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: 
Let's rewrite the desired inequality as

$${\theta\over2}\lt\sin\theta\quad\text{for }0\lt\theta\le{\pi\over2}$$

To prove this, draw the portion of the unit circle in the first quadrant, draw a typical angle $\theta$ from the origin $O$ to a point $P=(\cos\theta,\sin\theta)$, and let $Q=(1,0)$. The wedge of the circle for this angle has area $\theta/2$. It consists of two pieces: the triangle $\triangle OPQ$, of area ${1\over2}\sin\theta$, and a lens-shaped piece outside the triangle.

Here's the key point to convince yourself of: If you reflect the lens-shaped region across the line $PQ$, it *stays inside the triangle* $\triangle OPQ$, and therefore has smaller area. The upshot is that the area of the wedge is less than twice the area of the triangle, which is exactly what we want.

**Added later**: Here's a second proof based on the same picture.

In addition to the points $P=(\cos\theta,\sin\theta)$ and $Q=(1,0)$, let $R=(1,y)$ be the point of intersection of the two tangents to the circle at $P$ and $Q$. 

Now the length of the circular arc from $Q$ to $P$ is, by definition, $\theta$ (when the angle is measured in *radians*). But this length is less than the sum of the lengths $QR$ and $PR$. By symmetry, those two lengths are the same. Thus, since the length of $PR$ is obviously just $y$, we have $\theta\lt2y$. But $R$ clearly lies closer to the $x$-axis than $P$, hence $y\lt\sin\theta$.
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, $$x>\frac{x^3}{3}\\1>\frac{x^2}{3}\\ \text{At the max of x, show } 1>\pi^2/12$$