$\def\sp#1{\left\langle#1\right\rangle}$I am given $$ \Psi(x,0)=A_0 \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \cdot \exp\left(\frac{i}{\hbar}p_0x\right)\tag1$$
where $A_0=(\pi \sigma_0^2)^{-\frac{1}{4}}$
I am trying to find $\sp p$, which is $$\int\limits_{-\infty}^{+\infty}\Psi^*(x,0)\left(-i\hbar \frac{\partial \Psi(x,0)}{\partial x}\right)\,dx\tag2$$
I have $$\frac{\partial \Psi(x,0)}{\partial x}= A_0 \left[ \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x\right) \frac{i}{\hbar}p_0+ \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x]\right) \frac{2x}{2\sigma_0^2}2 \right]\tag3$$
Collecting terms:$$ \frac{\partial \Psi(x,0)}{\partial x}=A_0\cdot \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x\right)\left( \frac{i}{\hbar}p_0+\frac{2x}{\sigma_0^2}\right)\tag4$$
Multiplying by $-i\hbar$: $$ -i\hbar \frac{\partial \Psi(x,0)}{\partial x}=A_0\cdot \exp\left(-\frac{x^2}{2\sigma_0^2}\right) \exp\left(\frac{i}{\hbar}p_0x\right)\left(p_0-\frac{2i \hbar x}{\sigma_0^2}\right)\tag5$$
Premultiplying by $\Psi^*(x,0)$ and integrating:$$ \int\limits_{-\infty}^{+\infty} \Psi^*(x,0) \left(-i\hbar \frac{\partial \Psi(x,0)}{\partial x}\right)\,dx= \int\limits_{-\infty}^{+\infty} A_0 \exp\left( -\frac{x^2}{2\sigma_0^2}\right)\left(p_0-\frac{2i \hbar x}{\sigma_0^2}\right)\,dx\tag6$$
Now, it looks like I have an $i$ hanging around when I don't want one. $\sp p$ should be a real value, and I don't want any imaginary component. I'd appreciate it if someone could show me what I'm doing wrong. Thanks in advance, and apologies for the inconsistent formatting.
You are probably doing everything right. Look at the imaginary part of the integral that you have in the end: $$ \int_{-\infty}^{+\infty} A_0\exp\left(-\frac{x^2}{2\sigma_0^2}\right)\left(-\frac{2\hbar x}{\sigma_0^2}\right)dx. $$ The function under the integral is odd and quickly decreases when $x \to \infty$. Therefore this integral converges, and converges to $0$, because values on the positive half-axis "cancel out" with values on the negative half-axis. So your integral is real, you can just drop the imaginary part.