Close formula of Yukawa Potential in 3 dimension by integrating $\int_{0}^{\infty}{\frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}}dt$ In PDE, Yukawa potential can be calculate as
$$G^\mu(x) = \int_{0}^{\infty}  \frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}dt$$
When $d = 3$ we can get the close formula $G^\mu(x) = \frac{1}{4\pi|x|}e^{-\mu|x|}$.
How should we do the integral?
 A: Let
\begin{align}
I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right).
\end{align}
Then we see that
\begin{align}
I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{2}\left(\frac{1}{\sqrt{2t}}-\frac{\mu}{|x|} \sqrt{2t}\right)^2\right)\exp\left(-\mu |x|\right).
\end{align}
Set $u = \frac{1}{\sqrt{2t}}$ and $\alpha = \frac{\mu}{|x|}$ then it follows
\begin{align}
I(\mu, x) = \int^\infty_0 du\ \exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)(2\pi)^{-3/2}\exp(-\mu|x|) = \frac{e^{-\mu|x|}}{(2\pi)^{3/2}}J(\mu, x).
\end{align}
Hence it suffices to evaluate $J(\mu, x)$. Set $z=\alpha/u$ then we have
\begin{align}
J(\mu, x) = \int^\infty_0 \frac{dz}{z^2}\ \alpha\exp\left(-\frac{1}{2}x^2\left(z-\frac{\alpha}{z}\right)^2\right)
\end{align}
which means
\begin{align}
2J(\mu, x) =&\ \int^\infty_0 du\ \left(1+\frac{\alpha}{u^2} \right)\exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)\\
 =&\ \int^\infty_0 d\left(u-\frac{\alpha}{u} \right)\exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)\\
 =& \int^\infty_{-\infty} dw\ \exp\left(-\frac{1}{2}x^2w^2\right)= \sqrt{2\pi}\frac{1}{|x|}.
\end{align}
Hence it follows
\begin{align}
I(\mu, x) = \frac{e^{-\mu|x|}}{(2\pi)^{3/2}}J(\mu, x) = \sqrt{\frac{\pi}{2}}\frac{1}{|x|}\frac{e^{-\mu|x|}}{(2\pi)^{3/2}} = \frac{e^{-\mu |x|}}{4\pi |x|}.
\end{align}
Additional: Observe
\begin{align}
\int^\infty_0 dt\ \frac{1}{(4\pi t)^{5/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right) =&\ \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\left(-\frac{1}{2\pi x}\right)\frac{d}{dx}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right)\\
 =&\ \left(-\frac{1}{2\pi x}\right)\frac{d}{dx}I(\mu, x) = \frac{e^{-\mu|x|}(\mu|x|+1)}{8\pi^2 |x|^3}.
\end{align} 
Using this observation, we can recover all expression of $G^\mu(x)$ for odd dimension $d$. 
A: Using a CAS
$$G^\mu(x)  = \int_{0}^{\infty}  \frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t} dt$$ is given, before any simplication or assumptions by
$$G^\mu(x)= (2 \pi )^{-\frac d2} \left(\frac{\mu ^2}{x^2}\right)^{\frac{d-2}{4}}
   K_{\frac{d-2}{2}}\left(\sqrt{x^2 \mu ^2}\right)$$ which gives your result for $d=3$. 
Bessel functions only appear for even values of $d$.
