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