Why is the upper limit zero in this Laplace transformation problem?

[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:

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

 

[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
$$\int e^{(i-s)t}dt= \frac{1}{i-s}e^{(i-s)t}$$
IF s> i, then the limit as t goes to infinity will be 0. If $s\le i$ 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 $s \leq i$ mean, considering that the complex numbers are not ordered?

If $s = \sigma + i\omega$, then

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

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

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. $e^{(i-s)t}= e^{it}e^{-st}$. The $e^{it}$ 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.)