# Thread: How to verify a Laplace transform?

1. ## How to verify a Laplace transform?

$\displaystyle \mathcal{L}\left\{e^{at}\cos\!\left(kt\right)\righ t\}=\frac{s-a}{\left(s-a\right)^2+k^2}$

What is the proof of this? Or rather, could somebody get me started and point me in the right direction for this?

Thanks! Any help is appreciated.

2. One possible start is the computation of the integral...

$\displaystyle \mathcal{L} \{\cos kt\} = \int_{0}^{\infty} \cos kt\cdot e^{-st}\cdot dt$

... that can be done by parts...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. But what do you do with the e^at? Does that fit somewhere in the integral or no?

4. Originally Posted by tibetan-knight

But what do you do with the e^at? Does that fit somewhere in the integral or no?
yes...

$\displaystyle \int_{0}^{\infty}e^{at}e^{-st}cos(kt)dt=\int_{0}^{\infty}e^{-(s-a)t}cos(kt)dt$

Now just use integration by parts on this

Or note in general that

$\displaystyle \mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{at}e^{-st}f(t)dt$

$\displaystyle \mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{-(s-a)t}f(t)dt$

Let $\displaystyle s-a=m$ then we get

$\displaystyle \mathcal{L}(e^{at}f(t))=\int_{0}^{\infty}e^{-mt}f(t)dt=F(m)$
The above integral is the form of the laplace transform so we get
Where $\displaystyle F(m)=F(s-a)$

This is commonly know as the s-axis shift theorem.

So the above would give

$\displaystyle \mathcal{L}(e^{at}cos(kt))=\mathcal{L}(\cos(kt))\b igg|_{s\to s-a}=\frac{s}{s^2+k^2}\bigg|_{s \to s-a}=\frac{s-a}{(s-a)^2+k^2}$

5. Ooh. Shift theorem is helpful. Thank you.

But, to integrate the former by parts, which would be the 'u' part and which would be the 'dv' part?

6. Wat the "TheEmptySet" has showed is perfectly right but i like to show the same proof in another way ...We know by theorem Laplace {{e^at} f(t)}=F(s-a)
Proof of the theorem:
We know L{f(t)}=Intergral 0-infinity ({e^-st} f(t) dt)=F(s)
Therefore Intergral {e^at f(t)} = Intergral 0-infinity ({e^-st}{e^at} f(t) dt)
=Intergral 0-infinity ({e^-(s-a)t} f(t) dt)
=F(s-a) {Reason:Since Intergral 0-infinity ({e^-st} f(t) dt)=F(s)}
Hence Laplace {{e^at} f(t)}=F(s-a)..
In the qn here in the place of f(t) we have coskt..
Therefore we know laplace {coskt}=s/(s^2-k^2) = F(s)
Therefore F(s-a)=(s-a)/{(s-a)^2-k^2}
I think ive done my best to make the script understandable..
If not im really sorry about the inconvenience..