probability theory - Proof that $E(F_X(X/sigma))=frac12$ for every positive $sigma$

Define $X$ to be continuous random variable symmetric about zero with cdf $F_X$ and let $\sigma > 0$ denote a constant. Now show the following:
$$E\left[F_X\left(\frac{X}{\sigma}\right)\right] = 0.5$$
How can one prove this claim? Since the cdf $F_X$ isn't necessarily linear, we can't place the expectation into the cdf, which would render the problem trivial. Additionally, I've concluded that the cdf is convex from $-\infty$ to $0$ and concave from $0$ to $\infty$ as it is symmetric about zero. This means we can't use Jensen's inequality either. What am I missing?

Answer

Let $f$ be the PDF of $X$, which should be symmetric about the origin; i.e. $f(-x)=f(x)$. Then, the CDF is
$$F(x)=\int_{-\infty}^x f(t)\,\mathrm{d}t\tag{1}$$

The symmetry of $f$ means that
$$\begin{align} F(-x) &=\int_{-\infty}^{-x}f(t)\,\mathrm{d}t\\ &=\int_x^\infty f(-t)\,\mathrm{d}t\\ &=\int_x^\infty f(t)\,\mathrm{d}t\\ &=\int_{-\infty}^\infty f(t)\,\mathrm{d}t-\int_{-\infty}^xf(t)\,\mathrm{d}t\\[4pt] &=1-F(x)\tag{2} \end{align}$$
The expected value is
$$\begin{align} E(F(X/\sigma)) &=\int_{-\infty}^\infty f(x)F(x/\sigma)\,\mathrm{d}x\tag{3}\\ &=\int_{-\infty}^\infty f(-x)F(-x/\sigma)\,\mathrm{d}x\tag{4}\\ &=\int_{-\infty}^\infty f(x)(1-F(x/\sigma))\,\mathrm{d}x\tag{5}\\ &=\frac12\int_{-\infty}^\infty f(x)\,\mathrm{d}x\tag{6}\\ &=\frac12\tag{7} \end{align}$$
Explanation:
$(3)$: formula for expected value
$(4)$: substitute $x\mapsto-x$
$(5)$: symmetry of $f$ and $(2)$
$(6)$: average $(3)$ and $(5)$
$(7)$: $f$ is a PDF