Is it possible at all to make a Fourier Integral of a function such as

f(x) = x*exp(-x^2)

Any comments?

I guess it is (ik/2*2^1/2)exp(-k^2/4), if I rightly recall the Poisson's integral.

Yuriy, I think your 2^1/2 should be pi^1/2, and there should be a -1 in front of everything.


oxymoron, did you try to do the integration?

I have used the definition of Fourier-image F(k) of function F(x) as (1/(2pi)^1/2)*integral from -infinity to +infinity by dk from exp(+ikx)*F(x); so that Fourier-represantation of F(x) is (1/(2pi)^1/2)*integral from -infinity to +infinity by dk from exp(-ikx)*F(k). And I took Poisson's integral (from -infinity to +infinity) as (pi)^1/2.

Can you accomplish the same thing by another means of integration? Like say, take the natural log of both sides of the equation and then integrate by parts?

I do not understand your quastion.

Why not
∫ x*exp(-x^2)dx = -1/2 exp(-x^2)

Because
d/dx(-1/2 exp(-x^2)) = -1/2[d/dx(exp(-x^2))] = -1/2(-2x*exp(-x^2)) =x*exp(-x^2)

That was simple! I think I can do the Fourier Integral now.

Good luck.

I have used the definition of Fourier-image F(k) of function F(x) as (1/(2pi)^1/2)*integral from -infinity to +infinity by dk from exp(+ikx)*F(x); so that Fourier-represantation of F(x) is (1/(2pi)^1/2)*integral from -infinity to +infinity by dk from exp(-ikx)*F(k). And I took Poisson's integral (from -infinity to +infinity) as (pi)^1/2.

Ahh, right. I'm so used to exp(2*pi*i*k) as the kernel, I forgot about the -1 in the exp and the 1/(pi*2)^1/2 needed to make things symmetric when you drop the 2*pi in the exp.