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

I'm interested in the potential of such a technique. I got the idea from Moron's answer to this question, which uses the technique of differentiation under the integral.

Now, I'd like to consider this integral:

$$\int_{-\pi}^\pi \cos{(y(1-e^{i\cdot n \cdot t}))}\mathrm dt$$

I'd like to differentiate with respect to y. This will give the integral:

$$\int_{-\pi}^\pi -(1-e^{i\cdot n \cdot t})(\sin{(y(1-e^{i\cdot n \cdot t}))}\mathrm dt$$

...If I'm correct. Anyways, I'm interested in obtaining the results to this second integral, using this technique. So I'm wondering if solving the first integral can help give results for the second integral. I'm thinking of setting $y=1$ in the second integral. This should eliminate $y$ from the result, and give me the integral involving $x$.

The trouble is, I'm not sure I can use the technique of differentiation under the integral. I want to know how I can apply this technique to the integrals above. Any pointers are appreciated.

For instance, for what values of $y$ is this valid?

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– Fredrik Meyer Dec 3 '10 at 14:16
@Fredrik: Wikipedia's statement covers the Riemann integral case, but the statement I give below is for an integral with respect to an arbitrary measure (including the Lebesgue integral); in particular it covers the case of differentiating a series term-by-term. – Qiaochu Yuan Dec 3 '10 at 14:35

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Richard Feynman has remarks on solving problems this way in some of his books. As a physicist, all his functions were well behaved. – Ross Millikan Dec 3 '10 at 16:50

## 1 Answer

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 $y$).

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I'm wondering, additionally, if I can easily add a third variable to this. The idea is that I'd like to differentiate twice under the integral; once with respect to one variable, and then once with respect to a different variable. I'm wondering if any additional complications arise. (I'm debating on whether I should ask this in a seperate question) – Matt Groff Dec 3 '10 at 17:29

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Yes; apply the theorem twice. (in the first application E will be an open subset of R^2 instead of an open subset of R.) – Qiaochu Yuan Dec 3 '10 at 17:36
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