On the first paragraph of the Wikipedia page regarding the Leibniz integral rule,

we get the expression

$$\frac{d}{dx} \int_{y_0}^{y_1} f(x,y)dy = \int_{y_0}^{y_1} f_x(x,y)dy $$

and it says that "provided that both $f$ and $f_x$ are continuous over some region $[x_0,x_1]X[y_0,y_1]$

Can we relax this condition and just require that $f$ and $f_x$ be Riemann-integrable?

(I am not studying measure theory at the moment, so I don't want to discuss Lebesgue integrability.)

So that if I wanted to prove that some integral is differentiable, I just need to show that the derivative of the integrand (with respect to some variable x or y) is integrable, so that the differentiation under the integral sign is allowed to happen. This avoids using difference quotients to prove differentiability of some integral -- since the application of the Leibniz rule will already imply the differentiability. But showing the integrability of $f_x$ or $f_y$ does require some technical work too, usually I see that integration by parts is necessary here.

Is my thinking correct?

And one last question:

The Wikipedia page also gives a Leibniz rule for integrals with variable limits:

$$\frac{d}{dx} \int_{g(x)}^{h(x)}f(t)dt= f(h(x))h'(x)-f(g(x))g'(x)$$

this is a sort of easily acceptable claim, when thinking of differentiation and integration as (almost) inverses of each other and thinking about the chain rule.

However, prior to this statement, Wikipedia also gives this similar-looking claim:

$$\frac{d}{dx} \int_{g(x)}^{h(x)}f(x,t)dt= \int_{g(x)}^{h(x)}\frac{\partial f}{\partial x}dt + f(h(x))h'(x)-f(g(x))g'(x) $$

So, where did all the extra terms $f(h(x))h'(x)-f(g(x))g'(x)$ come from?

Shouldn't it just be

$$\frac{d}{dx} \int_{g(x)}^{h(x)}f(x,t)dt= \int_{g(x)}^{h(x)}\frac{\partial f}{\partial x}dt? $$

This is probably because $f$ is a function of two variables $(x,t)$ instead of just $t$, but I don't really see how it is justified.

Thanks so much,


1 Answer 1


For the first question: I'm afraid the answer is "in general, without continuity, the result is not true". Take a look at this link (Example 1) https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign#Examples

For the second question: notice that if you consider the incremental quotient definition you get something like (let me take $g=0$, but the general case is similar) $$ I=\frac{\int^{h(x+\delta)}_{0}f(x+\delta,t)dt-\int^{h(x)}_{0}f(x,t)dt}{\delta}, $$ $$ I=\frac{\int^{h(x+\delta)}_{0}f(x+\delta,t)dt\pm\int^{h(x)}_{0}f(x+\delta,t)dt-\int^{h(x)}_{0}f(x,t)dt}{\delta}. $$ Then, taking the limit $$ \lim I=\lim \frac{h(x+\delta)-h(x)}{\delta}\frac{1}{h(x+\delta)-h(x)}\int_{h(x)}^{h(x+\delta)}f(x+\delta,t)\pm f(x,t)dt+\int_0^{h(x)}f_x(x,t)dt, $$ For the last term, take a look at this result https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem

  • 1
    $\begingroup$ Hi @guacho, I have worked out the details of my second question and now believe in it :-) Regarding the first question, so if the function and it's derivative (partial derivative, if $f$ is a function of several variables) are continuous, then I can conclude immediately that the integral is differentiable, since the application of Leibniz's rule implies the differentiability, right? Thanks so much, $\endgroup$
    – User001
    Commented Dec 7, 2015 at 4:42

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