Prove or disprove questions?

Dec 2009
1,506
434
Russia
Let f,g continuous functions on [0,inf).

1. If |f(x)|<=|g(x)| for all x in [0,inf), and the integral int(0-->inf)g(x)dx converges so the ntegral int(0-->inf)f(x)dx converges.
(I think its not true...)

2. If ntegral int(0-->inf)g(x)dx converges so integral int(0-->inf)g(x)*sin(x)dx converges.
(Maybe with using Dirichelet?)

3. If ntegral int(0-->inf)g(x)dx converges so integral int(0-->inf)g(x)*e^-x^2dx converges.
(Abel test?)


Thank you!!!
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
Let f,g continuous functions on [0,inf).

1. If |f(x)|<=|g(x)| for all x in [0,inf), and the integral int(0-->inf)g(x)dx converges so the ntegral int(0-->inf)f(x)dx converges.
(I think its not true...)

2. If ntegral int(0-->inf)g(x)dx converges so integral int(0-->inf)g(x)*sin(x)dx converges.
(Maybe with using Dirichelet?)

3. If ntegral int(0-->inf)g(x)dx converges so integral int(0-->inf)g(x)*e^-x^2dx converges.
(Abel test?)
What sort of integration are you using? For the Lebesgue integral, the answer to all three questions is that the result is true. That is because the Lebesgue integral is an "absolute integral". It is defined in such a way that if f is integrable then so is |f|. The answers to questions 1.–3. then come from Lebesgue's Dominated Convergence Theorem.

But if these integrals are (improper) Riemann integrals then the answers will not all be positive. For example, \(\displaystyle \int_0^\infty\frac{\sin x}x\,dx\) converges, but \(\displaystyle \int_0^\infty\frac{|\sin x|}x\,dx\) does not. So you can get a counterexample to 1. by taking \(\displaystyle g(x) = \frac{\sin x}x\) and \(\displaystyle f(x) = \frac{|\sin x|}x\). Also for 2., \(\displaystyle g(x) = \frac{\sin x}x\) will give a counterexample.