Integral of exponential function with polynomial arguments with degree two $\int_{-\infty}^{\infty}e^{-ax^2}e^{-2bx}dx$.

Question

I have been using the following definite integral of exponential function for ages without knowing how to show it.

$$ \int_{-\infty}^{\infty}e^{-ax^2}e^{-2bx}dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}}. $$

Can anybody give some hints?

Answer 1

$$ \begin{align} I &=\int_{-\infty}^{+\infty}\exp\left[-ax^2-2bx\color{red}{-\frac{b^2}{a}+\frac{b^2}{a}}\right]\,dx \\ &=e^{\frac{b^2}{a}}\int_{-\infty}^{+\infty}\exp\left[-\left(\sqrt{a}x+\frac{b}{\sqrt{a}}\right)^2\right]\,dx \\ &\qquad\color{blue}{\left\{t=\sqrt{a}x+\frac{b}{\sqrt{a}} \implies dt=\sqrt{a}\,dx\right\}} \\ &=\frac{1}{\sqrt{a}}\,e^{\frac{b^2}{a}}\int_{-\infty}^{+\infty}e^{-t^2}\,dt = \color{red}{\sqrt{\frac{\pi}{a}}\,e^{\frac{b^2}{a}}} \end{align} $$ Where $\{\,\int_{-\infty}^{+\infty}e^{-t^2}\,dt=\sqrt{\pi}\,\}$ is Gaussian Integral.

Answer 2

Complete the square in the exponent. Convert the integrand into

$$\int e^k e^{(-(x-a)^2/2)}\ dx$$

then

$$e^k \int e^{x^2/2}\ dx$$

Answer 3

$$I=\int_{-\infty}^\infty e^{-ax^2}e^{-2bx}dx=\int_{-\infty}^\infty e^{-(ax^2+2bx)}dx$$ now: $$ax^2+2bx=\left(\sqrt{a}x+\frac{b}{\sqrt{a}}\right)^2-\frac{b^2}{a}$$ so our integral becomes: $$I=\int_{-\infty}^\infty e^{-\left(\left(\sqrt{a}x+\frac{b}{\sqrt{a}}\right)^2-\frac{b^2}{a}\right)}dx=e^{\frac{b^2}{a}}\int_{-\infty}^\infty e^{-\left(\sqrt{a}x+\frac{b}{\sqrt{a}}\right)^2}dx$$ now we let: $$u=\sqrt{a}x+\frac{b}{\sqrt{a}}$$ $$dx=\frac{du}{\sqrt{a}}$$ so now our integral becomes: $$I=\frac{e^{\frac{b^2}{a}}}{\sqrt{a}}\int_{-\infty}^\infty e^{-u^2}du=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}}$$ as this integral is a standard integral called the gaussian integral:

http://mathworld.wolfram.com/GaussianIntegral.html

and i mentioned how to solve it here: Determine whether the integral $ \int^{+\infty}_0\frac{e^{-t}} {\sqrt t} \, dt$ converges or diverges?