Can Integral Equality Solve for Real Values in Complex Integrals?

[eljose]

Integral equality...

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

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

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

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

 

[shmoe]

You've *somehow* decided that if your integrals are equal then x must have the same exponent? I don't even know where to begin with what's wrong here, but by your logic shouldn't the same be true for y then, so b=0? Please, tell me this is a joke.

Have you been reading some of the horrible "proofs" of the Riemann Hypothesis you find online? This looks awfully familiar.

 

[MalleusScientiarum]

Are you integrating over y in the second integral? x? A nonexistent variable?