We can see that the improper integral
$$\int_{0}^{\infty}\sin\left(\,{x^{2}}\,\right)
{\rm d}x
$$
converges as the transformation $u = x^{2}$ helps us to use the Dirichlet's test on
$$
\int_{0}^{\infty}\frac{\sin\left(\,{u}\,\right)}
{2\,\sqrt{\,{u}\,}\,}\,{\rm d}u.
$$
However, I need to know the absolute convergence of the same, whether it is conditionally convergence or not.
I feel that the area of contribution is eventually being demolished for $\left\vert\,{\sin\left(\,{x^{2}}\,\right)}\,\right\vert$ as similar to $\sin\left(\,{x^{2}}\,\right)$. But I couldn't certain the absolute convergence.
Thanks in advance.
Best Answer
Let $f(x) = |\sin x^2 | $, $a_n = \sqrt{n \pi+{1 \over 4} \pi}, b_n = \sqrt{n \pi+{3 \over 4} \pi}$. Suppose $n \ge 2$ is even, note that $f(x) \ge {1 \over \sqrt{2}} 1_{[a_n,b_n]}(x)$ for $x \in [a_n,b_n]$.
Hence $\int_{a_n}^{b_n} f(x) dx \ge {1 \over \sqrt{2}} (b_n-a_n) \ge {1\over \sqrt{2}}{b_n^2-a_n^2 \over b_n+a_n} \ge K {1 \over \sqrt{n+1}}$. Since the latter is not summable, it follows that $f$ is not integrable.