Signals Energy of 2 signals - Integral limits correct?

thomas49th
If signals x(t) and y(t) are orthogonal and if z(t) = x(t) + y(t) then
E_{z} = E_{x} + E_{y}:

Proof:

$$E_{z} => \int^{\infty}_{-\infty} {(x(t) + y(t))^{2}} dt => \int {(x(t) + y(t))^{2}}^{2} dt => \int (x^{2}(t)) + \int(y^{2}(t))dt + \int x(t)y(t)dt => E_{x} + E_{y}$$

because $$\int x(t)y(t)dt$$ = 0 because of integration by parts:

u = x(t) dv/dt = y(t)
u' = dx/dt, v = $$frac{y^{2}(t)}{2}$$

so $$x(t)\frac{y^{2}(t)}{2} - \int {\frac{y^{2}(t)}{2}\frac{dx}{dt}}dt$$
$$x(t)\frac{y^{2}(t)}{2} - \int {\frac{y^{2}(t)}{2}}dx$$
we can treat y^2(t) as a constant so:

$$x(t)\frac{y^{2}(t)}{2} - \int^{\infty}_{-\infty} {\frac{y^{2}(t)}{2}}dx$$
$$x(t)\frac{y^{2}(t)}{2} - } [{\frac{y^{2}(t)x}{2}}]^{\infty t}_{-\infty t}$$

but the problem is that the limits were destined for integrating with respect to time. I'm not integrating with respect to x.

Any suggestions?
Thanks
Thomas

Homework Helper
The integral of x(t)y(t) isn't zero because of some bogus 'integration by parts' argument. It's zero because that's what 'orthogonal' means.

thomas49th
Ofcourse! Execellent. May I ask, out of interest alone what the integral of x(t)y(t) with respect to t should be?

Thanks
Thomas