Leibniz integral rule

This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Leibniz integral rule" – news · newspapers · books · scholar · JSTOR (October 2016) (Learn how and when to remove this message)

Part of a series of articles about
Calculus
${\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}$
Fundamental theorem Limits Continuity Rolle's theorem Mean value theorem Inverse function theorem
Differential Definitions Derivative (generalizations) Differential infinitesimal of a function total Concepts Differentiation notation Second derivative Implicit differentiation Logarithmic differentiation Related rates Taylor's theorem Rules and identities Sum Product Chain Power Quotient L'Hôpital's rule Inverse General Leibniz Faà di Bruno's formula Reynolds
Integral Lists of integrals Integral transform Leibniz integral rule Definitions Antiderivative Integral (improper) Riemann integral Lebesgue integration Contour integration Integral of inverse functions Integration by Parts Discs Cylindrical shells Substitution (trigonometric, tangent half-angle, Euler) Euler's formula Partial fractions (Heaviside's method) Changing order Reduction formulae Differentiating under the integral sign Risch algorithm
Series Geometric (arithmetico-geometric) Harmonic Alternating Power Binomial Taylor Convergence tests Summand limit (term test) Ratio Root Integral Direct comparison Limit comparison Alternating series Cauchy condensation Dirichlet Abel
Vector Gradient Divergence Curl Laplacian Directional derivative Identities Theorems Gradient Green's Stokes' Divergence generalized Stokes Helmholtz decomposition
Multivariable Formalisms Matrix Tensor Exterior Geometric Definitions Partial derivative Multiple integral Line integral Surface integral Volume integral Jacobian Hessian
Advanced Calculus on Euclidean space Generalized functions Limit of distributions
Specialized Fractional Malliavin Stochastic Variations
Miscellanea Precalculus History Glossary List of topics Integration Bee Mathematical analysis Nonstandard analysis
v t e

In calculus, the Leibniz integral rule for differentiation under the integral sign, named after Gottfried Wilhelm Leibniz, states that for an integral of the form $${\displaystyle \int _{a(x)}^{b(x)}f(x,t)\,dt,}$$ where ${\displaystyle -\infty <a(x),b(x)<\infty }$ and the integrands are functions dependent on ${\displaystyle x,}$ the derivative of this integral is expressible as $${\displaystyle {\begin{aligned}&{\frac {d}{dx}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)\\&=f{\big (}x,b(x){\big )}\cdot {\frac {d}{dx}}b(x)-f{\big (}x,a(x){\big )}\cdot {\frac {d}{dx}}a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt\end{aligned}}}$$ where the partial derivative ${\displaystyle {\tfrac {\partial }{\partial x}}}$ indicates that inside the integral, only the variation of ${\displaystyle f(x,t)}$ with ${\displaystyle x}$ is considered in taking the derivative.^[1]

In the special case where the functions ${\displaystyle a(x)}$ and ${\displaystyle b(x)}$ are constants ${\displaystyle a(x)=a}$ and ${\displaystyle b(x)=b}$ with values that do not depend on ${\displaystyle x,}$ this simplifies to: $${\displaystyle {\frac {d}{dx}}\left(\int _{a}^{b}f(x,t)\,dt\right)=\int _{a}^{b}{\frac {\partial }{\partial x}}f(x,t)\,dt.}$$

If ${\displaystyle a(x)=a}$ is constant and ${\displaystyle b(x)=x}$, which is another common situation (for example, in the proof of Cauchy's repeated integration formula), the Leibniz integral rule becomes: $${\displaystyle {\frac {d}{dx}}\left(\int _{a}^{x}f(x,t)\,dt\right)=f{\big (}x,x{\big )}+\int _{a}^{x}{\frac {\partial }{\partial x}}f(x,t)\,dt,}$$

This important result may, under certain conditions, be used to interchange the integral and partial differential operators, and is particularly useful in the differentiation of integral transforms. An example of such is the moment generating function in probability theory, a variation of the Laplace transform, which can be differentiated to generate the moments of a random variable. Whether Leibniz's integral rule applies is essentially a question about the interchange of limits.

The right hand side may also be written using Lagrange's notation as: ${\textstyle f(x,b(x))\,b^{\prime }(x)-f(x,a(x))\,a^{\prime }(x)+\displaystyle \int _{a(x)}^{b(x)}f_{x}(x,t)\,dt.}$

Stronger versions of the theorem only require that the partial derivative exist almost everywhere, and not that it be continuous.^[2] This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. The (first) fundamental theorem of calculus is just the particular case of the above formula where ${\displaystyle a(x)=a\in \mathbb {R} }$ is constant, ${\displaystyle b(x)=x,}$ and ${\displaystyle f(x,t)=f(t)}$ does not depend on ${\displaystyle x.}$

If both upper and lower limits are taken as constants, then the formula takes the shape of an operator equation: $${\displaystyle {\mathcal {I}}_{t}\partial _{x}=\partial _{x}{\mathcal {I}}_{t}}$$ where ${\displaystyle \partial _{x}}$ is the partial derivative with respect to ${\displaystyle x}$ and ${\displaystyle {\mathcal {I}}_{t}}$ is the integral operator with respect to ${\displaystyle t}$ over a fixed interval. That is, it is related to the symmetry of second derivatives, but involving integrals as well as derivatives. This case is also known as the Leibniz integral rule.

The following three basic theorems on the interchange of limits are essentially equivalent:

• the interchange of a derivative and an integral (differentiation under the integral sign; i.e., Leibniz integral rule);
• the change of order of partial derivatives;
• the change of order of integration (integration under the integral sign; i.e., Fubini's theorem).

