Something about all this stuff I am not getting. Given the following wave function $$\psi = Ne^{\frac{-x^2}{\Delta^2}}e^{ik_0 x}$$ Where $N$ is the normalization constant, $N= \left(\frac{1}{\Delta\sqrt{\pi}}\right)^{1/2}$. I need to find the average momentum, $\langle p\rangle$, two ways. $$ \langle p\rangle = \int_{-\infty}^{\infty}~\mathrm dx\psi^*\left(-i\hbar\frac{\mathrm d}{\mathrm dx}\right)\psi=\int_{-\infty}^{\infty}\frac{~\mathrm dp}{2\pi}|\phi|^2p $$ Question: What is this second integral? I do not know what $\phi$ is. Any help for the context of this would be greatly appreciated. I do not know where to start with that integral.
Solving the first integral: For the first integral, which is using the momentum operator. I evaluated the derivative of the wave function first. $$\frac{\mathrm d}{\mathrm dx}e^{-\frac{x^2}{\Delta^2}+ik_0x}=\left(\frac{-2x}{\Delta^2}+ik_0\right)e^{-\frac{x^2}{\Delta^2}+ik_0x}$$ Taking note that $\psi^*\psi$ removes the complex part and multiplies a 2 to the real part of the exponent. I should get something like this: $$\langle p\rangle=(-i\hbar)N^2\int_{-\infty}^{\infty}~\mathrm dx \left(\frac{-2x}{\Delta^2}+ik_0\right)e^{-\frac{2x^2}{\Delta^2}}$$ Which I can split into two integrals. One of the integrals is odd and evaluates to $0$. The other remains. $$=(-i\hbar)N^2\int_{-\infty}^{\infty}~\mathrm dx(ik_0)e^{-\frac{2x^2}{\Delta^2}}$$ Which is a the Gaussian integral and the solution to this is $$\begin{align} &=N^2k_0\sqrt{\frac{\pi\Delta}{2}}=\frac{k_0}{\Delta\sqrt{\pi}}\sqrt{\frac{\pi\Delta}{2}}\\ &=\sqrt{\frac{k_0^2}{2\Delta}}=\langle p\rangle\end{align} $$
