?
If that is so, then the integral is 0, becoase integral of odd function from -infinity to +infinity is 0.
Here is one way.
INT.(-inf.-->+inf.)[x e^(-x^2)]dx
Let u = e^(-x^2)
So du = e^(-x^2)*(-2x dx)
INT.(-inf.-->+inf.)[x e^(-x^2)]dx
= INT.(-inf.-->+inf.)[e^(-x^2)](x dx)
= INT.(-inf.-->+inf.)[e^(-x^2)](-2x dx)*(-1/2)
= (-1/2)INT.(-inf.-->+inf.)[e^(-x^2)](-2x dx)
That is (-1/2)INT.[du]
which is (-1/2)(u) +C
so,
= (-1/2)[e^(-x^2)](-inf.-->+inf.)
= (-1/2)[e^(-(+inf.)^2) -e^(-(-inf.)^2)]
= (-1/2)[e^(-inf.) -e^(-inf)]
= 0 ---------answer.