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

Question

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.

Answer 1

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. $\square$

Your mistakes:

• 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 $x>0$.

Answer 2

$\int_{0}^{\epsilon} (x^{10} \ln x) dx<\int_{0}^{\epsilon} \frac{x^{10} \ln x}{\sqrt{x}} dx$ 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.