Question: I got the following integral 
$$
\int\_0^t\int\_{-\infty}^{+\infty}\frac{1}{\sqrt{4\pi (t-s)}}\exp\left({-\frac{(x-y)^2}{4(t-s)}}\right) y\ dy\ ds\tag{\*}
$$
when I solve the heat equation using the heat kernel on the whole real line:

> 
> $$
> u\_t=u\_{xx}+x, \quad u(x,0)\equiv 0.
> $$
> 
> 
> 

On the other hand, it is easy to check that $$u(x,t)=xt$$ is a solution. 

I'm interested in evaluating (\*). But I don't see an approachable way to do it.

Answer: 
Since $\sin(x)$ is concave on $[0,\pi]$, we have
$$
\begin{align}
\sin(x)
&\ge\sin(0)+(x-0)\frac{\sin(\pi/2)-\sin(0)}{\pi/2-0}\\
&=\frac{2x}{\pi}\\
&\ge\frac x2
\end{align}
$$

$\hspace{8mm}$![enter image description here](https://i.stack.imgur.com/gVxPM.png)
Since $\sin x > x-\frac{x^3}{6}$ for $x>0$, we can show that $x-\frac{x^3}{6}>x/2$ with your condition. That is, $$x>\frac{x^3}{3}\\1>\frac{x^2}{3}\\ \text{At the max of x, show } 1>\pi^2/12$$