# 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 $$0$$.

• Very broad hint: consider breaking the domain into chunks of length $\pi$ (maybe starting from $x=\pi$ rather than $x=1$ for simplicity's sake, since that clearly doesn't affect the convergence) and getting estimates on the integral on each interval. – Steven Stadnicki Jul 7 at 21:43
• (followup hint: the fact that $\alpha-\alpha^2/2 \leq \ln(1+\alpha) \leq \alpha$ may come in handy. Note that you can simplify the term in parentheses a bit...) – Steven Stadnicki Jul 7 at 21:56
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. $$=\sum_{k=1}^{\infty}\frac{1}{8}\int_{ k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{\sqrt{x}} \, dx$$ $$> \sum_{k=1}^{\infty}\frac{1}{8\sqrt{\pi}\sqrt{k}}\int_{ k\pi}^{(k+1)\pi} {|\sin(x)|} \, dx$$ $$=\sum_{k=1}^{\infty}\frac{1}{4\sqrt{\pi}\sqrt{k}}=\infty$$