Proof of $E(X)=a$ when $a$ is a point of symmetry I am trying to develop a proof of the following:
Given a random variable $X$ with symmetric probability density function $f(x)$, prove that $E(X)=a$ where $a$ is the point of symmetry.

A couple of thoughts:
It's easy to think of examples where this applies (e.g. normal) and doesn't (e.g. Cauchy). I am comfortable with calculating expectation, and even handling showing how 
$\int_{-\infty}^af(x)dx=\int_{a}^{\infty}f(x)dx$
But if the specific form of the pdf is not specified, is it possible to prove this in general? 
Also, how would one go about showing the existence of the expectation in general?

We give two proofs, the first by manipulation of the integral, and a second much shorter one that uses probabilistic language.
Proof 1: Symmetry about $a$ means that $f(a+z)=f(a-z)$ for all $z$. 
Suppose that the expectation $E(X)$ exists.  Then 
$$E(X)=\int_{-\infty}^\infty xf(x)\,dx.$$
 Rewrite as
$$\int_{-\infty}^\infty \left(a +(x-a)\right)f(x)\,dx,$$
and then as 
$$\int_{-\infty}^\infty \left(a +(x-a)\right)f(x)\,dx,$$
which is equal to
$$\int_{-\infty}^\infty af(x)\,dx+\int_{-\infty}^\infty (x-a)f(x)\,dx.\tag{1}$$
The first integral is $a$. We need to show the second integral is $0$.  This is essentially obvious by symmetry: $(x-a)f(x)$ is an odd function. But we do the details. 
In the second integral of (1), make the change of variable $y=x-a$. Then the integral becomes 
$$\int_{-\infty}^\infty yf(a+y)\,dy.$$
Break up at $y=0$. We get 
$$\int_{-\infty}^0 yf(a+y)\,dy +\int_0^\infty yf(a+y)\,dy.\tag{2}$$
For the first integral, make the change of variable $z=-y$. We get
$$\int_{z=\infty}^0 (-1)(-z)f(a-z)\,dz.$$
Using $f(a-z)=f(a+z)$, and some minor fooling with minus signs, we end up with
$$-\int_{z=0}^\infty zf(a+z)\,dz.$$
This cancels the second integral of (2).
Proof 2: A more probabilistic (and therefore better) way of doing the problem is to first let $X=a+Y$. Assume that $E(X)$ exists (it need not). Then $E(X)=a+E(Y)$. We need to show that $E(Y)=0$.
The density function $g(y)$ of $Y$ is symmetric about $y=0$.  It follows that 
that the random variable $Y$ has the same distribution as the random variable $-Y$. Thus $E(Y)=E(-Y)=-E(Y)$ and therefore each is $0$. 
