Examples of improper integrals conditionally convergent

Question

I've been looking for examples of improper integrals conditionally convergent, but I get the same result in every document I read: $$\int_0^\infty \frac{\sin x} x \, dx$$ Can you help me?.

Answer 1

You could mimic the same general idea by considering $$\int_0^\infty \frac{(-1)^{\lfloor x\rfloor}}{\lceil x\rceil}\,dx\ .$$

Answer 2

You can take a look at the Dirichlet test for improper integrals. Then for any odd and bounded periodic function $f$ and any monotone function $g$ such that $\lim_{x\to\infty}g(x)=0$ we have that the integral

$$\int_0^\infty f(x)g(x)\mathrm dx$$

converges. However in many cases we will have that

$$\int_0^\infty|f(x)g(x)|\mathrm dx=\infty$$

depending of the function $g$. That is: because $f$ is bounded then

$$\begin{align}\int_0^\infty|f(x)g(x)|\mathrm dx&=\sum_{k=0}^\infty|g(\xi_k)|\int_{kP}^{(k+1)P} |f(x)|\mathrm dx,\quad \xi_k\in[kP,(k+1)P]\\&=M\sum_{k=0}^\infty |g(\xi_k)|,\quad \xi_k\in[kP,(k+1)P]\\&\ge M\sum_{k=0}^\infty \inf\{|g(x)|:x\in [kP,(k+1)P]\}\end{align}$$

where we applied the MVT for integrals in the first step, $P$ is the period of $f$ and $M:=\int_{kP}^{(k+1)P}|f(x)|\mathrm dx$. By example: choosing $g(x)=x^{-\alpha}$ for $\alpha\in(0,1]$ the improper integral is conditionally convergent, however choosing $g(x)=x^{-\alpha}$ for $\alpha>1$ the improper integral is absolutely convergent.