Why is the position expectation value for this wave function independent of the parameter? For a homework problem we are given the wave function
$$ \Psi(x) = \frac{N}{x^2 + a^2},\ a > 0 $$
and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function
$$ \Psi(x) = \frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)} $$
Then, following my notes computed
$$ \int_{-\infty}^\infty \Psi(x)^* \langle x \rangle \Psi(x)\,\mathrm dx $$
Given that the wave equation is currently in the $x$ basis I then computed
$$ \int_{-\infty}^\infty
\left(\frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)}\right)^*
x
\left(\frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)}\right)\,\mathrm dx $$
by plugging it into Wolfram Alpha. Though this results in a solution of $0$.
This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$? 
 A: Josh's answer is sufficient. While he uses the final expectation value integral in his answer, you can also just look at the wavefunction. Since it is even, and since $|\psi|^2$ (also even) tells us the probability of measuring the particle to be within position $x$ and $x+dx$, we know that the particle "spends an equal amount of time" at positive and negative $x$. So it must be that the mean position is $0$.

I also wanted to address some of your notation and understanding of it.
It seems like you are familiar with bases. The expectation value of the position operator can be written without specifying any basis:
$$\langle X\rangle=\langle\psi|X|\psi\rangle$$
Since we are given the wavefunction in the position basis $\psi(x)=\langle x|\psi\rangle$, it makes sense to work in the position basis.
Therefore, we can use the identity $\int|x\rangle\langle x|dx=1$ to write our expectation value as
$$\langle X\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\langle\psi|x\rangle\langle x|X|x'\rangle\langle x'|\psi\rangle dxdx'$$
Then, knowing that $X$ in its own eigenbasis is $\langle x|X|x'\rangle=x'\delta(x'-x)$, the integral becomes 
$$\langle X\rangle=\int_{-\infty}^{\infty}\psi^*(x)x\psi(x)dx$$
This is the integral you want. You don't want your expectation value inside the integral (and you also need to specify which variable you are integrating over). Otherwise you have an integral that is not very useful:
$$\langle X\rangle=\int_{-\infty}^{\infty}\psi^*(x)\langle X\rangle\psi(x)dx=\langle X\rangle\int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx=\langle X\rangle$$
A: You've got the right answer! The wave-function is symmetric about $0$, while the operator $\langle x \rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$!
Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
Edit: I meant to write $\hat{x}  = x$, not $\langle x \rangle = x$. The correct expression for the expectation value is $\langle x \rangle = \int\psi^*\hat{x}\psi dx$.
