Show that $\int_1^\infty \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx$ is convergent only for $\lambda > \frac{1}{2}$

Question

Show that the improper integral $$\int_1^\infty \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx$$ is convergent only for $\lambda > \frac{1}{2}$.

We will show that the sequence of integrals on $[1,M]$ must be bounded for $\lambda>1/2$.

• Let $\lambda \leq 1/2$ thus for $x\geq e$ the result $$\frac{\ln x}{\left(1+x^2\right)^\lambda}\geq \frac{1}{\left(2x^2\right)^\lambda} = \frac{1}{2^\lambda x^{2\lambda}}\geq \frac{1}{2^\lambda x}$$ implies $$\int_e^\infty\frac{1}{x}\mathrm dx\text{ does not exist}\rightarrow\int_e^\infty \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx\text{ does not exist}.$$
• Let $\lambda >1/2$ and $\mu>0$ with $2\lambda-\mu >1$, hence $$\underbrace{\lim_{x\to\infty}\frac{\ln x}{x^\mu}}_{\color{red}(\color{red}*\color{red}1\color{red})}\overset{\color{red}(\color{red}*\color{red}2\color{red})}{=}\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{\mu}x^{\mu-1}}=\lim_{x\to\infty}\frac{\mu}{x^\mu}=0.$$ Therefore there must be a $M_0$ s.t. $\ln x\leq x^\mu$ on $[M_0,\infty)$. Furthermore $$f(x)=\frac{\ln x}{\left(1+x^2\right)^\lambda}\overset{\color{red}(\color{red}*\color{red}3\color{red})}{\leq}\frac{x^\mu}{x^{2\lambda}}=\frac{1}{x^{2\lambda-\mu}}=g(x)$$ with $g(x)$ being integrable on $[M_0,\infty)$ (because $2\lambda-\mu>1$) thus $f(x)$ as well. Since $$\int_1^{M_0} \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx$$ is finite so is $$\int_1^{M_0} \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx+\int_{M_0}^\infty \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx.$$

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I have some trouble understanding this proof in regards to the marked steps:

• $(*1)$ I don't understand what the motivation behind this expression is - is it just used to show that $\ln x=o\left(x^\mu\right)$?
• $(*2)$ This looked like l'Hôpital at first but the denominator seems to be wrong since $(x^\mu)'=\mu x^{\mu-1}$ or did I miss something else?
• $(*3)$ I guess this is the step where the result from $(*1)$ is used to make an estimation which feels really far-fetched. Can you give me an intuition why this is true?
• Furthermore I do see that $2\lambda-\mu>1$ for our $\mu$ was a decent pick after seing the final result as $g(x)$ is integrable due to that - how could I possibly make this assumption beforehand without knowing the solution?
• Finally: do you have alternative (maybe easier) solutions to this problem?

Answer 1

if $\lambda \gt \frac12$ $$ \int_1^{M} \frac{\ln x}{\left(1+x^2\right)^\lambda}\mathrm dx \lt \int_1^{M} \frac{\ln x}{x^{2\lambda}}\mathrm dx = \int_1^{M}\frac{\ln x}{x^{1+\mu}} \mathrm dx $$ where $\mu \gt 0$

integrating by parts the latter integral is: $$ \left[-\frac{\ln x}{\mu x^{\mu}} \right]_1^M +\int_1^{M}\frac1{\mu x^{1+\mu}} \mathrm dx $$ which remains finite as $M \to \infty$. in fact the first term is, apart from a constant, $M^{-\mu}\ln M$. considering this as a function of a positive real argument $M$, differentiation shows that it attains a maximum at $M=e^{\mu^{-1}}$.

Answer 2

Issue $(1)$

Yes, the motivation here is to prove that for any $\mu>0$, $\lim_{x\to \infty}\frac{\log x}{x^\mu}=0$. Note that another way to show this is to simply rely on the inequalities for $x>0$

$$\frac{x-1}{x}\le \log (x)\le x-1 \tag 1$$

Note that Equation $(1)$ implies that for $x>0$, we have $\frac{x^a-1}{x^a}\le \log x^a=a\log x\le x^a-1$. Therefore, if we take $\mu>a>0$, then we have

$$\frac{x^a-1}{x^\mu\,x^a}\le \frac{\log x}{x^\mu}\le \frac{x^{a-\mu}-x^{-\mu}}{a}$$

and from the squeeze theorem, we conclude that for all $x>0$ and any $\mu>0$, we have

$$\lim_{x\to \infty}\frac{\log x}{x^\mu}=0$$

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Issue $(2)$

This appears to by a careless error or typographical one. You are correct; this is a simple application of L'Hospital's Rule.

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Issue $(3)$

Yes, this is a consequence of the limit from Issue $(1)$. And , the set of inequalities in Equation $(1)$ provides intuition.

But note that we cannot write $\log x\le x^\mu -1 \le x^\mu$ for all $x$. We could write, however, $\log x\le \frac{x^\mu-1}{\mu}\le \frac{x^\mu}{\mu}$. Alternatively, for any $\mu$ and $x$ sufficiently large, we can write $\log x\le x^{\mu}$.

We also have $(1+x^2)^\lambda \ge x^{2\lambda}$ so that

$$\frac{\log x}{(1+x^2)^\lambda}\le \frac{1}{\mu x^{2\lambda -\mu}} \tag 2$$

We can choose any $\mu$ here and the inequality in Equation $(2)$ is valid. So, we are free to choose $\mu$ such that for $\lambda >1/2$, $2\lambda -\mu>1$.