This is an exercise from Oppenheim's course on signals and systems.
$$\int_{-\infty}^{+\infty}x^2(t)dt = \int_{-\infty}^{+\infty}x_e^2(t)dt + \int_{-\infty}^{+\infty}x_o^2(t)dt $$
where $x_e(t)$ and $x_o(t)$ are respectively the even and odd parts of $x(t)$
The way I did it was
$$\int_{-\infty}^{+\infty}x^2(t)dt = \int_{-\infty}^{+\infty}x_e^2(t)dt + \int_{-\infty}^{+\infty}x_o^2(t)dt + 2\int_{-\infty}^{+\infty}x_o(t)x_e(t)dt$$
Then $$ x_o(t)x_e(t) = \frac 12 [x(t) - x(-t)] * \frac 12 [x(t) + x(-t)] = \frac 14 [x^2(t) - x^2(-t)]$$
$$ \int_{-\infty}^{+\infty}x_o(t)x_e(t)dt = \frac 14 \int_{-\infty}^{+\infty}x^2(t)dt - \frac 14 \int_{-\infty}^{+\infty}x^2(-t)dt$$
Then using u substitution (and this is the part where I'm not sure what I did makes sense, namely can I flip the infinities by putting a minus signal before the integral? I know it works for definite integrals without infinities but I'm confused about integrals with infinities)
$$\frac 14 \int_{-\infty}^{+\infty}x^2(t)dt - \frac 14 \int_{+\infty}^{-\infty}-x^2(y)dy = \frac 14 \int_{-\infty}^{+\infty}x^2(t)dt + \frac 14 \int_{+\infty}^{-\infty}x^2(y)dy = \frac 14 \int_{-\infty}^{+\infty}x^2(t)dt - \frac 14 \int_{-\infty}^{+\infty}x^2(y)dy = 0$$