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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Jul 7 at 21:43
  • 1
    $\begingroup$ (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...) $\endgroup$ – Steven Stadnicki Jul 7 at 21:56
  • $\begingroup$ It only seems to be conditionally convergent (not absolutely). I don't have a proof of this though. $\endgroup$ – Varun Vejalla Jul 7 at 23:51

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. $$ =\sum_{k=1}^{\infty}\frac{1}{8}\int_{ k\pi}^{(k+1)\pi} \frac{|\sin(x)|}{\sqrt{x}} \, dx $$ $$ > \sum_{k=1}^{\infty}\frac{1}{8\sqrt{\pi}\sqrt{k}}\int_{ k\pi}^{(k+1)\pi} {|\sin(x)|} \, dx $$ $$ =\sum_{k=1}^{\infty}\frac{1}{4\sqrt{\pi}\sqrt{k}}=\infty $$

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