<s>[INST] 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
 [/INST] 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
$$u(x,t)=\frac{\sqrt{\frac{\pi }{2}} e^{-\frac{(x +i)^2}{16t}} \left(e^{\frac{2 i x }{8t}} \text{erfc}\left(\frac{1+i x }{\sqrt{8t} }\right)-\text{erf}\left(\frac{-1+ix}{\sqrt{8t}}\right)+1\right)}{16t }.$$
</s>