Sunday, October 27, 2019

integration - How to evaluate the following integral involving a gaussian?



I want to evaluate the following integral:



0xsinpxexp(a2x2)dx



Now I am unsure how to proceed. I know that this is an even function so I can extend the limit terms to , and then divide by 2. I have tried to evaluate this on Wolfram Alpha, but it only shows the answer while I am interested in the procedure.


Answer




Start with:
I(p)=0cos(px)exp(a2x2)dx
We can use differentiation under the integral sign:



I(p)=0xsin(px)exp(a2x2)dx
Integration by parts using u=sin(px)anddv=xexp(a2x2)dx
I(p)=sin(px)exp(a2x2)2a2|0p2a20cos(px)exp(a2x2)dx
The first term on the right vanishes, and we have the first-order differential equation:
I(p)I(p)=p2a2ln(I(p))=p24a2+C
Using I(0)=πa

We can find C=ln(πa)
hence
ln(I(p))=p24a2+ln(πa)
So
I(p)=πaexp(p24a2)
Finally the integral in question equals
I(p)=ddp(πaexp(p24a2))


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