(Hopefully easy) integration question

[lotm]

Heyo,

I'm having difficulty seeing how these two lines follow. I'm fairly sure I'm being an eejit and the answer's straightforward, but would appreciate a quick explanation of what's going on.

$$\frac{1}{2\pi}\int d^3p e^{-i(\emph{p}^2/2m)t} \\ \times e^{i\emph{p.(x-x_0)}} \\ = (\frac{m}{2 \pi it})^{3/2}e^{\frac{im(\emph{x-x_0}^2}{2t}}$$

Thanks in advance.

 

[CompuChip]

Which two lines do you mean exactly?
I only see a single expression.

 

[lotm]

Yeah, I'm struggling a bit to get the tex code right. Screw it, I'll come back once I've figured out how to ask the question properly.

 

[homology]

Do you mean:

$$\frac{1}{2\pi}\int d^3p e^{-i(\vec{p}^2/2m)t} e^{i\vec{p}\cdot(\vec{x}-\vec{x_0})} = \left (\frac{m}{2 \pi it}\right )^{3/2}e^{\frac{im(\vec{x-x_0})^2}{2t}}$$

 

[homology]

If that's the case then you can click on the equation to see the markup.

What you've got here, is a Fourier Transform of a Gaussian returning a Gaussian. It is easy though it might look daunting. Complete the square and use a little substitution and you'll be all set.