Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove the following version of Liebniz's Rule:

Let $f:[a,b]\times [c,d]\to \mathbb{R}$ be integrable with respect to the first variable, $\phi,\psi:[c,d]\to [a,b]$ be differentiable and let $F:[c,d]\to \mathbb{R}$, \begin{equation}F(y)=\int_{\phi(y)}^{\psi (y)}f(x,y)\, dx\end{equation} If $f$ is partially differentiable with respect to $y\in [c,d]$ and there exists an integrable functions $g:[a,b]\to \mathbb{R}$ so that \begin{equation}\forall x\in [a,b]\ \left|\partial_y f(x,y)\right|\le g(x)\end{equation} then $F$ is differentiable in $[c,d]$ and it holds \begin{equation}F'(y)=f(\psi(y),y)\psi'(y)-f(\phi(y),y)\phi'(y)+\int_{\phi(y)}^{\psi(y)}\partial_yf(x,y)\, dx\end{equation} almost everywhere in $[c,d]$.

Proof: By linearity it suffices to deal with constant $\phi(y)=a$. Now let $y_n\to y$ in $[c,d]$ with $y_n\neq y$. Then,

\begin{equation}\frac{F(y_n)-F(y)}{y_n-y}=\frac1{y_n-y}\left(\int_{a}^{\psi (y_n)}f(x,y_n)\, dx-\int_{a}^{\psi (y)}f(x,y)\, dx\right)=\\ \frac1{y_n-y}\left(\int_{a}^{\psi (y_n)}f(x,y_n)-f(x,y)\, dx\right)+\frac1{y_n-y} \int_{\psi(y)}^{\psi (y_n)}f(x,y)\, dx\end{equation}

The limit of the second term is essentialy the derivative of $\int_a^{\phi(t)}f(x,y) dx$ with respect to $t$ at $y$. By Lebesgue's differentiation theorem, \begin{equation}\lim_{y\to +\infty}\frac1{y_n-y}\int_{\psi(y)}^{\psi (y_n)}f(x,y)\, dx=f(\psi(y),y)\psi'(y)\end{equation} almost everywhere.

For the first term, consider the integrable function \begin{equation}h_n(x)=\frac{f(x,y_n)-f(x,y)}{y_n-y}\end{equation} By the Mean Value Theorem there exists some $\xi_{n,x}$ between $y_n$ and $y$ with \begin{equation}h_n(x)=\frac{f(x,y_n)-f(x,y)}{y_n-y}=\partial_yf(x,\xi_{n,x})\end{equation} By the Dominated Convergence Theorem, as $h_n(x)\to \partial_yf(x,y)$, \begin{equation}\int_a^{\psi(y_m)}\partial_yf(x,t)\, dx=\lim_{n\to+\infty}\frac1{y_n-y}\left(\int_{a}^{\psi (y_m)}f(x,y_n)-f(x,y)\, dx\right) \end{equation} and so by the continuity of the integral, \begin{equation}\lim_{m\to+\infty}\lim_{n\to+\infty}\frac1{y_n-y}\left(\int_{a}^{\psi (y_m)}f(x,y_n)-f(x,y)\, dx\right)=\int_a^{\psi(y)}\partial_yf(x,y)\, dx\end{equation} and similarly, \begin{equation}\lim_{n\to+\infty}\lim_{m\to+\infty}\frac1{y_n-y}\left(\int_{a}^{\psi (y_m)}f(x,y_n)-f(x,y)\, dx\right)=\int_a^{\psi(y)}\partial_yf(x,y)\, dx\end{equation}

As I asked here earlier today we can't deduce that \begin{equation}\lim_{n\to+\infty}\frac1{y_n-y}\left(\int_{a}^{\psi (y_n)}f(x,y_n)-f(x,y)\, dx\right)=\int_a^{\psi(y)}\partial_yf(x,y)\, dx\end{equation} So how do we proceed?

share|cite|improve this question
up vote 2 down vote accepted

First, of all note that if you consider φ and ψ as variables which are dependent on y, then you obtain: $$F(y)=\int_\phi^\psi (x,y)dx=H(\phi,\psi,y)$$ Now differentiate, and by application of Leibniz's rule you obtain: $$F'(y)= \frac{\partial H}{\partial \phi}\frac{d \phi}{dy}+ \frac{\partial H}{\partial \psi}\frac{d \psi}{dy}+\frac{\partial H}{\partial y}$$

It's easy now to see that : $$\frac{\partial H}{\partial \psi}=f(\psi,y), \frac{\partial H}{\partial \phi}=-f(\phi,y)$$ also since $|\frac{\partial f(x,y)}{\partial y}|\leq g(x)$, by applying dominated convergence theorem we can put the derivative inside the integral:$$\frac{\partial H}{\partial y}=\int_\phi^\psi\frac{\partial f(x,y)}{\partial y}dx$$ and combining the above relations we obtain the desired result.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.