Consider $f$ and $g$, two scalar functions of one variable defined over a the interval [-$a$,$a$], with $a$ real number. $f$ is symmetric, $f(x)=f(-x)$ and $g$ antisymmetric $g(-x)=-g(x)$. We consider the following integral: $$\int_{-a}^a \mathrm{d}x\int_{-a}^a \mathrm{d}y\; f(x)^2f(y)V(|x-y|)g(y),$$
where $V$ is just another scalar function and $|x|$ is the absolute value of $x$.
I am under impression that this kind of integrals are always equal to zero. I am working with a numerical program that has to do this kind of integrals many times with different "$g,f$", and I always get a negligible value (possibly zero).
My questions are:
- Is this always zero? If so why? is this the result of some theorem?
- If it is not zero for that interval, would it be zero for the domain $(-\infty,\infty)$?
- If any of this is false, would any other condition make it true?
I have manipulated a few simple sinusoidal functions to see if it is true (and it seems to be) but I am under impression that there is something more general. There has to be a manipulation that would make everything more evident.
Edit: $f$, $g$ and $V$ are bounded
$$I=\int_{-a}^a \left(\int_{-a}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x$$ $$I=I_1+I_2+I_3+I_4\qquad\begin{cases} I_1= \int_{0}^a \left(\int_{0}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\ I_2= \int_{-a}^0 \left(\int_{0}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\ I_3= \int_{0}^a \left(\int_{-a}^0 V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\ I_4= \int_{-a}^0 \left(\int_{-a}^0 V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\ \end{cases}$$ Change of variable $x$ to $-x$ in $I_2$ : $$I_2= \int_{a}^0 \left(\int_{0}^a V(|-x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}(-x)$$ $$I_2= \int_{0}^a \left(\int_{0}^a V(|-x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x$$
Change of variable $y$ to $-y$ in $I_3$ : $$I_3= \int_{0}^a \left(\int_{0}^{-a} V(|x+y|)f(-y)(g(-y))\mathrm{d}(-y)\right)f(x)^2\mathrm{d}x$$ $$I_3= \int_{0}^a \left(\int_{0}^a V(|x+y|)f(y)(-g(y))\mathrm{d}y\right)f(x)^2\mathrm{d}x$$ $$I_3= -\int_{0}^a \left(\int_{0}^a V(|x+y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x$$ Since $V(|x+y|)=V(|-x-y|)$
$$\boxed{I_3=-I_2}$$ Change of variables $x$ to $-x$ and $y$ to $-y$ in $I_4$ : $$I_4= \int_{a}^0 \left(\int_{a}^0 V(|-x+y|)f(-y)g(-y)\mathrm{d}(-y)\right)f(-x)^2\mathrm{d}(-x)$$ $$I_4= \int_{0}^a \left(\int_{0}^a V(|x-y|)f(y)(-g(y))\mathrm{d}y\right)f(x)^2\mathrm{d}x$$ $$I_4= -\int_{0}^a \left(\int_{0}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x$$ Since $V(|x-y|)=V(|-x+y|)$ $$\boxed{I_4=-I_1}$$ Finally : $$I=I_1+I_2+I_3+I_4=I_1+I_2-I_2-I_1=0$$ $$\int_{-a}^a \left(\int_{-a}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x=0$$