calculus - How can I deduce the value of $frac{1}{sqrt{4pi t}}int_{-infty}^{infty}sin(y)e^{-frac{(x-y)^2}{4t} } dy$ 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?

Answer

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

$$e^{-t}\sin(x) = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty e^{-\frac{(x-y)^2}{4t}}\sin(y) dy$$