Will moving differentiation from inside, to outside an integral, change the result?

Solution 1:

Wikipedia doesn't seem to have a precise statement of this theorem. Here's a very general statement.

Theorem (Differentiation under the integral sign): Let $U$ be an open subset of $\mathbb{R}$ and let $E$ be a measure space (which you can freely take to be any open subset of $\mathbb{R}^n$ if you want). Let $f : U \times E \to \mathbb{R}$ have the following properties:

• $x \mapsto f(t, x)$ is integrable for all $t$,
• $t \mapsto f(t, x)$ is differentiable for all $x$,
• for some integrable function $g$, for all $x \in E$, and for all $t \in U$,

$$\left| \frac{\partial f}{\partial t}(t, x) \right| \le g(x).$$

Then the function $x \mapsto \frac{\partial f}{\partial t}(t, x)$ is integrable for all $t$. Moreover, the function $F : U \to \mathbb{R}$ defined by

$$F(t) = \int_E f(t, x) \mu(dx)$$

is differentiable, and

$$F'(t) = \int_E \frac{\partial f}{\partial t}(t, x) \mu(dx).$$

In practice the only condition that isn't easily satisfiable is the third one. It is satisfied in this case, so you're fine (for all $y$).

Solution 2:

The general theorem wroten above by Qiaochu Yuan is formulated in the German and French Wikipedias with proofs.

They also give links to some literature, but the French book I wasn't able to find, while in the German books I havent found the statement in its generality.