Is my solution for divergence of $\int_{0}^{\infty} \frac{\sin^{10}x \ln x}{\sqrt{x}}$ correct?

I have big doubts whether what I did was legal or not.

Investigate for convergence: $$\int_{0}^{\infty} \frac{\sin^{10}x \ln x}{\sqrt{x}} dx$$

First evaluate: $$\int_{0}^{\epsilon} ( \ln x) dx<\int_{0}^{\epsilon} (x^{10} \ln x) dx<\int_{0}^{\epsilon} \frac{x^{10} \ln x}{\sqrt{x}} dx<\int_{0}^{\infty} \frac{\sin^{10}x \ln x}{\sqrt{x}} dx$$ Work with the smallest one: $$\int_{0}^{\epsilon} ( \ln x)dx=\int_{-\infty}^{\ln\epsilon} \ln e^u de^u= \int_{-\infty}^{\ln\epsilon} u de^u$$ $$\int ude^u=ue^u-\int e^udu=ue^u-e^u=e^u(u-1)$$ $$\int_{-\infty}^{\ln\epsilon} u de^u=\epsilon\cdot(\ln \epsilon-1)-e^{A}\cdot(A-1) ,$$ where $$A\to-\infty$$. The first addendum is not interesting, since it's a finite number. In the second addendum, we have indeterminacy of type $$0\cdot\infty$$. If we use $$e^x=1+x+\cdots$$, we will see it tends to $$-\infty$$. So the original integral is divergent.

• The result is correct but your first inequalities are wrong – Maximilian Janisch May 22 at 13:04
• @MaximilianJanisch Thank you, could you please explain why (or maybe how to do the exercise differently)? I tried to see what's wrong, but couldn't find the mistake. (Epsilon is close to zero.) – fragileradius May 22 at 13:08

TLDR: Your solution is not correct, but your result is correct nonetheless.

Correct proof:

Claim. Your integral diverges to $$+\infty$$.

Proof. Note that $$\sin^{10}(x)\geq0$$ for all $$x\in\mathbb R$$ and that $$\sin^{10}(x)>c$$ for all $$x\in\bigcup_{n\in\Bbb N} [\frac\pi4+2n,\frac{3\pi}4+2n]$$ (where $$c>0$$ is some constant). Thus, by "$$\sigma$$-additivity" of the integral, we have (note that $$\frac{\ln x}{\sqrt x}$$ is positive for all $$x\geq 1$$) $$\begin{equation}\label1\tag1 \int_{1}^{\infty} \frac{\sin^{10}x \ln x}{\sqrt{x}} \,\mathrm dx \geq c \sum_{n=1}^\infty\int_{\frac\pi4+2n}^{\frac{3\pi}4+2n} \frac{\ln x}{\sqrt x} \,\mathrm dx \geq c\sum_{n=1}^\infty\int_{\frac\pi4+2n}^{\frac{3\pi}4+2n} \frac1x\,\mathrm dx. \end{equation}$$

The right-hand side of \eqref{1} equals $$c\sum_{n=1}^\infty \ln(\frac{3\pi}4+2n)-\ln(\frac\pi4+2n)$$. If we can show that the sum of the last expression diverges, we have thus shown that the integral diverges. Let me thus show that the sum diverges:

We have by this question $$\begin{equation} \lim_{n\to\infty} \frac{\ln(\frac34 \pi + 2n)-\ln(\frac\pi4+2n)}{\ln(2n+2)-\ln(2n)} = \frac\pi4 > 0. \end{equation}$$ We also have $$\begin{equation}\sum_{n=1}^\infty \ln(2n+2)-\ln(2n) = \lim_{n\to\infty}\ln(2n)=\infty.\end{equation}$$

Hence, by the comparison test, the sum that should diverge actually does diverge.

This achieves a proof of the claim. $$\square$$

• Proving that $$\int_{0}^{\epsilon} ( \ln x) \,\mathrm dx = -\infty$$ does not show that our integral diverges (you would have to bound it the other way around, i.e. you would need "our integral < $$\int_{0}^{\epsilon} ( \ln x) \,\mathrm dx$$" for that to be correct.)
• The above fact is actually wrong (see also the answer by mihaild). We know that $$\int_{0}^{\epsilon} ( \ln x) \,\mathrm dx = \epsilon\ln(\epsilon)-\epsilon-\lim_{x\to 0}(x\ln(x)-x)$$. By writing the last limit as $$\frac{\ln(x)}{\frac1x}$$, we find that your integral is actually a finite number for every $$\epsilon$$!
• $${x^{10} \ln x}<\frac{x^{10} \ln x}{\sqrt{x}}$$ is clearly wrong for $$x<1$$. In fact, the opposite is the case.
• $$\frac{x^{10} \ln x}{\sqrt{x}}<\frac{\sin^{10}x \ln x}{\sqrt{x}}$$ is also wrong for small $$x>0$$.

$$\int_{0}^{\epsilon} (x^{10} \ln x) dx<\int_{0}^{\epsilon} \frac{x^{10} \ln x}{\sqrt{x}} dx$$ is wrong. Also $$e^A \cdot (A - 1) \to 0$$ as $$A \to -\infty$$. And $$\int_{0}^\varepsilon \ln x\, dx = (x \ln x - x)\rvert_0^\varepsilon = \varepsilon(\ln \varepsilon - 1)$$, so this integral converges.

To prove that your integral diverges, you can use that $$\sin^{10} x > a$$ for some positive $$a$$ if $$x \in [2 \pi k + \frac{\pi}{4}, 2 \pi k + \frac{3\pi}{4}]$$ for some integer $$k$$, $$\frac{\sin^{10} x \ln x}{\sqrt{x}} > \frac{a}{\sqrt x}$$ if $$x > 2\pi$$ and so $$\int\limits_{2\pi}^{2\pi(m + 1)} \frac{\sin^{10} x \ln x}{\sqrt{x}}\, dx > \sum\limits_{k = 2}^{m} \int\limits_{2 \pi k + \frac{\pi}{4}}^{2 \pi k + \frac{3\pi}{4}}\frac{1}{\sqrt x}\, dx > \sum\limits_{k = 2}^{m} \frac{1}{\sqrt{2\pi k + \frac{3\pi}{4}}}$$

As this series diverges (by comparsion with harmonic series, for example - $$\frac{1}{\sqrt{2 \pi k + \frac{3 \pi}{4}}} < \frac{1}{k}$$ for large enough $$k$$) - so does your integral.

• It's essentially the same as @Maximilian Janisch's solution, but with a bit tighter lower bound for integration term and a bit simpler proof for divergence (it's enough to estimate integral over segment by lower bound of function, it's not necessary to calculate it exactly). – mihaild May 22 at 18:02
• This is a bit smarter than my solution basically (+1) 😃 – Maximilian Janisch May 22 at 18:18