Solve the integral of heat equation with only boundary condition $u\left(x,0\right)=\dfrac{1}{x^2+1}$ with $x \in R$ The most thing I want is solve this integral
$$\frac{1}{{16t\sqrt \pi  }}\int\limits_{ - \infty }^{ + \infty } {\frac{{\exp \left( { - {a^2}} \right)}}{{{{\left( {a + \frac{x}{{4\sqrt t }}} \right)}^2} + \frac{1}{{16t}}}}} da$$
I have the PDE that
$
\begin{cases}
u_t=4u_{xx} & -\infty < x < + \infty\\
u\left(x,0\right)=\dfrac{1}{x^2+1} &
\end{cases}
$
When $-\infty < x < + \infty$, I use Poisson's equation, so that
$$
\begin{aligned}
u\left( {x,t} \right) = \frac{1}{{4\sqrt {\pi t} }}\int\limits_{ - \infty }^{ + \infty } {{e^{ - \dfrac{{{{\left( {\xi  - x} \right)}^2}}}{{16t}}}}.\frac{1}{{{\xi ^2} + 1}}d\xi } 
\end{aligned}
$$
with $\dfrac{{\xi  - x}}{{4\sqrt t }} = a \Rightarrow \left\{ \begin{array}{l}
d\xi  = 4\sqrt t da\\
\xi  = x + 4a\sqrt t 
\end{array} \right.$
I have
$$
\begin{aligned}
u\left( {x,t} \right) &= \frac{1}{{\sqrt \pi  }}\int\limits_{ - \infty }^{ + \infty } {\exp \left( { - {a^2}} \right).\frac{1}{{16t{a^2} + 8xa\sqrt t  + {x^2} + 1}}da} \\
 &  = \frac{1}{{\sqrt \pi  }}\int\limits_{ - \infty }^{ + \infty } {\exp \left( { - {a^2}} \right).\frac{1}{{16t\left( {{a^2} + \frac{{xa}}{{2\sqrt t }} + \frac{{{x^2} + 1}}{{16t}}} \right)}}da} \\
 &  = \frac{1}{{\sqrt \pi  }}\int\limits_{ - \infty }^{ + \infty } {\exp \left( { - {a^2}} \right).\frac{1}{{16t\left( {{a^2} + \frac{{xa}}{{2\sqrt t }} + \frac{{{x^2} + 1}}{{16t}}} \right)}}da} \\
 &  = \frac{1}{{16t\sqrt \pi  }}\int\limits_{ - \infty }^{ + \infty } {\frac{{\exp \left( { - {a^2}} \right)}}{{{{\left( {a + \frac{x}{{4\sqrt t }}} \right)}^2} + \frac{1}{{16t}}}}} da
\end{aligned}
$$
I dont know how to solve this integral, anyone can help me for this integral or other solution of this PDE. Many thanks

We convolve the initial condition with the fundamental solution:
$$u(x,t)=\int_{\mathbb{R}}\frac{1}{1+y^2}\frac{1}{\sqrt{\pi 16t}}\exp\bigg\{-\frac{(x-y)^2}{16t}\bigg\}dy=\mathbb{E}_{Y \sim\mathcal{N}(x,8t)}\bigg[\frac{1}{1+Y^2}\bigg]$$
This  has been solved on stats.stackexchange. The solution to the PDE is
$$u(x,t)=\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{(x +i)^2}{16t}} \left(e^{\frac{2 i x }{8t}} \text{erfc}\left(\frac{1+i x }{\sqrt{8t} }\right)-\text{erf}\left(\frac{-1+ix}{\sqrt{8t}}\right)+1\right)}{16t }.$$
