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Upper limit gives zero 
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#1
Nov1907, 12:27 PM

P: 511

I was performing a Laplace transformation problem,where I happened to face:
{lower limit constant and upper limit infinity} exp [(is)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? 


#2
Nov1907, 01:19 PM

P: 511

OK,I am writing it in LaTeX:
[tex]\int^\infty_{2\pi/3}[/tex] [tex]\ e^{(is)t} [/tex] [tex]\ dt [/tex] 


#3
Nov1907, 01:22 PM

P: 511

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



#4
Nov1907, 01:57 PM

Math
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Thanks
PF Gold
P: 39,682

Upper limit gives zero
Well, obviously, the antiderivative is
[tex]\int e^{(is)t}dt= \frac{1}{is}e^{(is)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. 


#5
Nov1907, 07:13 PM

P: 511

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


#6
Nov1907, 08:12 PM

Sci Advisor
PF Gold
P: 1,594

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. 


#7
Nov1907, 08:28 PM

P: 511

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 


#8
Nov2007, 06:51 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682

My mistake. I had not realized that "i" was the imaginary unit! In that case, separate it into two parts. [itex]e^{(is)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 antiderivative goes to 0.
(The "ghost" of e^{it} I like that.) 


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