m!/s^m+1, unfortunately I can not understand why integral from 0 to inf of (t^m)(e^-st)dt = (-d/ds)^m. integral 0 to inf of e^-stdt... ?
 
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You can exchange the order, that is, take -d/ds inside the integral, since they are both linear operations. It takes down a factor t from the exponential. Do it m times and get t^m.
 
If m is a positive integer, use induction together with integration by parts:
\int_0^\infty t^m e^{-st}dt
Let u= tm, dv= e-stdt. Then du= m tm-1 and v= -1/s e-st. The integral becomes
-\frac{m}{s}\int_0^\infty t^{m-1}e^{-st}dt
since uv= 0 at both ends.
 

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