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?

Question

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 1

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

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

Answer 2

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 $$ \int_{-\infty}^{+\infty}\cos(a\:u)\,e^{-u^2 } du=\sqrt{\pi}\:e^{-a^2/4}. $$