I have improper integral: $$\int \limits_{1}^{+ \infty} \frac{x \sin(\log^2(x))}{\log(x+1)}dx$$ I know, that it diverges, but I can't prove it. I tried to use the negation of Cauchy criterion for improper integrals, but I failed, because this integral doesn't have representation in elementary functions. So, how to prove the divergence?

  • $\begingroup$ Hint. It suffices to prove that $$ \int_{e^{\sqrt{\pi n}}}^{e^{\sqrt{\pi (n+1)}}} \frac{x \sin(\log^2 x)}{\log(x+1)} \, dx$$ diverges. $\endgroup$ – Sangchul Lee May 20 '17 at 22:35
  • $\begingroup$ @SangchulLee I know about this thing, but the main problem is in computation of this integral. $\endgroup$ – MOVZBL May 20 '17 at 22:38
  • $\begingroup$ You do not need an exact value, as it seems impossible to do. But even a very crude estimation will be enough for proof. For instance, use the substitution $x \mapsto \exp(\sqrt{x})$ to write $$\int_{e^{\sqrt{\smash[b]{\pi n}}}}^{e^{\sqrt{\smash[b]{\pi (n+1)}}}} \frac{x \sin(\log^2 x)}{\log(x+1)} \, dx = \int_{\pi n}^{\pi(n+1)} \frac{e^{2\sqrt{x}} \sin x}{2\sqrt{x}\log(1+e^{\sqrt{x}})} \, dx $$ $\endgroup$ – Sangchul Lee May 20 '17 at 22:48
  • $\begingroup$ @SangchulLee Can you give, please, an example of such estimation? It's not clear for me how to do that. $\endgroup$ – MOVZBL May 21 '17 at 7:45

Let $F: [1,\infty) \to \Bbb{R}$ be defined by

$$ F(x) = \int_{1}^{x} \frac{t \sin(\log^2 t)}{\log(1+t)} \, dt. $$

The improper integral is then defined as the limit of $F(x)$ as $x\to\infty$ if it exists. So if the improper would have existed, then we would also have $F(e^{\sqrt{\smash[b]{\pi (n+1)}}}) - F(e^{\sqrt{\smash[b]{\pi n}}}) \to 0$ as $n\to\infty$. We show that this is not the case.

With the substitution $t = e^{\sqrt{x}}$, we find that

\begin{align*} \left| F(e^{\sqrt{\smash[b]{\pi (n+1)}}}) - F(e^{\sqrt{\smash[b]{\pi n}}}) \right| &= \left| \int_{\pi n}^{\pi(n+1)} \frac{e^{2\sqrt{x}} \sin x}{2\sqrt{x}\log(1+e^{\sqrt{x}})} \, dx \right| \\ &= \int_{0}^{\pi} \frac{e^{2\sqrt{x+n\pi}} \sin x}{2\sqrt{x+n\pi}\log(1+e^{\sqrt{x+n\pi}})} \, dx \\ &\geq \int_{0}^{\pi} \frac{e^{2\sqrt{n\pi}} \sin x}{2\sqrt{(n+1)\pi}\log(1+e^{\sqrt{\smash[b]{(n+1)\pi}}})} \, dx \\ &=\frac{e^{2\sqrt{n\pi}}}{\sqrt{(n+1)\pi}\log(1+e^{\sqrt{\smash[b]{(n+1)\pi}}})}. \end{align*}

But notice that the denominator is $\sim n\pi$ as $n\to\infty$, while the numerator diverges much faster that polynomials. So

$$\lim_{n\to\infty} \frac{e^{2\sqrt{n\pi}}}{\sqrt{(n+1)\pi}\log(1+e^{\sqrt{\smash[b]{(n+1)\pi}}})} = \infty $$

and the desired conclusion follows.


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