Integration in Laplace Transform

[juan.]

Hello everyone, I have a question about integrating in Laplace Transform. For example, if I have:
f(t)=e^{i.t}
I have to solve this equation:
\int_{0}^{\infty}e^{i.t}.e^{-s.t}dt

If I do like this, it's very simple:
\int_{0}^{\infty}e^{i.t}.e^{-s.t}dt=\int_{0}^{\infty}e^{-t.(s-i)}dt=\frac{-1}{s-i}.(0-1)=\frac{1}{s-i}

But, if I do like this, I can't solve it:
\int_{0}^{\infty}e^{i.t}.e^{-s.t}dt=\int_{0}^{\infty}e^{t.(i-s)}dt=\frac{1}{i-s}.(?? - 1)= ??

Where it says '??', I don't know what to write.

Any help? Thanks!

 

[da_nang]

You're technically dealing with a complex variable s and an improper integral. So what you really want to calculate is
\lim_{n \rightarrow \infty}\int_{0}^{n}e^{it}e^{-(a+b i)t}dt = \lim_{n \rightarrow \infty}\int_{0}^{n}e^{(-a+(1-b)i)t}dt = \lim_{n \rightarrow \infty} \frac{1}{-a+(1-b)i} \left ( e^{(-a+(1-b)i)n} - 1 \right) .

 

[Hawkeye18]

The Laplace transform here is defined for $s>0$. Integrating you get $$\int_0^\infty e^{t(i-s)} dt = \frac{e^{t(i-s)}}{i-s} \Bigm|_{t=0}^{t=\infty} = \frac1{i-s} (0-1)= \frac1{s-i} .$$ Here by the value at $t=\infty$ you should understand the limit as $t\to\infty$, and this limit is $0$ if $s>0$. You can also consider complex $s$, the same computation works and the limit as $t\to\infty$ is still zero if $\operatorname{Re}s>0$. In your computations you skipped on step (did not write the antiderivative), maybe that was the source of your confusion.

 

[juan.]

Thanks for your answers!
I can see now that e^{t.(i-s)}, with t->\infty is equal to 0, but why?
I can separate that exponential in 2 parts: real and immaginary. In the real part, of course I have e^{-t.s}=0 but in the immaginary part I have: e^{i.t} and that's not 0.
So, I have 0.\infty, with the infinity in the immaginary. Isn't that strange? Why do you say that is 0?

 

[Hawkeye18]

For real $t$ we have $|e^{it}|=1$, so we have something bounded ($e^{it}$) times something with limit $0$. The resulting limit is then $0$.

 

[juan.]

Oh, that's true!
ThanksssssssssssssssssssssSorry, Euler, I forgot about you haha