Remark 2

Tag 089D, permanent. Cite it as [Bourbaki, Tag 089D].

Functions of a Real Variable, chapter II, § 3. DERIVATIVES AND INTEGRALS OF FUNCTIONS DEPENDING ON A PARAMETER

When, in an integral $\int_{a(\alpha)}^{b(\alpha)} f(t, \alpha) \, dt$ , the endpoints of the interval are finite functions of the parameter, the study of this integral as a function of $\alpha$ can be related to that of an integral over $[0, 1]$ ; indeed, by the change of variable $t = a(\alpha)(1 - u) + b(\alpha)u$ , one has

\int_{a(\alpha)}^{b(\alpha)} f(t, \alpha) \, dt = \int_0^1 f(a(\alpha)(1 - u) + b(\alpha)u, \alpha) (b(\alpha) - a(\alpha)) \, du.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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