Proposition 2 ("continuity of an integral with respect to a parameter")

Let $F$ be a compact space, let $I = [a,\ b]$ be a compact interval in $\mathbf{R}$, and let $f$ be a continuous map of $I \times F$ into a complete normed space $E$ over $\mathbf{R}$; then the function $h(\alpha) = \int_a^b f(x, \alpha)\, dx$ is continuous on $F$.

Indeed, since $f$ is uniformly continuous on the compact space $I \times F$, the functions $f(x, \alpha)$ converge uniformly to $f(x, \alpha_0)$ on $I$, when $\alpha$ tends to an arbitrary point $\alpha_0 \in F$.

Here is an application of this proposition: the function $(x, \alpha) \mapsto x^\alpha$ is continuous on the product $I \times J$, where $I = [a,\ b]$ is a compact interval such that $0 < a < b$, and $J$ is any compact interval in $\mathbf{R}$; one concludes that $\int_a^b x^\alpha\, dx$ is a continuous function of $\alpha$ on $\mathbf{R}$; now, for $\alpha$ rational and $\neq -1$, this function is equal to $\frac{b^{\alpha+1} - a^{\alpha+1}}{\alpha + 1}$, and the function $\alpha \mapsto \frac{b^{\alpha+1} - a^{\alpha+1}}{\alpha + 1}$ is continuous on every interval of $\mathbf{R}$ not containing $-1$; one thus has (extension of identities) $\int_a^b x^\alpha\, dx = \frac{b^{\alpha+1} - a^{\alpha+1}}{\alpha + 1}$ for all real $\alpha \neq -1$; this again means that, for all real $\alpha$, the derivative of $x^\alpha$ is $\alpha x^{\alpha-1}$ (cf. III, p. 94).