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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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    $\begingroup$ Wikipedia has a nice article about this. en.wikipedia.org/wiki/Differentiation_under_the_integral_sign You may differentiate under the integral sign provided your function is nice enough (in this case; continious function + continous derivative) $\endgroup$ – Fredrik Meyer Dec 3 '10 at 14:16
  • $\begingroup$ @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. $\endgroup$ – Qiaochu Yuan Dec 3 '10 at 14:35
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    $\begingroup$ Richard Feynman has remarks on solving problems this way in some of his books. As a physicist, all his functions were well behaved. $\endgroup$ – Ross Millikan Dec 3 '10 at 16:50
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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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  • $\begingroup$ 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) $\endgroup$ – Matt Groff Dec 3 '10 at 17:29
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    $\begingroup$ 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.) $\endgroup$ – Qiaochu Yuan Dec 3 '10 at 17:36
  • $\begingroup$ How is this result proven? $\endgroup$ – Brian Bi Apr 20 '14 at 22:08
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    $\begingroup$ @BrianBri: You basically use the definition of the derivative, then convert the limit $(F(t+h)-F(t))/h$ to a sequential limit $(F(t+h_n)-F(t))/h_n$, use the linearity of the integral and apply the dominated convergence theorem. One thing I find important is that often one can not find a function $g$ as above. But most of the time, this is only due to the fact that $U$ is "too large"; consider e.g. the (trivial) example $f(0,\infty)\times\Bbb{N}, (y,n)\mapsto 1/(y \cdot n^2)$. But because differentiability is a local property, we can shrink $U$ (e.g. to (\epsilon,\infty) above). $\endgroup$ – PhoemueX Dec 20 '14 at 9:35
  • $\begingroup$ The theorem given in the German Wikipedia (see Yaroslav's answer below) requires $f$ to be continuously differentiable which is because, I assume, they use the mean-value theorem to be able to apply the dominated convergence theorem and taking the limit of the derivative inside the integral then requires continuity. Is the theorem above really true if $f'$ is not necessarily continuous? $\endgroup$ – balu Feb 3 '16 at 20:54
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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.

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  • $\begingroup$ A theorem formulation in a foreign language may not be the best for the reference, but if one doesn't know the language, I've realised that even a word-by-word translation may be easier than proving the theorem yourself. $\endgroup$ – Yaroslav Nikitenko Oct 24 '14 at 15:15

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