Is $\int_0^{+\infty}\frac{\sin \ln x}{\sqrt x}\,dx$ convergent or absolutely convergent? Is
$$\int_0^\infty\frac{\sin\ln x}{\sqrt x}\,dx$$
convergent? Absolutely convergent?
I found, that in fact it converges, but the only way I found is to say that $$\int_0^{+\infty}\frac{\sin \ln x}{\sqrt x} \, dx \leq \int_0^{+\infty}\frac{1}{\sqrt x }\,dx,$$ but this integral doesn't converge, so this way I can't say something about convergence of this integral.
 A: Use substitution $x=e^u$ to get
$$I=\int_0^{+\infty}\frac{\sin\ln x}{\sqrt x}\,dx = \int_{-\infty}^{+\infty}\frac{\sin u}{e^{u/2}}e^u\,du = \int_{-\infty}^{+\infty}e^{u/2}\sin u\,du$$
and then use partial integration:
\begin{align}I &=\int_{-\infty}^{+\infty}e^{x/2}\sin x\,dx \\ &=\left[\begin{array}{ll} u = \sin x,&du=\cos x\\ v =2e^{x/2},&dv = e^{x/2}\end{array}\right] \\
&= \left.2e^{x/2}\sin x\,\right|_{-\infty}^{+\infty} - 2\int_{-\infty}^{+\infty}e^{x/2}\cos x\,dx \\
&= \left[\begin{array}{ll}u = \cos x,&du=-\sin x\\ v =2e^{x/2},&dv = e^{x/2}\end{array}\right] \\
&= \left.2e^{x/2}\sin x\,\right|_{-\infty}^{+\infty} - 2\left(\left.2e^{x/2}\cos x\,\right|_{-\infty}^{+\infty} + 2\int_{-\infty}^{+\infty}e^{x/2}\sin x\,dx\right) \\
&= \left.2e^{x/2}(\sin x-2\cos x)\,\right|_{-\infty}^{+\infty}-4I\end{align}
We conclude that $I = \frac 25 \left.e^{x/2}(\sin x-2\cos x)\,\right|_{-\infty}^{+\infty}$ which doesn't converge.
A: Hint: For $n=1,2,\dots,$ estimate the integral over $[e^{2\pi n +\pi/4},e^{2\pi n +\pi/2}].$ Note that $\sin (\ln x) \ge 1/\sqrt 2$ on this interval.
A: You could solve the indefinite integral,
\begin{align*}
\int\frac{\sin(\ln(x))}{\sqrt x}\,dx &= \int\frac{\sqrt x \cdot\sin(\ln(x))}{x}\,dx \\ &= \int e^{\ln\sqrt x}\sin(\ln x)\,d(\ln x) \\ &= \int e^{\frac12 \ln x}\sin(\ln x)\,d(\ln x) \\ &= \frac25\sqrt x\left(2\cos(\ln x))-\sin(\ln x))\right)\\
\end{align*}
and then evaluate the limits.
