# Integration, change of variables

• February 16th 2013, 02:11 PM
bobred
Integration, change of variables
Hi

I have the standard integral

${\displaystyle \int}_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$

But need to find what

${\displaystyle \int}_{-\infty}^{\infty} e^{-\alpha x^2} dx$ is using the expression above

I have found on the internet that it is $\sqrt{\frac{\pi}{\alpha}}$ but can't see how to get it.

I am assuming the use of changing the variable but can't see how to go about it.

Thanks, James
• February 16th 2013, 05:00 PM
HallsofIvy
Re: Integration, change of variables
Try $y= \sqrt{2}x$.
• February 16th 2013, 11:56 PM
hollywood
Re: Integration, change of variables
I think he means to substitute $y=\sqrt{\alpha}x$, which should transform your integral into the standard integral.

- Hollywood
• February 17th 2013, 12:57 AM
bobred
Re: Integration, change of variables
Thanks.

I can see how we get the substitution when we know the integral equals $\sqrt{\frac{\pi}{\alpha}}$, but how would we go about it not knowing what it equalled?

Thanks
• February 17th 2013, 02:09 AM
Prove It
Re: Integration, change of variables
\displaystyle \begin{align*} \int_{-\infty}^{\infty}{e^{-\alpha \, x^2}\,dx} &= \int_{-\infty}^{\infty}{e^{-\left( \sqrt{\alpha} \, x \right) ^2 } \, dx} \\ &= \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{\sqrt{\alpha}\, e^{ -\left( \sqrt{\alpha} \, x \right) ^2 } \, dx } \end{align*}

Now make the substitution \displaystyle \begin{align*} u = \sqrt{\alpha} \, x \implies du = \sqrt{\alpha} \, dx \end{align*} and note that \displaystyle \begin{align*} u \end{align*} still goes to \displaystyle \begin{align*} -\infty \end{align*} and \displaystyle \begin{align*} \infty \end{align*} when \displaystyle \begin{align*} x \end{align*} does. Then the integral becomes

\displaystyle \begin{align*} \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{\sqrt{\alpha}\, e^{-\left( \sqrt{\alpha} \, x \right) ^2 } \, dx} &= \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{ e^{-u^2} \, du } \\ &= \frac{1}{\sqrt{\alpha}} \, \sqrt{\pi} \\ &= \frac{\sqrt{\pi}}{\sqrt{\alpha}} \\ &= \sqrt{\frac{\pi}{\alpha}} \end{align*}
• February 17th 2013, 02:48 AM
bobred
Re: Integration, change of variables
Got it, thanks all.
James