Finding fourier Transform of $x_i x_j e^{-\pi|x|^2}$ I have a function $f : x \in \mathbb{R}^d \rightarrow x_i x_j e^{-\pi |x|^2}$ and need to find its Fourier Transform for any $i,j=1,...,d$. 
I thought of making two cases $i=j$ and $i\neq j$ and since I know that $\widehat{e^{-\pi|x|^2}}(\xi)=e^{-\pi |\xi|^2}$ it shouldn't be to hard to calculate the cases independent of $i,j$, but I don't see how to evaluate the two appearing integrals which depend on $i,j$:
for $i=j$: $\int_\mathbb{R}x_i^2 e^{-2\pi i x_i\xi_i}e^{-\pi x_i^2}dx_i$ and
for $i\neq j$: $\int_\mathbb{R}\int_\mathbb{R}x_i x_j e^{-2\pi i (x_i\xi_i+x_j\xi_j)}e^{-\pi (x_i^2+x_j^2)}dx_idx_j$
Can anybody help me with this, or is there a more elegant way to calculate the Fourier Transform of this function?
 A: Hint: for $i\ne j$ you wish to Fourier-transform $-\frac{1}{4\pi^2}\partial_i\partial_j e^{-\pi\sum_i x_i^2}$. I'm sure you can express the FT of $\partial_i f$ in terms of the FT of $f$. The case $i=j$ is similar.
A: We can change variables carefully: firstly, complete the square in the exponent:
$$ -\pi\lVert x \rVert^2 -2\pi i \langle x, \xi \rangle = -\pi \lVert x+i\xi \rVert^2 -\pi\lVert \xi \rVert^2 $$
So we can take out the $e^{-\pi\lVert \xi \rVert^2}$ as usual, and now, let $y=x-i\xi$. Then the integral becomes
$$ e^{-\pi\lVert \xi \rVert^2} \int (y_i-i\xi_i)(y_j-i\xi_j) e^{-\pi \lVert y \rVert^2} \, dy. $$
Expanding the brackets gives you three types of term: $-\xi_i\xi_j \int e^{-\pi \lVert y \rVert^2} \, dy = -\xi_i\xi_j $, two linear terms that look like $-\xi_i \int y_j e^{-\pi \lVert y \rVert^2} \, dy $, which are zero using integration by parts, and
$$ \int y_i y_j e^{-\pi \lVert y \rVert^2} \, dy, $$
which is zero if $i \neq j$, and by homogeneity,
$$ \int y_i y_i e^{-\pi \lVert y \rVert^2} \, dy $$
This last can be found by calculating $ \int_{-\infty}^{\infty} u^2 e^{-\pi u^2} du $ and multiplying it by the integrals of the other components, which turn out to all be $1$, so this term is $1/2\pi$. So we find
$$ \int x_i x_j e^{-\pi\lVert x \rVert^2 -2\pi i \langle x, \xi \rangle} dx = \left(\frac{1}{2\pi}\delta_{ij} - \xi_i \xi_j \right)e^{-\pi\lVert \xi \rVert^2}. $$
