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Integral of e^(-as^2)cos(Bs)ds from 0-infinity

by schattenjaeger
Tags: 0infinity, eas2cosbsds, integral
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schattenjaeger
#1
Apr18-06, 03:51 PM
P: 178
I can do the gaussian function fine on 0-infinity, but I dunno how to do this. Integration by parts doesn't really help, so I tried expressing the cos as exponentials and piddled and fiddled and floundered, and here I am. I know what the answer is, but I can't get there
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nrqed
#2
Apr18-06, 07:12 PM
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P: 2,964
Quote Quote by schattenjaeger
I can do the gaussian function fine on 0-infinity, but I dunno how to do this. Integration by parts doesn't really help, so I tried expressing the cos as exponentials and piddled and fiddled and floundered, and here I am. I know what the answer is, but I can't get there
I would write the cos as a sum of two imaginary exponentials. In the second integral, I would make the change of variable s-> -s. Then the sum of the two integrals can now be combined into an integral from minus infinity to plus infinity. Now you complete the square of the total exponential and after a simple redefinition of variable, you end up with something proportional to the integral from minus infinity to plus infinity of e^(- C x^2) dx, which is well known.

Unless I missed something, this seems to be the way to go.

Pat
Jameson
#3
Apr18-06, 07:57 PM
P: 787
I know that the integral of something like [tex]\int e^xcos(x)dx[/tex] can be done by integration by parts twice, which results in the original integral being repeated. The using basic algebra the answer can be found. I don't know if this method will hold true for this integral though.

schattenjaeger
#4
Apr18-06, 08:05 PM
P: 178
Integral of e^(-as^2)cos(Bs)ds from 0-infinity

Thanks Jameson, the problem is if you apply integration by parts across that definite interval, you end up with an only marginally different integral, and applying it again gets you back where you started(which was kinda neat)that's the first thing I tried, also with the cos in exponential form(which I should've known just gets me the same effect)

and nrged, I think you got it, lemme run with it after I'm done eating and see if I get it. I got the whole thing down into two different integrals and tried to think of a useful change of variable I could do, but I was looking at the whooooole thing, not just a single integral after breaking it into two. Good eyes laddy ^_^
FireWalker877
#5
Mar2-09, 10:30 PM
P: 2
Hey guys, I'm bringing this one back from the dead. Would someone be so kind as to show the steps nrged has suggested. My integral is 2*integral{0,inf}( exp(-s^2)*cos(w*s)*ds ) I realize that the integral of e^(-s^2) = SQRT(pi). Would it be easier to evaluate the integral{-inf,inf}( exp(-s^2)*exp(-j*w*s)*ds ). This is one tricky fourier transform! Thanks!
FireWalker877
#6
Mar2-09, 10:47 PM
P: 2
Nevermind! I got it! That was a toughie. Thanks for the suggestions, guys from two years ago! :)

One of the important tricks is to use differentiation in the frequency domain before integration by parts.


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