Laplace Transform of a periodic function

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SUMMARY

The Laplace transform of a periodic function f(t) with period T is given by the formula L(f) = \frac{1}{1-e^{-sT}}\int_0^T f(t)e^{-st} dt. The discussion highlights the initial steps in deriving this formula using the definition of the Laplace Transform, L(f) = \int_0^\infty f(t)e^{-st}dt, and the periodicity of the function. The user attempts to express the integral as a geometric series but encounters difficulties. A substitution method, t = nT + u, is suggested to facilitate the transformation.

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Homework Statement


Suppose f(t) is a periodic function with period T. Show that the Laplace transform of f is:

L(f) = \frac{1}{1-e^{-sT}}\int_0^T f(t)e^{-st} dt

The Attempt at a Solution


I started with the definition of a Laplace Transform for f:

L(f) = \int_0^\infty f(t)e^{-st}dt

Using the periodicity of the function this becomes:

\sum_{K=0}^\infty \int_{KT}^{(K+1)T} f(t)e^{-st}dt

At this point I have been trying to get this in the form of a geometric series, since the fraction in the final result leads me to look for a geometric series, but this has been without success. Any hints into the right direction to move from here on out would be appreciated. Thank you for any help you can offer.
 
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Try a substitution: t = nT+u with u as new variable.
 

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