0

$\int _0^{\infty }\:f\left(x\right)\cdot g\left(x\right)dx$ converge.

$\int _0^{\infty }\:f\left(x\right)dx\:or\:\int _0^{\infty \:}g\left(x\right)dx$ Converges?

I tried of thinking of some functions, like to use $\sin(x)$ and such. But I cant find any counter for it... is it possible it is true?

I am thinking it might be true only becauses of limit laws, if $f$ and $g$ converges - so is $fg$. But I dont know if its also the opposite...

Gary
  • 31,845
  • 1
    Take $f(x)=\sin x$ and $g(x)=\frac{1}{x}$. – Gary May 16 '22 at 09:04
  • 2
    You can easily find examples where both $\int_0^\infty f(x)dx$ and $\int_0^\infty g(x)dx$ diverge, but $ f(x) g(x) = 0$ everywhere. – Martin R May 16 '22 at 09:05
  • I thought of it, but $\frac {sin(x)} {1/x}$ does not converge at that range. –  May 16 '22 at 09:05
  • You get $\frac{\sin x}{x}$ which is not $\frac{\sin x}{1/x}=x\sin x$. – Gary May 16 '22 at 09:05
  • Oh yea, sorry, my bad, anyway, it doesnt converge. I thought of it. –  May 16 '22 at 09:06
  • Are you sure? https://math.stackexchange.com/questions/5248/evaluating-the-integral-int-0-infty-frac-sin-x-x-mathrm-dx-frac-pi (Hint: Dirichlet test) – Gary May 16 '22 at 09:06
  • 1
    $f(x) = \max(\sin(x), 0)$ and $g(x) = \max(-\sin(x), 0)$. – Martin R May 16 '22 at 09:07
  • @Gary wow, weird.. it didnt seem like that to me before... Thankss! –  May 16 '22 at 09:07
  • Simpler example, giving absolute convergence of the integral of the product: $f(x)=g(x)=\frac1{1+x}$ – David C. Ullrich May 16 '22 at 10:36
  • oh yea, I actually thought of this later, took two functions $f(x)$ and $g(x)$ of: $\frac 1 x$ and it is also good. THanks! –  May 16 '22 at 19:14

0 Answers0