proof that if $\int_a^\infty f(x)$ converges then $\int_a^\infty f(x^2)$ converges

Question

juat as the title says. i would like to know first if this is even true. If it is can i get a proof and if not then a counter example?(please keep it as simple as possible) thank you very much in advance. repeat of the question: prooff or counter example that if $$\int_1^\infty f(x) dx$$ converges then $$\int_1^\infty f(x^2) dx$$ converges

$f$ is a riemann integral

edit: changed lower limit from a to 1

What are the assumptions on $f$? Lebesgue or Riemann integral? What are your thoughts on the problem? — Martin R, Jul 20 '18 at 08:02
@KaviRamaMurthy Hmmm, how do you get rid of the term $1/\sqrt y$ when the sign of $f$ is not constant? — Did, Jul 20 '18 at 08:11
ok so to answar your questions,1. f is reimann integral(i havn't learned about lebesgue).2. i'm not sure why the implication fails when a=0? the assumption is that f(x) converges.3. i have not idea what you mean by that. but f is a function, if f was a constant the integral couldn't converge unless it was 0. in wich case the entire claim is trivial. — Nadav Matityahu, Jul 20 '18 at 08:26

Zeekless · Accepted Answer · 2018-07-20T11:35:57.640

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Counterexample:

$$ a = \frac{1}{2}, \quad f(x) = \frac{1}{x^2(x-\frac{1}{4})}. $$

EDIT Although, the claim is true if $a \ge 1$, because then

$$ \int_{a}^{\infty}f(x)dx {\; \; \rm \text converges} \quad \Rightarrow \quad \int_{a^2}^{\infty}f(t)dt {\; \; \rm \text converges}, $$

and for the integral

$$ \int_a^{\infty}f(x^2)dx = \int_{a^2}^{\infty}\frac{f(t)}{2\sqrt{t}}dt = \int_{a^2}^{\infty}f(t)g(t)dt $$ the Abel-Dirichlet test is positive, since

$$ \begin{aligned} &1) \;\int_{a^2}^{\infty}f(t)dt {\; \; \rm \text converges;} \\ &2) \; g(t) = \frac{1}{2\sqrt{t}} {\; \; \rm \text is \; monotonic\; and\; bounded \; on \;}[a^2, \infty). \end{aligned} $$

Therefore, $\int_a^{\infty}f(x^2)dx$ also converges.

P.S. Abel-Dirichlet test has two faces. You can use the following premises as well:

$$ \begin{aligned} &1) \;\int_{a^2}^{b}f(t)dt {\; \; \rm \text is \; finite \; for \; any\;} b>a^2; \\ &2) \; g(t) = \frac{1}{2\sqrt{t}} {\; \; \rm \text is \; monotonic,\; differentiable \; and \; tends\; to \; zero\; on \;}[a^2, \infty). \end{aligned} $$

The result will be the same.

edited Jul 20 '18 at 11:35

answered Jul 20 '18 at 10:00

Zeekless

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so... that's not a counterexample. ehat you just gave is a converging $f(x)$ where $f(x^2)$ also converges. that is the claim – Nadav Matityahu Jul 20 '18 at 10:06
@NadavMatityahu $\int_a^{\infty}f(x^2)dx$ diverges. Try it. – Zeekless Jul 20 '18 at 10:08
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right you are my friend. thank you! – Nadav Matityahu Jul 20 '18 at 10:27
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so i have another question.wha if $a>=1$ but the function could be negative? do you have a counterexample for that too? – Nadav Matityahu Jul 20 '18 at 10:31
@NadavMatityahu, updated my answer. Abel-Dirichlet works. – Zeekless Jul 20 '18 at 11:07
umm could you explain the second line of equations? just before you use the abel-dirivhlet test? i'm not sure how those equations are true/work – Nadav Matityahu Jul 20 '18 at 11:25
@NadavMatityahu, we change the variable: $x^2=t$. Therefore $x=\sqrt{t}$ and $dx = \frac{dt}{2\sqrt{t}}$. And $g(t)$ is just a new name for $\frac{1}{2\sqrt{t}}$ to visually meet the conditions of Abel-Dirichlet test. – Zeekless Jul 20 '18 at 11:31
thank you :) just figured it out a second ago. you just gave me peace of mind that i might have passed my test. thank you :) – Nadav Matityahu Jul 20 '18 at 11:32
@NadavMatityahu, you are welcome:) Please accept the answer, if it satisfies you. – Zeekless Jul 20 '18 at 11:34

proof that if $\int_a^\infty f(x)$ converges then $\int_a^\infty f(x^2)$ converges

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