10
$\begingroup$

How is Leibniz Integral rule derived?

$$\frac {\mathrm{d}}{\mathrm{d}x}\left(\int_{a(x)}^{b(x)}f(x, t) \,\mathrm{d}t\right)= f(x,b(x))\frac{\mathrm{d}}{\mathrm{d}x}b(x)- f(x, a(x))\dfrac{\mathrm{d}}{\mathrm{d}x}a(x)+ \displaystyle\int_{a(x)}^{b(x)}\dfrac{\partial f(x,t)}{\partial x} \,\mathrm{d}t.$$

Also, what is the intuition behind this formula?

$\endgroup$
2

3 Answers 3

11
$\begingroup$

First consider the simplest case where $a(x)=a$ and $b(x)=b$ for all $x$. Then the Leibniz formula becomes $$\frac{d}{dx}\left(\int_a^bf(x,t)dt\right)=\int_a^b\frac{\partial }{\partial x}f(x,t)dx $$ i.e. it is reduced to moving the derivative inside the integral. In this special case, the formula may be proven using the uniform bound on $\frac{\partial}{\partial x}f(x,t)$ which is amongst the hypotheses of Leibniz's rule.

Another thing to notice is that by the fundamental theorem of calculus, if we differentiate with respect to the extrema of integration, we have $$\frac{d}{db}\int_a^bf(x,t)dt=f(x,b),\qquad \frac{d}{da}\int_a^bf(x,t)dt=-f(x,a) $$

In the general case, I like to see it as a consequence of the chain rule (i.e. differentiation of a composition of multivariate functions). Suppose $f(x,t)$ is defined for $x\in [\alpha,\beta]$, and let $I:=a([\alpha,\beta])$, $J:=b([\alpha,\beta])$, so that $f(x,t)$ is defined for all $t\in I\cup J$. consider the map \begin{align*}F: [\alpha,\beta]\times I \times J &\to \mathbb{R}\\ (x,a,b)&\mapsto\int_a^bf(x,t)dt \end{align*} as well as the curve \begin{align*}\gamma: [\alpha,\beta]&\mapsto [\alpha,\beta]\times I\times J\\ x&\mapsto (x,a(x),b(x)) \end{align*} Which (by assumption) is differentiable, with derivative given by $$\gamma'(x)=(1,a'(x),b'(x)) $$ Finally, using the chain rule, as well as the special cases considered at the beginning: \begin{align*}&\frac{d}{dx}\left(\int_{a(x)}^{b(x)}f(x,t)dt\right)=\frac{d}{dx}(F\circ \gamma)(x)={\nabla F}(\gamma(x))\cdot\gamma'(x)= \\ &=\frac{\partial F}{\partial x}(\gamma(x))+a'(x)\frac{\partial F }{\partial a}(\gamma(x))+b'(x)\frac{\partial F}{\partial b}(\gamma(x))=\\ &=\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}dt-f(x,a(x))a'(x)+f(x,b(x))b'(x) \end{align*} As desired.

$\endgroup$
1
  • $\begingroup$ Hi friends, Is it possible to use the same rule to derivate determine $u_x$ and $u_y$, where $$u(x,y)=\int_0^x \int_0^y f(t)dtds?$$ I suppose it is possible but I'm having problems on doing it! Thanks $\endgroup$ Dec 14, 2021 at 21:46
6
$\begingroup$

We have a function $\Phi$of three variables, namely $$\Phi(u,v,w):=\int_u^v f(w,t)\>dt\ ,$$ with the necessary continuity assumptions when $(u,v,w)$ range in some three-dimensional domain $\Omega$. With these givens the function$$g(x):=\Phi\bigl(a(x),b(x),x\bigr)$$ is defined, and we are told to compute its derivative $g'(x)$. By the chain rule we have $$g'(x)=\Phi_{.1}\bigl(a(x),b(x),x\bigr)a'(x)+\Phi_{.2}\bigl(a(x),b(x),x\bigr)b'(x)+\Phi_{.3}\bigl(a(x),b(x),x\bigr)x'(x)\ .$$ Since $$\Phi_{.1}\bigl(u,v,w\bigr)=-f(w,u),\quad \Phi_{.2}\bigl(u,v,w\bigr)=f(w,v),\quad \Phi_{.3}\bigl(u,v,w\bigr)=\int_u^v f_{.1}(w,t)\>dt\ ,$$ whereby only the third part really needs some work, we arrive at the stated formula by plugging everything in.

It remains to prove the "vanilla" Leibniz rule $${d\over dw}\int_u^v f(w,t)\>dt=\int_u^v f_{.1}(w,t)\>dt\ .$$ For this we need that under suitable continuity assumptions the convergence $$\lim_{h\to0}{f(w+h,t)-f(w,t)\over h}=f_{.1}(w,t)$$ is uniform in $t\in[u,v]$.

$\endgroup$
2
  • $\begingroup$ Isn't the vanilla Leibniz rule proved with the dominated convergence theorem (DCT)? See the section of https://en.wikipedia.org/wiki/Leibniz_integral_rule that talks about the DCT. I am trying to understand what's the relationship between the DCT and the Leibniz integral rule. $\endgroup$
    – user168764
    Jul 24, 2020 at 14:09
  • $\begingroup$ @nbro: I'm sure there are various proofs. I don't know whether the OP wanted the weakest assumptions for the formula. $\endgroup$ Jul 24, 2020 at 14:53
1
$\begingroup$

\begin{align} f(u,v) &= \frac{\partial}{\partial v} F(u,v) \\ g(u,v) &= \frac{\partial}{\partial u} F(u,v) \\ \lambda (u,v) &= \frac{\partial}{\partial v} g(u,v) \\ &= \frac{\partial^2}{\partial v \, \partial u} F(u,v) \\ &= \frac{\partial^2}{\partial u \, \partial v} F(u,v) \\ &= \frac{\partial}{\partial u} f(u,v) \\ \int f(u,v) \, dv &= F(u,v) \\ \frac{d}{dx} \int f(u,v) \, dv &= u'(x) \frac{\partial}{\partial u} F(u,v)+ v'(x) \frac{\partial}{\partial v} F(u,v) \\ &= u'(x) g(u,v)+v'(x)f(u,v) \\ \frac{d}{dx} \int_{a(x)}^{b(x)} f(x,t) \, dt &= b'(x)f[x,b(x)]-a'(x)f[x,a(x)]+g[x,b(x)]-g[x,a(x)] \\ &= b'(x)f[x,b(x)]-a'(x)f[x,a(x)]+\int_{a(x)}^{b(x)} \lambda (x,t) \, dt \\ &= b'(x)f[x,b(x)]-a'(x)f[x,a(x)]+ \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) \, dt \\ \end{align}

See also the journal article here.

$\endgroup$
1
  • 2
    $\begingroup$ -1. I think that an answer like that, although providing an elegant "wordless proof", is not helpful for the OP, who is seeking for the intuition behind the formula, which was explained by Lorenzo's answer, rather than a rigorous proof. $\endgroup$
    – rafa11111
    Jul 4, 2018 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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