[SOLVED] integral

**akane** · May 30th 2010, 07:44 AM

$\int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} = \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz + z^2}{2 \sigma^2}}du} = e^{-\frac{z^2}{2 \sigma^2}} \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz}{2 \sigma^2}}du}$

$= e^{-\frac{z^2}{2 \sigma^2}} \int_{\infty}^{\infty} {e^{-\frac {u^2 - uz}{ \sigma^2}}du}$

what now?

sorry, I didn't mean to put this thread here...

**Anonymous1** · May 30th 2010, 11:37 AM

Originally Posted by akane

$\int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} = \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz + z^2}{2 \sigma^2}}du} = e^{-\frac{z^2}{2 \sigma^2}} \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz}{2 \sigma^2}}du}$

$= e^{-\frac{z^2}{2 \sigma^2}} \int_{\infty}^{\infty} {e^{-\frac {u^2 - uz}{ \sigma^2}}du}$

what now?

sorry, I didn't mean to put this thread here...

You want to use the fact that $\int f_X\text{ dx }= 1.$

Turn your integrand into a normal pdf.

**akane** · May 30th 2010, 12:32 PM

Originally Posted by Anonymous1

You want to use the fact that $\int f_X\text{ dx }= 1.$

Turn your integrand into a normal pdf.

I have this:

$g_{Z}(z) = \int_{-\infty}^{\infty} f(u,u-z) du = \int_{-\infty}^{\infty} \frac{1}{2 \pi \sigma^2} e^{-\frac{u^2+(u-z)^2}{2 \sigma^2}} du$

Z should also be a normal variable, I don't know how to get a function of only one variable under the integral

**Anonymous1** · May 30th 2010, 05:22 PM

Have you ever heard of the error function?

$\int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} =\left(\frac{1}{2}\sqrt{\pi\sigma^2} e^{-\frac{z^2}{4\sigma^2}}\right)\cdot\left[\text{erf}(-\frac{z-2u}{2\sigma})|_{-\infty}^{+\infty}\right]$

Also, $\text{erf}(\infty)=1$ and $\text{erf}(-\infty)=-1$

**akane** · May 30th 2010, 08:50 PM

Originally Posted by Anonymous1

Have you ever heard of the error function?

$\int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} =\left(\frac{1}{2}\sqrt{\pi\sigma^2} e^{-\frac{z^2}{4\sigma^2}}\right)\cdot\left[\text{erf}(-\frac{z-2u}{2\sigma})|_{-\infty}^{+\infty}\right]$

Also, $\text{erf}(\infty)=1$ and $\text{erf}(-\infty)=-1$

thanks for help

I've heard of it, but this is actually the first time it appeared in my assignment so I wasn't able to recognize it

**CaptainBlack** · May 30th 2010, 09:30 PM

Originally Posted by akane

$\int_{-\infty}^{\infty} {e^{-\frac {u^2 + (u-z)^2}{2 \sigma^2}} du} = \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz + z^2}{2 \sigma^2}}du} = e^{-\frac{z^2}{2 \sigma^2}} \int_{-\infty}^{\infty} {e^{-\frac {2u^2 - 2uz}{2 \sigma^2}}du}$

$= e^{-\frac{z^2}{2 \sigma^2}} \int_{\infty}^{\infty} {e^{-\frac {u^2 - uz}{ \sigma^2}}du}$

what now?

sorry, I didn't mean to put this thread here...

You have:

$\int_{-\infty}^{\infty} e^{-\frac{u^2+(u-z)^2}{2\sigma^2}}\;du$

To do this we complete the square of the integrand:

$u^2+(u-z)^2=\left(\sqrt{2}u-\frac{z}{\sqrt{2}}\right)^2+\frac{z^2}{2}$

So:

$\int_{-\infty}^{\infty} e^{-\frac{u^2+(u-z)^2}{2\sigma^2}}\;du<br /> =e^{\frac{z^2}{4\sigma^2}} \int_{-\infty}^{\infty}$ $<br /> <br /> e^{-\frac{\left(\sqrt{2}u-z/\sqrt{2}\right)^2}{2\sigma^2}}\;du<br />$

Now make the change of variable:

$v=\frac{\sqrt{2}u-z/\sqrt{2}}{\sigma}$

and see what happens.

CB

Thread: [SOLVED] integral

LinkBack

Thread Tools

Display

[SOLVED] integral

Similar Math Help Forum Discussions

[SOLVED] integral of (t)(cos(3t^2))dt

[SOLVED] integral

[SOLVED] Another integral

[SOLVED] Line integral, Cauchy's integral formula

[SOLVED] Line integral, Cauchy's integral formula

Search Tags