Convergence of an improper integral (with a parameter)

Question

I'm currently trying to solve an exercise that asks for which values of $\alpha \in \mathbb{R}$ the integral

$\int_0^{+\infty}f(x)\,dx$ is convergent, where $f(x)=\frac{1}{x^\alpha} \log\left(1+\frac{1}{x}\right)\arctan(x)$.

I first split the integral in $\int_0^1 f(x) \, dx$ and $\int_1^{+\infty}f(x)\,dx$. I found that $$f(x)\sim \frac{\log\left(1+\frac{1}{x}\right)\frac{\pi}{2}}{x^{\alpha+1}\frac{1}{x}} \sim \frac{\frac{\pi}{2}}{x^{\alpha+1}} \text{ as } x \rightarrow +\infty.$$

So $\int_1^{+\infty} f(x)\,dx$ converges iff $\alpha > 0$.

And now I'm stuck here. How should I continue?

Thanks.

Answer 1

As $x \to 0^+ $, you have $$\frac{1}{x^{\alpha}} \ln\left(1+\frac{1}{x}\right)\arctan(x) \sim -\frac{x\:\ln x}{x^{\alpha}}=-\frac{\ln x}{x^{\alpha-1}}$$ and, by comparison, $\displaystyle \int_0^1\frac{1}{x^{\alpha}} \ln\left(1+\frac{1}{x}\right)\arctan(x) \:dx$ is of the same nature than $$-\int_0^1\frac{\ln x}{x^{\alpha-1}}dx=\frac{1}{(\alpha -2)^2} \quad (\alpha<2).$$

Thus

$$ \int_0^{\infty}\frac{1}{x^{\alpha}} \ln\left(1+\frac{1}{x}\right)\arctan(x) \:dx \quad \text{is convergent iff} \quad 0<\alpha<2.$$

Answer 2

HINT:

For the convergence at the upper limit, note that

$$\left|\frac{\log(1+1/x)}{x^a}\right|=\left|\frac{\log(1+x)-\log x}{x^a}\right|\le \frac{1/x}{x^a}$$

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HINT 2:

For convgence at the lower limit, note

$$\frac{\log(1+1/x)\arctan(x)}{x^a}=\frac{-x\log x+O(x^2)}{x^a}$$