Find functions $f$ and $\alpha$ such that $\int_1^{\infty}|f|d\alpha$ converges, but $\int_1^{\infty}fd\alpha$ does not exist?

Question

Find functions $f$ and $\alpha$ such that the improper Riemann-Stieltjes integral $\int_1^{\infty}|f|d\alpha$ converges, but $\int_1^{\infty}fd\alpha$ does not exist?

I'm really not sure how to start this problem, and I haven't been able to find another post on here that has considered this.

EDIT: I know that $\alpha$ needs to be some function which is not increasing or differentiable on $[1,\infty )$ since then absolute convergence implies conditional convergence

Thank you,

Answer 1

How about the function:

$$f(x)=\begin{cases}\frac{1}{\sqrt{n}} \text{ for } x \in [n,n+1) \text{ and odd } n\\ -\frac{1}{\sqrt{n}} \text{ for } x \in [n,n+1) \text{ and even } n \end{cases}$$

and let $\alpha(x)=xf(x)$.

Then for odd $n$,

$$\int_{n}^{n+1} f(x) d\alpha(x)=\int_{n}^{n+1} f(x)\alpha'(x) dx=\int_{n}^{n+1} \frac{1}{n} dx=\frac{1}{n}$$

$$\int_{n}^{n+1} |f(x)| d\alpha(x)=\int_{n}^{n+1} \frac{1}{n} dx=\frac{1}{n}$$

For even $n$:

$$\int_{n}^{n+1} f(x) d\alpha(x)=\frac{1}{n}$$ $$\int_{n}^{n+1} |f(x)| d\alpha(x)=-\frac{1}{n}$$

So then $$\int_{1}^{\infty} |f(x)| d\alpha(x)=\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n}<\infty$$ $$\int_{1}^{\infty} f(x) d\alpha(x)=\sum\limits_{n=1}^{\infty} \frac{1}{n}=\infty$$

Answer 2

I see this is the same basic idea as another answer, but here it is anyway: Set $\alpha (x) = -\cos x.$ Then $d\alpha (x) = \sin x \,dx.$ Define $f(t)= 1/t$ on $(0,\pi), (2\pi,3\pi), \dots, f(t) = -1/t$ on $ (\pi,2\pi),(3\pi,4\pi), \dots$ Then

$$\int_0^\infty |f(t)|\,d\alpha(t) = \int_0^\infty \frac{\sin t}{t}\,dt,$$

which converges. But

$$\int_0^\infty f(t)\,d\alpha(t) = \int_0^\infty \frac{|\sin t|}{t}\,dt = \infty.$$