Evaluating the function

1. Sep 24, 2003

eljose79

Let be the function

f(t)=L(-1)1/(exp(-s)-1) where L(-1) means the laplace inverse transform..my doubt is to know what is the value of

f(0),f(1),f(2)...f(n) being n an integer.

2. Sep 24, 2003

Tom Mattson

Staff Emeritus
I'll take a stab at this.

First, we need to develop the Laplace transform of a periodic function:

f(t)=f(t+nT) for all integers n.

L{f(t)}=&int;0&infin;e-stf(t)dt

Let's break this up over intervals of width T:

L{f(t)}=&int;0Te-stf(t)dt+&int;T2Te-stf(t)dt+&int;2T3Te-stf(t)dt+...

Now perform substitutions on each integral such that the limits of each integral are [0,T]:

L{f(t)}=&int;0Te-stf(t)dt+&int;0Te-s(t+T)f(t+T)dt+&int;0Te-s(t+2T)f(t+2T)dt
+&int;0Te-s(t+3T)f(t+3T)dt+...

Noting that f(t)=f(t+T)=f(t+2T)=f(t+3T)=..., and factoring &int;0Te-stf(t)dt out of each factor yields:

L{f(t)}=(1+e-sT+e-2sT+e-3sT+...)&int;0te-stf(t)dt

The first factor on the right is a geometric series whose sum is:

(1-e-sT)-1

So, I have finally:

L{f(t)}=(1-e-sT)-1&int;0Tf(t)dt

Now, we get your function if we let f(t)=d(t+n+1/2). That is, a periodic delta function whose period is 1. I used the half integer n+1/2 so that there is only one delta function in each interval. I could just as easily have chosen n+1/3, n+1/4, or whatever. So, it seems that the function is not unique.

Does that help?

edit: fixed a variety of bracket errors

Last edited: Sep 24, 2003