- #1

eljose

- 492

- 0

**Integral equality...**

let be a and b real numbers..and let be the integral...

[tex]\int_{-\infty}^{\infty}dxf(x)x^{a}\int_{-\infty}^{\infty}g(x+y)y^{ib}=0 [/tex] so if this is zero also will be its conjugate:

[tex]\int_{-\infty}^{\infty}dxf(x)x^{a}\int_{-\infty}^{\infty}g(x+y)y^{-ib}=0 [/tex] now let,s suppose we would have that (1-a,-b) is also a zero so:

[tex]\int_{-\infty}^{\infty}dxf(x)x^{1-a}\int_{-\infty}^{\infty}g(x+y)y^{-ib}=0 [/tex] then my conclusion is that 1-a=a a=1/2 and there is no other solution.. :zzz: