How do I convert an integral from LHS to RHS?

[Ratzinger]

I have a problem with a integral. Could someone tell me how to get from the LHS to RHS? It must be very easy, but as so often I don't get it.

Since I have not mastered Latex yet, I attached a pdf file.

thanks

 

[HallsofIvy]

In LaTex what you have is
\frac{1}{k^2+ m^2}\int e^{i\vec{k}\dot\vec{r}}d\vec{k}= \int k d\phi \intk sin(\theta)d\theta \int \frac{e^{ik|r|cos(/theta)}}{k^2+ m^2}dk
(Click on that for a pop-up box showing the code.)

I had to stare at that a while until I realized- those blasted engineers have mixed up \theta and [/itex]\phi[/itex] again! Also, it is using "k" for the length of \vec{k}. I take it that "\vec{k}" is the position vector of a point in the space being integrated over (here all of R³).

The right hand side is in spherical coordinates, using k instead of the more standard \rho. In spherical coordinates (mathematics version!)
dV= \rho^2 sin(\phi)d\rho d\phi d\theta
In "engineer speak", as here, it is
dV= \k^2 sin(\theta)dk d\theta d\phi

That has been separated into
(k d\phi)(k sin(\theta)d\theta)(dk)
Of course, since \vec{u}\dot\vec{v}= |u||v|cos(\theta), that
i\vec{k}\dot\vec{r}= ik|r|cos(\theta)[/tex]<br /> <br /> It&#039;s not clear to me that breaking it up that way is valid, because clearly k is a variable (dk in one integral) and so putting k into the d\theta and d\phi integrals doesn&#039;t make sense.

 

[Galileo]

Basically, all that's happened is writing the integral over all space in spherical coordinates d^3\vec k=k^2\sin \theta dk d\theta d\phi and using \vec k \cdot \vec r=|r|k\cos \theta.
But the way it's written it looks like a product of three integrals, which it is not!
And unless \vec r points along the z-axis you must not confuse the angle between \vec k and \vec r with the angle \theta over which you are integrating.

 

[Ratzinger]

Thanks HallsofIvy, thanks Galileo!

Good to know that this time it was not all my fault.