# Laplace transformation

1. Jan 14, 2006

### Tony11235

laplace transform

Could anyone should me how it is you arrive to $$\frac{n!}{(s - a)^{n+1}}$$ as being the laplace transform of $$t^n e^{at}$$ ? Or at least provide a link?
I've done integration by parts but the term $$e^{(s-a)t}$$ will not go away.

Last edited: Jan 14, 2006
2. Jan 14, 2006

### benorin

If n is an integer the $$e^{(a-s)t}$$ term will go away after n iterations of integration by parts and one final integration of the term itself.

Last edited: Jan 14, 2006
3. Jan 14, 2006

### Tony11235

So far I have $$\int t^n e^{(s-a)t} dt = \frac{t^n e^{(s-a)t}}{s-a} - \frac{n t^{n-1} e^{(s-a)t}}{(s-a)^{2}} + \frac{(n^2-n) t^{n-2} e^{(s-a)t}}{(s-a)^{3}} . . . . . . .$$. I'm just not seeing how $$e^{(s-a)t}$$ goes away.

Last edited: Jan 14, 2006
4. Jan 14, 2006

### qbert

I think the easiest way is to do it in pieces.

prove L[ t^n ] = n!/s^(n+1).

This is easy if you know
$$n! = \Gamma ( n+1 ) = \int_0^{\infty} t^n exp(-t) dt$$
otherwise an inductive proof on n is pretty straight forward.

Then i would show if L[f(t)] = F(s) then L[f(t)exp(at)] = F(s-a). Since this is just a relable-ing operation it should
be easy to show.

Last edited: Jan 14, 2006
5. Jan 14, 2006

### benorin

an inductive proof

EDIT: changed the limits of integration to be from 0 to infinity (instead of -infinity to infinity).

The Laplace transform of $t^n e^{at}$ is

$$\int_{t=0}^{\infty} t^n e^{at} e^{-st}dt=\int_{t=0}^{\infty} t^n e^{(a-s)t} = \frac{n!}{(s - a)^{n+1}}$$

an inductive proof goes as follows:

i. For n=0, $$\int_{t=0}^{\infty} t^0 e^{(a-s)t}= \frac{1}{s - a} \left[ e^{(a-s)t} \right]_{t=0}^{\infty} = \frac{0!}{(s - a)^{0+1}}$$

ii. Assume that

$$\int_{t=0}^{\infty} t^n e^{(a-s)t} = \frac{n!}{(s - a)^{n+1}}$$

holds for some fixed integer n >0, then

$$\int_{t=0}^{\infty} t^{n+1} e^{(a-s)t} dt= \left[ -\frac{t^{n+1}e^{(a-s)t}}{s-a}\right]_{t=0}^{\infty} + \frac{n+1}{s-a}\int_{t=0}^{\infty} t^{n} e^{(a-s)t} dt$$

which, by hypothesis,

$$= \frac{n+1}{s-a}\frac{n!}{(s - a)^{n+1}} = \frac{(n+1)!}{(s - a)^{(n+1)+1}}$$

and so

$$\int_{t=0}^{\infty} t^n e^{(a-s)t} = \frac{n!}{(s - a)^{n+1}}$$

holds for every non-negative integer n by induction.

Last edited: Jan 15, 2006
6. Jan 14, 2006

### benorin

Remember that you differentiate the t^n term each iteration of integration by parts, so the exponent of t should decrese by one for successive terms in the above sum, after n such terms the exponent of t will be 0 leaving only the exponential to integrate.

7. Jan 14, 2006

### Tony11235

Nice advice, but I was instructed to derive the laplace transform from integration by parts, or else I would have used the property, If L{f(x)} = F(s), then L{(e^at)f(x)} = F(s-a).

8. Jan 14, 2006

### Tony11235

Sorry, that was a typo.