# Laplace transform

1. Feb 5, 2016

### hotjohn

for the right part of the notes, why the integral of (e^-su)f(u) from 0 to T will become integral of (e^-st)f(t) from 0 to T suddenly ? why not integral of (e^-s (t-nT) )f(t-nT) from 0 to T ? as we can see, u = t +nTd given/known data

2. Relevant equations

3. The attempt at a solution

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2. Feb 5, 2016

### Samy_A

If so, that's just renaming the (dummy) integration variable from u to t.

3. Feb 5, 2016

### hotjohn

yes , why thgere is no need to turn u into t -nT ? since it is given at the left part of the notes

4. Feb 5, 2016

### Samy_A

It's just a name. It doesn't matter whether the integration variable is called u or t (or something else).

Example: the two following integrals are equal
$\displaystyle \int_a^b e^x dx = \int_a^b e^y dy$

Last edited: Feb 5, 2016
5. Feb 7, 2016

### vela

Staff Emeritus
The author was a little sloppy. The earlier $t$ (the one related to $u$) and the $t$ in the last integral aren't the same $t$.