Proof of certain Gaussian integral form

Question

I am having trouble understanding where the following integral form comes from: $$\int_{-\infty}^{\infty} e^{-a x^2 }e^{-bx}=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}}$$ I see and understand that the value of the improper Gaussian integral is there, and I surmise that a change of variables is necessary when integrating, but I cannot figure out exactly how to prove this answer.

Answer 1

$$ \begin{aligned} \int_{-\infty}^{\infty} e^{-a x^2 }e^{-bx} \ dx &=\int_{-\infty}^{\infty} e^{-a\left( x+\frac{b}{2a} \right)^2 } e^{\frac{b^2}{4a}} \ dx \\ &=e^{\frac{b^2}{4a}}\int_{-\infty}^{\infty} e^{-a y^2 } \ dy \\ &= e^{\frac{b^2}{4a}} \sqrt{\frac{\pi}{a}} \end{aligned} $$

So there is a typo in the question, I think.

Answer 2

The best thing to do is to complete the square in the exponential term. Observe that

$$ ax^2 + bx \;\; =\;\; ax^2 + bx + \frac{b^2}{4a} - \frac{b^2}{4a} \;\; =\;\; a \left ( x + \frac{b}{2a} \right )^2 - \frac{b^2}{4a}. $$

We therefore obtain that

$$ \int_{-\infty}^\infty e^{-ax^2} e^{-bx} dx \;\; =\;\; e^{\frac{b^2}{4a}}\int_{-\infty}^\infty e^{-a \left ( x + \frac{b}{2a} \right )^2} dx \;\; =\;\; e^{\frac{b^2}{4a}} \sqrt{\frac{\pi}{a}} $$

where a change of variables $u = x + \frac{b}{2a}$ doesn't change the value of the integral. Can you check to see whether there is a $4$ in the denominator of your exponent in the answer you're supposed to obtain?