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|∞0−p2a2∫∞0cos(px)exp(−a2x2)dx
The first term on the right vanishes, and we have the first-order differential equation:
I′(p)I(p)=−p2a2⇒ln(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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