Leibniz rule derivation

Question

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?

Answer 1

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.

Answer 2

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 $t\in[u,v]$.

Answer 3

\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.