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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.

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  • $\begingroup$ This is very similar to the Laplace transform or the moment generating function of a normal random variable. Those calculations will show your integral. You could try completing the square too. $\endgroup$ – jdods Mar 7 '15 at 23:59
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$$ \begin{aligned} \int_{-\infty}^{\infty} e^{-a x^2 }e^{-bx} \ dx &=\int_{-\infty}^{\infty} e^{-a\left( x+\frac{b}{2a} \right)^2 } e^{\frac{b^2}{4a}} \ dx \\ &=e^{\frac{b^2}{4a}}\int_{-\infty}^{\infty} e^{-a y^2 } \ dy \\ &= e^{\frac{b^2}{4a}} \sqrt{\frac{\pi}{a}} \end{aligned} $$

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

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  • $\begingroup$ Thank you a lot, and yes, that was a typo. $\endgroup$ – Ice Felix Mar 8 '15 at 0:21
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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?

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    $\begingroup$ I got the same thing as you. $\endgroup$ – jdods Mar 8 '15 at 0:06
  • $\begingroup$ Yes, it was a typo. Thank you very much, I would put an up arrow if my reputation allowed me too (I gave the answer to jdods because he answered first). $\endgroup$ – Ice Felix Mar 8 '15 at 0:24
  • $\begingroup$ well, technically Mnifldz "answered first"... but I did comment that completing the square was the trick first... lol. Can we split the points? ;) $\endgroup$ – jdods Mar 8 '15 at 0:26
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    $\begingroup$ It's fine either way. @jdods Enjoy the points. IceFelix, glad to help. $\endgroup$ – Mnifldz Mar 8 '15 at 0:28

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