Im trying to evaluate the following integral
$$ \frac{1}{4 \pi \sqrt{t} } \int\limits_{-\infty}^{\infty} \exp\left( - \frac{(y+2)^2 }{4} - \frac{(x-y)^2 }{4t} \right) - \exp\left( - \frac{(y-2)^2 }{4} - \frac{(x-y)^2 }{4t} \right) dy$$
now, clearly, we can use the fact that $\int_{-\infty}^{\infty} e^{-y^2} dy = \sqrt{ \pi }$.
Looks messy. I tried to see if we can rewrite this in a more suggestive form: for instance take the first exp: we have in the exponent
$$ -\frac{1}{4} \left( y^2 + 4 y + 4 + x^2 - 2xy + y^2 \right) $$
but, this doesnt seem to help too much. Can someone help me finding the correct substitution to simplify this intergal as much as possible?
HINT
Assuming $x$ is constant with respect to $y$, you have 2 different integrals like $$ \int_{-\infty}^\infty \exp\left(-\frac{(y+2)^2}{2^2}\right)dy $$ and using $u = (y+2)/2$ transforms the integrand into $e^{-u^2}$...
UPDATE
$$ \begin{split} \frac{(y+2)^2 }{4} + \frac{(x-y)^2 }{4t} &= \frac{ty^2+2ty+4t+y^2+x^2-2xy}{4t} \\ &= \frac{(t+1)y^2+2y(t-x)+4t+x^2}{4t} \\ &= \frac{y^2+2y\frac{t-x}{t+1}+\frac{4t+x^2}{t+1}}{4t(t+1)} \\ &= \frac{(y+A)^2}{2B}+C \end{split} $$ So if $u = (y+A)/\sqrt{B}$ and $dy = du/\sqrt{B}$, we have $$ \int \exp\left(-\frac{(y+A)^2}{2B}-C\right) dy = e^{-C} \int \exp \left(-\frac12\left(\frac{y+A}{\sqrt{B}}\right)^2\right) dy = \frac{1}{e^C\sqrt{B}} \int e^{-u^2/2}du $$