Integral of some anti/symmetric functions of two variables 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
 A: $$I=\int_{-a}^a \left(\int_{-a}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x$$
$$I=I_1+I_2+I_3+I_4\qquad\begin{cases}
I_1= \int_{0}^a \left(\int_{0}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\
I_2= \int_{-a}^0 \left(\int_{0}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\
I_3= \int_{0}^a \left(\int_{-a}^0 V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\
I_4= \int_{-a}^0 \left(\int_{-a}^0 V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x\\
\end{cases}$$
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 :
$$I=I_1+I_2+I_3+I_4=I_1+I_2-I_2-I_1=0$$
$$\int_{-a}^a \left(\int_{-a}^a V(|x-y|)f(y)g(y)\mathrm{d}y\right)f(x)^2\mathrm{d}x=0$$
