How can I deduce the value of $\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}\sin(y)e^{-\frac{(x-y)^2}{4t} } dy$ without actually evaluating it? How can I deduce that
$$
\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}\sin(y)\,e^{-\frac{(x-y)^2}{4t} } dy = e^{-t} \sin(x) 
$$
without actually evaluating the definite integral?
 A: I will assume the following result:

The solution to the Heat Equation
$$\frac{df}{dt} = \nabla^2f$$
with initial condition $f(x,0) = g(x)$ can be written
$$f(x,t) = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty e^{-\frac{(x-y)^2}{4t}}g(y) dy$$

Now by inserting
$$f(x,t) = e^{-t}\sin(x)$$
into the Heat Equation we find that it does satisfy it with the initial condition $f(x,0) = \sin(x)$. From the result above it therefore follows that
$$e^{-t}\sin(x) = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty e^{-\frac{(x-y)^2}{4t}}\sin(y) dy$$
A: Hint. By the change of variable $\displaystyle u=\frac{y-x}{2\sqrt{t}}$, you easily get
$$
\begin{align}
&\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{+\infty}\sin(y)\,e^{-\frac{(x-y)^2}{4t} } dy \\\\ &=\frac{1}{\sqrt{\pi} }\int_{-\infty}^{+\infty}\sin(2\sqrt{t}\: u+x)\,e^{-u^2 } du \\\\
&=\frac{1}{\sqrt{\pi} }\int_{-\infty}^{+\infty}\cos x\sin(2\sqrt{t}u)\,e^{-u^2 } du+\frac{1}{\sqrt{\pi} }\int_{-\infty}^{+\infty}\sin x\cos(2\sqrt{t}u)\,e^{-u^2 } du \\\\
&=0+\frac{\sin x}{\sqrt{\pi} }\int_{-\infty}^{+\infty}\cos(2\sqrt{t}\:u)\,e^{-u^2 } du  \\\\
&= \sin x \:e^{-t} 
\end{align}
$$ where we have just used the parity of one integrand and the classic gaussian evaluation
$$
\int_{-\infty}^{+\infty}\cos(a\:u)\,e^{-u^2 } du=\sqrt{\pi}\:e^{-a^2/4}.
$$
