Peskin & Schroeder: Free particle propagation In Peskin & Schroeder Ch. 2, p. 14, in proving that the NRQM propagation amplitude for a free particle is nonzero everywhere, they move from
\begin{equation}
U(t)~=~ \frac{1}{(2\pi)^3} \int d^3p \hspace{2pt} e^{-i(\mathbf{p}^2/2m)t} \cdot e^{i\mathbf{p}\cdot(\mathbf{x} - \mathbf{x_0})}
\end{equation}
to the end result:
\begin{equation}
U(t)~=~  \left(\frac{m}{2\pi i t}\right)^{3/2} \hspace{2pt} e^{im(\mathbf{x} - \mathbf{x_0})^2/2t}.
\end{equation}
I don't quite get all of the steps in between. In evaluating the first integral, I first put it in polar coordinates, with z along $x - x_0$, but then I eventually end up with a gaussian integration that looks like it should be zero. How do I get from the first equation to the second?
EDIT:
The next step I do after the above is to rewrite the integral:
\begin{equation}
\frac{1}{(2\pi)^2} \int_0^{\infty} \int_{1}^{-1} dp \hspace{2pt} d\cos \theta \hspace{2pt} p^2 e^{-i(\mathbf{p}^2/2m)t} \cdot e^{ip\Delta x \cos \theta}
\end{equation}
where $\Delta x \equiv |x - x_0|$ and I did the integration over $\phi$. From here, I get 
\begin{equation}
\frac{1}{(2\pi)^2i\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} \left( e^{-ip\Delta x} - e^{ip\Delta x} \right).
\end{equation}
It seems like this integration should give $0$, unless I'm making a mistake somewhere. Where's my mistake?
 A: Note the integral of an arbitrary Gaussian function, 
\begin{equation}
\int_{-\infty}^{\infty} e^{-ax^2 + bx} dx = \sqrt{\frac{\pi}{a}} ~e^{\frac{b^2}{4a}}
\end{equation} 
A: You seem to think that 
\begin{equation}
\frac{1}{(2\pi)^2i\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} \left( e^{-ip\Delta x} - e^{ip\Delta x} \right) = 0
\end{equation}
probably because the exponential functions kinda look like they cancel, but this is not so.
Note that 
$$ e^{-ip\Delta x} - e^{ip\Delta x} = -2i\sin(p\Delta x)$$
which means that you want to compute 
\begin{equation}
\frac{-2}{(2\pi)^2\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t}\sin(p\Delta x) 
\end{equation}
and it is quite obvious that this will not vanish. In fact we can do a little work on the second term in the original integral 
\begin{equation}
\frac{-1}{(2\pi)^2i\Delta x} \int_{0}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t} e^{ip\Delta x} 
\end{equation}
substitute $p' = -p$, then this equals
\begin{equation}
\frac{1}{(2\pi)^2i\Delta x} \int_{-\infty}^{0}dp' \hspace{2pt}p'\hspace{2pt}e^{-i(\mathbf{p'}^2/2m)t} e^{-ip'\Delta x} 
\end{equation}
so your original integral is just
\begin{equation}
\frac{1}{(2\pi)^2i\Delta x} \int_{-\infty}^{\infty}dp \hspace{2pt}p\hspace{2pt}e^{-i(\mathbf{p}^2/2m)t}  e^{-ip\Delta x} = \frac{1}{(2\pi)^2i\Delta x} i \frac{d}{d\Delta x}\int_{-\infty}^{\infty}dp \hspace{2pt}e^{-i(\mathbf{p}^2/2m)t}  e^{-ip\Delta x}
\end{equation}
this is just the derivative of a normal gaussian integral. Using the general formula for gaussian integrals
\begin{equation}
\int_{-\infty}^{\infty} e^{-ax^2 + bx} dx = \sqrt{\frac{\pi}{a}} ~e^{\frac{b^2}{4a}}
\end{equation}
which R.G.J. has already provided in his answer we immediately obtain
\begin{equation}
 (\frac{m}{2\pi it})^{3/2} \exp(i \frac{\Delta x^2 m}{2t})
\end{equation}
which is your desired result. 
A: Hint :
Make the hypothesis that the integral of an arbitrary Gaussian function (see R.G.J's answer)
\begin{equation}
\int_{-\infty}^{\infty} e^{\boldsymbol{-}ax^2\boldsymbol{+} bx} dx = \sqrt{\frac{\pi}{a}} ~e^{(b^2/4a)}
\tag{01}\label{eq01}    
\end{equation}
is valid for $\;a,b\;$ pure imaginary numbers and for our case
\begin{align}
a &=i\left(\dfrac{t}{2m}\right)
\tag{02.1}\label{eq02.1}\\
b_k &=i\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_k, \quad k=1,2,3
\tag{02.2}\label{eq02.2}
\end{align}
Then
\begin{align} 
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a p_1^2\boldsymbol{+}b_1 p_1}\mathrm dp_1 &=\sqrt{\dfrac{\pi}{a}}\,e^{(b_1^2/4a)}=\sqrt{\dfrac{2\pi m}{i t}}\,e^{im\vert\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_1\vert^2/2t}
\tag{03.1}\label{eq03.1}\\
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a p_2^2\boldsymbol{+}b_2 p_2}\mathrm dp_2 &=\sqrt{\dfrac{\pi}{a}}\,e^{(b_2^2/4a)}=\sqrt{\dfrac{2\pi m}{i t}}\,e^{im\vert\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_2\vert^2/2t}
\tag{03.2}\label{eq03.2}\\
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a p_3^2\boldsymbol{+}b_3 p_3}\mathrm dp_3 &=\sqrt{\dfrac{\pi}{a}}\,e^{(b_3^2/4a)}=\sqrt{\dfrac{2\pi m}{i t}}\,e^{im\vert\left(\mathbf{x}\boldsymbol{-}\mathbf{x}_0\right)_3\vert^2/2t}
\tag{03.3}\label{eq03.3}
\end{align} 
Multiplying above 3 equations side by side we have
\begin{equation}
\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}\int\limits_{\boldsymbol{-\infty}}^{\boldsymbol{+\infty}}e^{\boldsymbol{-}a \Vert\mathbf{p}\Vert^2\boldsymbol{+}\mathbf{b}\boldsymbol{\cdot}\mathbf{p}}\mathrm dp_1\mathrm dp_2\mathrm dp_3=\left(\dfrac{\pi}{a}\right)^{3/2}e^{(\Vert\mathbf{b}\Vert^2/4a)}=\left(\dfrac{2\pi m}{i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t}
\tag{04}\label{eq04}    
\end{equation}
or
\begin{equation}
\int\limits_{\mathbb{R}^3} e^{\boldsymbol{-}i (\Vert\mathbf{p}\Vert^2/2m)t\boldsymbol{+}i\mathbf{p}\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}_0\right)}\mathrm d^3\mathbf{p}=\left(\dfrac{2\pi m}{i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} 
\tag{05}\label{eq05}    
\end{equation}
and
\begin{equation}
\frac{1}{(2\pi)^3}\int\limits_{\mathbb{R}^3} e^{\boldsymbol{-}i (\Vert\mathbf{p}\Vert^2/2m)t\boldsymbol{+}i\mathbf{p}\boldsymbol{\cdot}\left(\mathbf{x}-\mathbf{x}_0\right)}\mathrm d^3\mathbf{p}=\left(\dfrac{m}{2\pi i t}\right)^{3/2}e^{im\Vert\mathbf{x}\boldsymbol{-}\mathbf{x}_0\Vert^2/2t} 
\tag{06}\label{eq06}    
\end{equation}
So try to prove the hypothesis.
