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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.

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up vote 1 down vote accepted

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.

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Could you explain "the integral I is actually two-dimensional and over the space {u3=0}". I'v been working on this for a while so its a bit hard to break free of the way I'v been visualizing the problem. – Anode Dec 14 '13 at 20:51
Sure, but you first: please define what you mean by a 3D integral with the Dirac delta in it. – Did Dec 14 '13 at 20:52
The integral only gives non zero values when k dot u =0 where u is the integration variable and k is some arbitrary direction. – Anode Dec 14 '13 at 20:54
I dont know if that answers your question? – Anode Dec 14 '13 at 21:13
Not at all. $ $ – Did Dec 15 '13 at 8:09

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.

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Could you explain your edit a little bit more? I guess I don't see how this is much different than aligning k and u along w. – Anode Dec 14 '13 at 20:47
shouldn't you have $\delta(\mathbf{k}\cdot(\mathbf{u}'+\mathbf{w})$? – Anode Dec 14 '13 at 20:57
Basically, you want to rotate your coordinate system so that you're only integrating in the plane orthogonal to k. In this system k = (0,0,1), u' = (u'_1,u'_2,u'_3). But since the delta is only non-zero when u'_3 != 0. Therefore, we can drop any contribution from u'_3. Did's answer is a very compact way of saying, translate (to get rid of w), then rotate (to get rid of the $\delta$). Make sure you prove that the integral remains invariant under these changes! – nimish Dec 14 '13 at 20:59
Oops, yes. Hence the nasty algebra... – nimish Dec 14 '13 at 21:00
what about the k dot w in the delta function? – Anode Dec 14 '13 at 21:03

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