Please tell me where I am going wrong in this integral

[bluepilotg-2_07]

Homework Statement

Given the wavefunction, determine the expectation value of the particle's position

Relevant Equations

$\Psi=\frac{1}{\sqrt{\pi}(b)^{1/4}}\exp(\frac{ip_0x}{\hbar}-\frac{(x-x_0)^2}{2b^2})$

$|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).$
$\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).$
$y=x-x_0 \quad x=y+x_0 \quad dy=dx.$
The boundaries remain infinite, I believe.
$\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).$
$\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).$
I then resolved the two integrals separately using the relations for Gaussian integrals in the back cover of Introduction to Quantum Mechanics by Griffiths.
$\frac{b}{\sqrt{\pi}}+x_0.$
This is not correct. The result should be $x_0$. I have retraced my steps multiple times and have redone the calculation using a different substitution but get the same result. Would appreciate someone pointing out the problem.

 

[TSny]

In the second line, the first integral has an integrand that is odd in $y$. You cannot replace $\int_{-\infty}^{\infty}$ by $2\int_{0}^{\infty}$.

 

[bluepilotg-2_07]

Hello, thank you for the response! I should have known better, but wasn't considering even and oddness. I realize now that the first integral goes to zero since it is odd and being integrated over symmetric boundaries and the rest is trivial.
Thanks again!