Example

i) The integral $\int_1^{+\infty} dt/t^2$ is convergent and equal to 1, for

2) The integral $\int_0^1 dt / \sqrt{t}$ is convergent and equal to 2, for

3) Let $(\mathbf{u}_n)_{n \geq 1}$ be an infinite sequence of points of E, and let $\mathbf{f}$ be the step function defined on the interval $[1, +\infty[$ by the conditions: $\mathbf{f}(x) = \mathbf{u}_n$ for $n \leq x < n+1$. Then for the integral $\int_1^{+\infty} \mathbf{f}(t)\,dt$ to be convergent it is necessary and sufficient that the series with general term $\mathbf{u}_n$ be convergent in E; indeed, one has

so the condition is necessary; conversely, if the series with general term $\mathbf{u}_n$ converges in E, then $\lim_{n \to \infty} \mathbf{u}_n = 0$; now, if $n \leq x \leq n+1$, one has $\int_1^n \mathbf{f}(t)\,dt = \sum_{p=1}^{n-1} \mathbf{u}_p + \mathbf{u}_n(x-n)$. so this integral has the limit $\sum_{n=1}^{\infty} \mathbf{u}_n$ when $x$ tends to $+\infty$.

It is immediate that if a piecewise regulated function $\mathbf{f}$ admits an integral over I then the formulae (4) to (9) of II, p. 59 remain valid. Similarly, formula (10) of II, p. 59 extends in the following manner: $\mathbf{f}$ and $\mathbf{g}$ are assumed to be primitives of the regulated functions $\mathbf{f}'$ and $\mathbf{g}'$ on $]a,\ b[$, and one denotes by $[\mathbf{f}.\mathbf{g}]|_a^b$ the limit (if it exists) of $[\mathbf{f}.\mathbf{g}]|_x^y$ as $(x,\ y)$ tends to $(a,\ b)$ (with $a < x \leq y < b$); then, if two of the (three) expressions $[\mathbf{f}.\mathbf{g}]|_a^b$, $\int_a^b [\mathbf{f}(t).\mathbf{g}'(t)]\,dt$, and $\int_a^b [\mathbf{f}'(t).\mathbf{g}(t)]\,dt$ have a meaning, then so has the third, and the formula (10) of II, p. 59 is valid.

Finally, let $f$ be a real function which is defined and continuous on $I = ]a,\ b[$, and is the primitive of a regulated function $f'$ on $]a,\ b[$; let on the other hand $g$ be a continuous vector function on an open interval J containing $f(I)$; if the function $g(f(x))f'(x)$ admits an integral over I, and if $f$ tends to a limit (finite or not) at the points $a$ and $b$, then $g$ admits an integral from $f(a+)$ to $f(b-)$, and one has the formula

Indeed, if $(x,\ y)$ tends to $(a,\ b)$, then $(f(x),\ f(y))$ tends to $(f(a+),\ f(b-))$ by hypothesis; it suffices to apply formula (12) of II, p. 60 between $x$ and $y$, and to pass to the limit to obtain (1).

Given a regulated function $\mathbf{f}$ on a non-compact interval $I \subset \mathbf{R}$, with endpoints $a$ and $b$ ($a < b$), the condition for $\mathbf{f}$ to have an integral over I can be presented in the following manner. The compact intervals $J \subset I$ form a directed set $\mathscr{K}(I)$ with respect to the relation $\subset ^1$, for if $[ \alpha , \beta ]$ and $[ \gamma , \delta ]$ are two compact intervals contained in $I$, and if one puts $\lambda = \min(\alpha, \gamma)$, $\mu = \max(\beta, \delta)$, then the interval $[ \lambda , \mu ]$ is contained in $I$ and contains the two intervals considered. For each compact interval $J = [ \alpha , \beta ]$ contained in $I$, let us put

for $f$ to admit an integral over $I$ it is necessary and sufficient that the map $J \mapsto \int_J f(t) \, dt$ have a limit with respect to the directed set $\mathfrak{K}(I)$; this limit is then the integral $\int_a^b f(t) \, dt$, which we again denote by $\int_I f(t) \, dt$.