Upper limit gives zero

neelakash

I was performing a Laplace transformation problem,where I happened to face:

{lower limit constant and upper limit infinity} exp [(i-s)t] the variable being t.

I am not sure if the upper limit gives zero,but if I assume that the answer becomes correct.

Can anyone please tell me why the upper limit is zero here?

neelakash

OK,I am writing it in LaTeX:

[tex]\int^\infty_{2\pi/3}[/tex] [tex]\ e^{(i-s)t} [/tex] [tex]\ dt [/tex]

neelakash

Friends,I do not know why the exponential dd not appear.Please assume this and let me know:

HallsofIvy

Well, obviously, the anti-derivative is
[tex]\int e^{(i-s)t}dt= \frac{1}{i-s}e^{(i-s)t}[/tex]
IF s> i, then the limit as t goes to infinity will be 0. If [itex]s\le i[/itex] the integral does not exist.

neelakash

I see...Thank you.

Can you please tell me why the integral does not exist for i<s?

Ben Niehoff

Moreover, what does an expression like [itex]s \leq i[/itex] mean, considering that the complex numbers are not ordered?

If [itex]s = \sigma + i\omega[/itex], then

[tex]\begin{array}{rcl}\frac 1 {i - s} e^{(i -s)t} & = & \frac1 {i - \sigma - i\omega} e ^{(i- \sigma - i\omega)t}\\&&\\ &=& \frac 1 {-\sigma + i(1-\omega)} e^{i(1-\omega)t} e^{-\sigma t}\end{array}[/tex]

which has a finite limit at infinity only if [itex]\sigma > 0[/itex], and hence [itex]\Re(s) > 0[/itex].

Hmm...maybe that's what was meant originally, then.

neelakash

Yes,I afree.
If you take the modulus,the ghost of exp[it] runs away and the thing goes to zero as t tends to infinity

HallsofIvy

My mistake. I had not realized that "i" was the imaginary unit! In that case, separate it into two parts. [itex]e^{(i-s)t}= e^{it}e^{-st}[/itex]. The [itex]e^{it}[/itex] part oscilates (it is a sin, cos combination) while e^{-st} will go to 0 as t goes to infinity because of the negative exponential. The entire product goes to 0 very quickly so the anti-derivative goes to 0.

(The "ghost" of e^it- I like that.)