-1
$\begingroup$

I came across Gaussian integrals, and was trying to prove them myself. I proved the basics, but am stuck on the following

$$\int\limits_{-\infty}^{\infty}xe^{-a(x-b)^2}dx = b \sqrt{\frac{\pi}{a}}$$

I am having trouble thinking of how to attack this problem. I am currently thinking integration by parts to get rid of the $x$, and then try and coerce the exponent into the general Gaussian integral form but was wondering if there was a better solution.

$\endgroup$
1
  • 2
    $\begingroup$ Substitution with $u=x-b$. $\endgroup$ – Randall Sep 11 '18 at 3:04
2
$\begingroup$

HINT

If we make the substitution $y = x - b$, we get: \begin{align*} \int_{-\infty}^{+\infty} xe^{-a(x-b)^{2}}\mathrm{d}x & = \int_{-\infty}^{+\infty}(y+b)e^{-ay^{2}}\mathrm{d}y \\ & = \int_{-\infty}^{+\infty}ye^{-ay^{2}}\mathrm{d}y + \int_{-\infty}^{+\infty}be^{-ay^{2}}\mathrm{d}y\\ & = \int_{-\infty}^{+\infty}be^{-ay^{2}}\mathrm{d}y \end{align*}

once the function $ye^{-ay^{2}}$ is odd. Thus we have: \begin{align*} \left(\int_{-\infty}^{+\infty}be^{-ay^{2}}\mathrm{d}y\right)^{2} & = \left(\int_{-\infty}^{+\infty}be^{-ay^{2}}\mathrm{d}y\right)\times\left(\int_{-\infty}^{+\infty}be^{-az^{2}}\mathrm{d}z\right)\\ & = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}b^{2}e^{-a(y^{2} + z^{2})}\mathrm{d}y\mathrm{d}z \end{align*}

Hence, if you make the change of variable $y = \rho\cos(\theta)$ and $z = \rho\sin(\theta)$, you are able to obtain the sought result. Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.