“Differentiating under the Integral” is a useful trick, and here we describe and prove a sufficient condition where we can use the trick. This is the Measure-Theoretic version, which is more general than the usual version stated in calculus books.

Let be an open subset of , and be a measure space. Suppose satisfies the following conditions:

1) is a Lebesgue-integrable function of for each .

2) For almost all , the derivative exists for all .

3) There is an integrable function such that for all .

Then for all ,

**Proof:**

By definition,

Let be a sequence tending to 0, and define

It follows that is measurable.

Using the Mean Value Theorem, we have for each .

Thus for each , by the Dominated Convergence Theorem, we have which implies

That is,

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*Related*

Prof Feynman (Nobel Physics) was very good at this technique learned during his high school for complicated integral :

https://tomcircle.wordpress.com/2013/04/23/differentiating-under-integral/

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