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Let $f:mathbb{R}^d rightarrow mathbb{R}^d$ be the function with $f(x) = exp(-frac{1}{2}mid x mid^2)$. Show that the fourier transform of $f$ is given by $hat f = (sqrt{2 pi})^d f$.

The fourier transform is given by $hat f (xi ) := displaystyleint_{mathbb{R}^d} f(x) exp(-ixi cdot x) mu(dx)$.

I would like to start with looking at the case $d = 1$ but I dont know how to proceed.

It's a Gaussian integral---complete the square up in the exponential to get $(2pi)^{d/2} e^{-|xi|^2/2}$. Standard result on multivariate Gaussians (and no need to worry about Lebesgue).

For $d=1$, we have $f(x)=e^{-x^2/2}$ and therefore (remember that $(ixi+x)^2=-xi^2+2ixi x+x^2$)
begin{equation*}
begin{split}
hat{f}(xi) & = int_{-infty}^{infty} e^{-1/2(2ixi x + x^2)} {rm d}mu(x) \
& = int_{-infty}^{infty} e^{-1/2((ixi+x)^2+xi^2)} {rm d}mu(x) \
& = f(xi) int_{-infty}^{infty} e^{-(ixi+x)^2/2} {rm d}mu(x) \
& =f(xi) int_{-infty}^{infty} e^{-y^2/2} {rm d}mu(y) \
& = sqrt{2pi} f(xi).
end{split}
end{equation*}

It's pretty similar for $d>1$.