Integral equality

  • Thread starter eljose
  • Start date
  • #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.. :biggrin: :zzz: :biggrin: :biggrin: :biggrin: :biggrin:
 

Answers and Replies

  • #2
shmoe
Science Advisor
Homework Helper
1,994
1
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.
 
  • #3
Are you integrating over y in the second integral? x? A nonexistent variable?
 

Suggested for: Integral equality

Replies
35
Views
2K
  • Last Post
Replies
0
Views
5
  • Last Post
Replies
2
Views
459
Replies
1
Views
378
  • Last Post
Replies
3
Views
388
  • Last Post
Replies
1
Views
519
  • Last Post
Replies
4
Views
469
  • Last Post
Replies
1
Views
90
  • Last Post
Replies
5
Views
398
Replies
4
Views
355
Top