Approach to the integral $\int d^3u \exp(-\alpha|\mathbf{w}-\mathbf{u} |^2)\delta(\mathbf{k}\cdot\mathbf{u})$ I am trying to evaluate the integral
$\int d^3u \exp(-\alpha|\mathbf{w}-\mathbf{u} |^2)\delta(\mathbf{k}\cdot\mathbf{u})$
where I have 3 vectors w,u, and k, a constant alpha, and the integral is taken over all u (3 spatial coordinates). Usually for an integral of this form of the exponential I would just use a coordinate to make the exponential symmetric about the orgin. The issue here is that the delta function prevents (It makes the problem seemingly worse) me from effectively making this shift. I have played around with using the gaussian delta function sequence to integrate and then do the limit afterwards but this did not go so well. 
I was thinking that if I align $ \mathbf{u}$ along $\mathbf{w}$ and $\mathbf{k}$ along $\mathbf{w}$ so that
$\mathbf{k}=k_\perp cos(\theta)\hat{x}+k_\perp sin(\theta)\hat{y}+k_\parallel \hat{w}$
$\mathbf{u}=u_\perp cos(\theta)\hat{x}+u_\perp sin(\theta)\hat{y}+u_\parallel \hat{w}$
then I'd have $\mathbf{k}\cdot\mathbf{u}=k_\perp u_\perp+k_\parallel u_\parallel$
and I could attempt the integral in cylindrical coordinates.
Any suggestions on an approach to deal with the delta function would be great. 
Also If anyone has encountered a similar but slightly different integral Id be curious to know what approach you've used. 
 A: The Dirac delta distribution is non-zero only when its argument is 0. So integrating over all space is really not precise. Since  k is constant, the only contribution to the integral is in the space $\mathbf{k}\cdot \mathbf{u} = 0$ which you should be able to prove is a plane orthogonal to k. You should then be able to do a change of variables which lowers the dimension by 1, and you can go from there.
Alternatively, you can in fact do the linear shift--it should simplify things a bit.
Let $\mathbf{u}^\prime  = \mathbf{u} - \mathbf{w}$. Changing variables should get us $\int_{\mathbb{R}^3} \exp(-\alpha |\mathbf{u}^\prime|^2)\delta(\mathbf{k}\cdot\mathbf{u}^\prime + \mathbf{k}\cdot\mathbf{w}) d\mathbf{u}^\prime$
edit: To be clearer, express k, u, w in terms of orthonormal coordinates that are parallel and orthogonal to k.
A: If $k=(0,0,1)$, then the integral $I$ is actually two-dimensional and over the space $\{u_3=0\}$ and the integrand is $\exp(-\alpha(w_1-u_1)^2)\cdot\exp(-\alpha(w_2-u_2)^2)$. The change of variable 
$$(u'_1,u'_2)=(u_1-w_1,u_2-w_2)
$$ 
shows that $I$ is the square of
$$
\int_\mathbb R\mathrm e^{-\alpha v^2}\mathrm dv=\sqrt{\frac{\alpha}\pi},
$$
hence
$$
I=\frac{\alpha}\pi.
$$
The general case follows, by a homothety and a rotation.
