If there is a sequence $x_n \to \infty$ such that $\lim_{n \to \infty} \int_0^{x_n} f(x)dx$ converges, does $\int_0^{\infty} f(x)dx$ converge? 
State whether the sentence below are true or false. If it's true, proof it. if it's false, give a counterexample.


(A) Let $f : [0, \infty) \to \mathbb R$ be a continuous function. If there is a sequence $x_n \to \infty$ such that $\lim_{n \to \infty} \int_0^{x_n} f(x)dx$ converges, then $\int_0^{\infty}f(x)dx$ converges.


State whether converge or diverge and justify.


($a_1$) $\int_0^{\infty} \frac{1}{(1+x)\sqrt x}dx$


($a_2$) $\int_0^{\infty} \sin(x^2)dx$

My attempt
$$\int_0^{\infty} \frac{dx}{(1+x) \sqrt x}$$
We separate the integral into two
$$\int_0^{\infty} \frac{dx}{(1+x) \sqrt x}=\int_0^{1} \frac{dx}{(1+x) \sqrt x}+\int_1^ {\infty} \frac{dx}{(1+x) \sqrt x}$$
and we analyze the convergence of each of them, first to the infinite.
$$\int_1^{\infty} \frac{dx}{(1+x) \sqrt x} \lt \int_1^{\infty} \frac{dx}{x \sqrt x}$$.
Where the one on the right converges, then the one on the left also converges. In the integral
($a_1$)
$$\int_0^{1} \frac{dx}{(1+x) \sqrt x}$$ we make the transformation $y= \frac{1}{x}$ where we can arrive at the integral $$ \int_1^ {\infty} \frac{dx}{(1+x) \sqrt x}=\int_0^{1} \frac{dx}{(1+x) \sqrt x}$$ We've already seen that it converges.
($a_2$)
$\sin(x^2)dx$ converges, but not absolutely. The convergence of $\int_1^{\infty}\sin(x^2)dx$ , making the change $y = x^2$ the integral becomes $$\int_1^{\infty} \frac{\sin(y)}{2 \sqrt y}dy$$ I fell in the case of the integral $$\int_1^{\infty} \frac{\sin(y)}{y^r}dy$$  with $0 \lt r = \frac{1}{2} \lt 1$, which converges however not absolutely.
I managed to make $a_1$ and $a_2$ (is good enough?)but A couldn't make a very convincing argument.
I thought like that.
Since $ f $ is continuous, then equivalently $ x_n $ is continuous and $ x_n \to \infty $. And as per hypothesis $\lim_{n \to\infty} \int_0^{x_n}f(x)dx$ converge then $\int_0^{x_n}f(x)dx$.
 A: Statement (A) is false.  For example if $x_n=2n\pi$ for $n\in\Bbb N$ and $f(x)=\cos x$ then $\int_0^{x_n}f(x)dx=0$ but $\int_0^{(2n+1)\pi}f(x)dx=2$.
For ($a_2$): For $n\in\Bbb N$ let $h(n)=\int_{n\pi}^{(n+1)\pi}\frac {|\sin x|}{\sqrt x}dx.$
Note that $0<h(n)<\frac {\pi}{\sqrt {n\pi}}$ so $h(n)\to 0.$
Let $H(n)=\sum_{j=1}^n(-1)^j\cdot h(j)=\int_{\pi}^{(n+1)\pi}\frac {\sin x}{\sqrt x}dx.$
For $z>0$ let $[z/\pi]$ be the greatest integer not exceeding $z/\pi.$ Then for $z\ge \pi$ we have $$0\le \left|H([z/\pi])-\int_{\pi}^z\frac {\sin x}{\sqrt x}dx\right|< h([z/\pi]).$$ $(\bullet)$.  Now $h(n)\to 0$ so $\int_{\pi}^z\frac {\sin x}{\sqrt x}dx$ converges if $H(n)$ converges. We have $$0<h(n+1)=\int_{y=(n+1)\pi}^{y=(n+2)\pi}\frac {|\sin y|}{\sqrt y}dy=\int_{x=n\pi}^{x=(n+1)\pi} \frac {|\sin (x+\pi)|}{\sqrt {x+\pi}}d(x+\pi)=$$ $$=\int_{n\pi}^{(n+1)\pi} \frac {|\sin x|}{\sqrt {x+\pi}}dx<$$ $$<\int_{n\pi}^{(n+1)\pi} \frac {|\sin x|}{\sqrt x}dx=h(n).$$ So $H(n)=\sum_{j=1}^n(-1)^j\cdot h(j)$ is an alternating series with the absolute values of its terms going monotonically  to $0.$ So $H(n)$ converges.
