Determine the convergence of $ \int_1^\infty (\ln(x+\sin x)-\ln(x))\sqrt{x} \, dx $ I need to determine if the integral $$ \int_1^\infty (\ln(x+\sin x) - \ln(x))\sqrt{x} \, dx $$ is convergent/divergent, and if it's convergent then to check if it's absolutely convergent.
I tried to do integration by parts and to check what will be the limit, but it was too complicated.
Also, we cant use any comparison test, because $ \ln\left(x+\sin x\right)-\ln\left(x\right) $ changes the sign, we cant use Dirichlet's test, because $ \sqrt{x} $ doesn't tend to $ 0 $.
Any idea would be helpful, thanks in advance.
 A: Combining the $\ln$ terms, $$ I=\int_1^\infty (\ln(x+\sin x) - \ln(x))\sqrt{x} \, dx $$
$$= \int_1^\infty \ln\left(1+\frac{\sin(x)}{x}\right)\sqrt{x} \, dx $$Note that $\sin(x)/x<1$, since equality is only reached at $x=0$. By Taylor's Theorem, for $|z|<1$ $\ln(1+z) = z + O(z^2)$; then
$$
I = \int _1^{\infty} \frac{\sin(x)}{x}\sqrt{x}\,dx + \int _1^{\infty} O\left(\frac{\sin^2(x)}{x^2}\right)\sqrt{x}\,dx
$$The first integral converges by comparison with the usual Fresnel sine integral and the second integral converges by comparison with $x^{-3/2}$. So $I$ exists, i.e. the integral converges.
That being said, the convergence is only conditional, not absolute. To see this, note we can bound $|\ln(1+\sin(x)/x)|>\sin(x)/(8x)$. Then
$$
\int_{ 1}^{\infty } \ln\left(1+\frac{\sin(x)}{x}\right)\sqrt{x} \, dx
$$
$$
> \int_{ \pi}^{\infty } \left|\frac{\sin(x)}{8x}\right|\sqrt{x} \, dx
$$
$$
=\frac{1}{8}\int_{ \pi}^{\infty } \frac{|\sin(x)|}{\sqrt{x}} \, dx
$$Now break the integration into intervals of length $\pi$ and the result follows.
$$
=\sum_{k=1}^{\infty}\frac{1}{8}\int_{ k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{\sqrt{x}} \, dx
$$
$$
> \sum_{k=1}^{\infty}\frac{1}{8\sqrt{\pi}\sqrt{k}}\int_{ k\pi}^{(k+1)\pi} {|\sin(x)|} \, dx
$$
$$
=\sum_{k=1}^{\infty}\frac{1}{4\sqrt{\pi}\sqrt{k}}=\infty
$$
