The following Leibniz integral rule is well-known

**Theorem**. Let be a function such that both and its partial derivative are continuous in and in some region of the -plane, including , . Also suppose that the functions and are both continuous and both have continuous derivatives for . Then, for ,

The purpose of this note is to show that, in fact, it is not is not necessary to assume the function to be continuous. We note that this is indeed the case in which the limits of the integral do not depend on the parameter . For convenience, it is routine to assume the continuity, which immediately implies that all integrals are well-defined.

As mentioned above, we want to show that this is also the case for integrals of the form above.

To this purpose, we first establish the following result:

**Lemma**. Let be an integrable function such that is continuous in both and in some region of the -plane, including , . Also suppose that the function is continuous and has continuous derivative for . Then, for ,

Here is a proof for this lemma.

*Proof of Lemma*. By definition, it is necessary to consider the quotient

Fixing and letting to zero, it is suffices to show that the above tends to the right hand side of the identity in the lemma. To this end, we write

We observe that the second term on the right hand side of the preceding identity is a good term as it tends to

as . For the first term, we shall make use of the mean value theorem in the following way. We write

For the first term on the right hand side, we apply the Lagrange theorem, for each fixed and fixed to get

where lies between and . Since is continuous in the region , , it is bounded in it. Hence, there is some constant such that

for all . In particular, we estimate

to get

due to the continuity of at . For the second term, we apply the mean value theorem to get

for some lying between and . Since exists, , as a function of , is continuous. Hence taking the limit as gives

thanks to the fact that as . The present proof is complete.

We now apply the above lemma to conclude the Leibniz rule in the following way. If we want to prove that the integral is differentiable at the point and its derivative is nothing but

we decompose as follows

From this and after taking the limit as , it is not hard to realize that

as claimed.

I thank Dr. Trinh Viet Duoc for useful discussion on this rule.

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Thank you so much for your hard work and the beautiful proof presented here!

Comment by Tiến — September 15, 2018 @ 6:28