Integration, change of variables

**bobred** · February 16th 2013, 02:11 PM

Hi

I have the standard integral

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

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

**HallsofIvy** · February 16th 2013, 05:00 PM

Try $y= \sqrt{2}x$ .

**hollywood** · February 16th 2013, 11:56 PM

I think he means to substitute $y=\sqrt{\alpha}x$ , which should transform your integral into the standard integral.

- Hollywood

**bobred** · February 17th 2013, 12:57 AM

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

**Prove It** · February 17th 2013, 02:09 AM

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{e^{-\alpha \, x^2}\,dx} &= \int_{-\infty}^{\infty}{e^{-\left( \sqrt{\alpha} \, x \right) ^2 } \, dx} \\ &= \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{\sqrt{\alpha}\, e^{ -\left( \sqrt{\alpha} \, x \right) ^2 } \, dx } \end{align*}$

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

$\displaystyle \begin{align*} \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{\sqrt{\alpha}\, e^{-\left( \sqrt{\alpha} \, x \right) ^2 } \, dx} &= \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{ e^{-u^2} \, du } \\ &= \frac{1}{\sqrt{\alpha}} \, \sqrt{\pi} \\ &= \frac{\sqrt{\pi}}{\sqrt{\alpha}} \\ &= \sqrt{\frac{\pi}{\alpha}} \end{align*}$

**bobred** · February 17th 2013, 02:48 AM

Got it, thanks all.
James

Thread: Integration, change of variables

LinkBack

Thread Tools

Display

Integration, change of variables

Re: Integration, change of variables

Re: Integration, change of variables

Re: Integration, change of variables

Re: Integration, change of variables

Re: Integration, change of variables

Similar Math Help Forum Discussions

Is there a change of variables formula for integration using a generic measure?

Change of Variables

change of variables

change of variables

PDE; change of variables

Search Tags