Improper Integrals Between Finite Limits

Assume that the function $f(x)$ is monotone on the interval $(0, 1)$. It need not be bounded at the points $x = 0$, $x = 1$, we assume however that the improper integral $\int_{0}^{1} f(x) \, dx$ exists. Under these conditions

(continued). If $\varphi (x)$ is properly integrable over $\left[0, 1\right]$ we have

Prove in a way different from Problem 71 of Part I that for $\alpha > 0$

We put

Then

exists and is non-zero. (If $0 < \alpha < 1$, $0 < \beta < 1$, $\alpha + \beta \geq 1$, $z = -1$, the product series is divergent although the two factors are convergent.)

The converse of the statement of Problem 20 is not valid: There are functions, monotone in the interval $(0, 1)$, for which $\lim_{n \to \infty } \sum_{k=1}^{n-1} \frac{1}{n} f\left(\frac{k}{n}\right)$ exists but not the integral.

The integral $\int_{0}^{1} f(x) \, dx$ exists if $f(x)$ is monotone in the interval $(0, 1)$, finite at $x = 0$ or $x = 1$ and if the following limit is finite

Let the monotone function $f(x)$ be defined on $(0, 1)$. Then the equation

holds under the condition that the integral at right exists.

If $\alpha > 0$

If $f(x)$ is properly integrable over $\left[0, 1\right]$ we have obviously

Show that this is true also if $f(x)$ is improperly integrable but monotone.

If $f(x)$ is monotone for $x > 0$, $\lim_{n \to \infty } \varepsilon_{n} = 0$, $c > 0$, $\varepsilon_{n} > \dfrac {c}{n}$ we find

provided that the integral at right exists and $f(x)$ is finite at $x = 1$.