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
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.