Exponential definite integral

Question

Please can anybody tell the exact solution or the way to solve the integral of form $$\int_{-\infty}^{\infty} e^{-2b^2(x^2+a^2)}dx$$

Answer 1

$$ \int_{-\infty}^{\infty} e^{-2b^2(x^2+a^2)}dx=e^{-2b^2a^2}\int_{-\infty}^{\infty} e^{-2b^2 x^2}dx=e^{-2b^2a^2}\sqrt{\frac{\pi}{2b^2}} $$ done using $$ \int_{\mathbf R} e^{-\alpha x^2} {\rm dx}=\sqrt{\frac{\pi}{\alpha}} $$

Answer 2

after taking out $e^{-2b^2a^2}$, substitute $\sqrt 2bx=t$. this leaves us with $\int_{-\infty}^{\infty}e^{-t^2}\text dt$ with some factor multiplied. Now we substitute $t = \sqrt { y}$ leaves us with $\Gamma({\frac 12})$.