A Leibniz integral rule for a two dimensional surface moving in three dimensional space is^[3][4]

$${\displaystyle {\frac {d}{dt}}\iint _{\Sigma (t)}\mathbf {F} (\mathbf {r} ,t)\cdot d\mathbf {A} =\iint _{\Sigma (t)}\left(\mathbf {F} _{t}(\mathbf {r} ,t)+\left[\nabla \cdot \mathbf {F} (\mathbf {r} ,t)\right]\mathbf {v} \right)\cdot d\mathbf {A} -\oint _{\partial \Sigma (t)}\left[\mathbf {v} \times \mathbf {F} (\mathbf {r} ,t)\right]\cdot d\mathbf {s} ,}$$

where:

• F(r, t) is a vector field at the spatial position r at time t,
• Σ is a surface bounded by the closed curve ∂Σ,
• dA is a vector element of the surface Σ,
• ds is a vector element of the curve ∂Σ,
• v is the velocity of movement of the region Σ,
• ∇⋅ is the vector divergence,
• × is the vector cross product,
• The double integrals are surface integrals over the surface Σ, and the line integral is over the bounding curve ∂Σ.

The Leibniz integral rule can be extended to multidimensional integrals. In two and three dimensions, this rule is better known from the field of fluid dynamics as the Reynolds transport theorem: $${\displaystyle {\frac {d}{dt}}\int _{D(t)}F(\mathbf {x} ,t)\,dV=\int _{D(t)}{\frac {\partial }{\partial t}}F(\mathbf {x} ,t)\,dV+\int _{\partial D(t)}F(\mathbf {x} ,t)\mathbf {v} _{b}\cdot d\mathbf {\Sigma } ,}$$

where ${\displaystyle F(\mathbf {x} ,t)}$ is a scalar function, D(t) and ∂D(t) denote a time-varying connected region of R³ and its boundary, respectively, ${\displaystyle \mathbf {v} _{b}}$ is the Eulerian velocity of the boundary (see Lagrangian and Eulerian coordinates) and dΣ = n dS is the unit normal component of the surface element.

The general statement of the Leibniz integral rule requires concepts from differential geometry, specifically differential forms, exterior derivatives, wedge products and interior products. With those tools, the Leibniz integral rule in n dimensions is^[4] $${\displaystyle {\frac {d}{dt}}\int _{\Omega (t)}\omega =\int _{\Omega (t)}i_{\mathbf {v} }(d_{x}\omega )+\int _{\partial \Omega (t)}i_{\mathbf {v} }\omega +\int _{\Omega (t)}{\dot {\omega }},}$$ where Ω(t) is a time-varying domain of integration, ω is a p-form, ${\displaystyle \mathbf {v} ={\frac {\partial \mathbf {x} }{\partial t}}}$ is the vector field of the velocity, ${\displaystyle i_{\mathbf {v} }}$ denotes the interior product with ${\displaystyle \mathbf {v} }$, d_xω is the exterior derivative of ω with respect to the space variables only and ${\displaystyle {\dot {\omega }}}$ is the time derivative of ω.

The above formula can be deduced directly from the fact that the Lie derivative interacts nicely with integration of differential forms $${\displaystyle {\frac {d}{dt}}\int _{\Omega (t)}\omega =\int _{\Omega (t)}{\mathcal {L}}_{\Psi }\omega ,}$$ for the spacetime manifold ${\displaystyle M=\mathbb {R} \times \mathbb {R} ^{3}}$, where the spacetime exterior derivative of ${\displaystyle \omega }$ is ${\displaystyle d\omega =dt\wedge {\dot {\omega }}+d_{x}\omega }$ and the surface ${\displaystyle \Omega (t)}$ has spacetime velocity field ${\displaystyle \Psi ={\frac {\partial }{\partial t}}+\mathbf {v} }$. Since ${\displaystyle \omega }$ has only spatial components, the Lie derivative can be simplified using Cartan's magic formula, to $${\displaystyle {\mathcal {L}}_{\Psi }\omega ={\mathcal {L}}_{\mathbf {v} }\omega +{\mathcal {L}}_{\frac {\partial }{\partial t}}\omega =i_{\mathbf {v} }d\omega +di_{\mathbf {v} }\omega +i_{\frac {\partial }{\partial t}}d\omega =i_{\mathbf {v} }d_{x}\omega +di_{\mathbf {v} }\omega +{\dot {\omega }}}$$ which, after integrating over ${\displaystyle \Omega (t)}$ and using generalized Stokes' theorem on the second term, reduces to the three desired terms.

Let ${\displaystyle X}$ be an open subset of ${\displaystyle \mathbf {R} }$, and ${\displaystyle \Omega }$ be a measure space. Suppose ${\displaystyle f\colon X\times \Omega \to \mathbf {R} }$ satisfies the following conditions:^[5][6][2]

1. ${\displaystyle f(x,\omega )}$ is a Lebesgue-integrable function of ${\displaystyle \omega }$ for each ${\displaystyle x\in X}$.
2. For almost all ${\displaystyle \omega \in \Omega }$ , the partial derivative ${\displaystyle f_{x}}$ exists for all ${\displaystyle x\in X}$.
3. There is an integrable function ${\displaystyle \theta \colon \Omega \to \mathbf {R} }$ such that ${\displaystyle |f_{x}(x,\omega )|\leq \theta (\omega )}$ for all ${\displaystyle x\in X}$ and almost every ${\displaystyle \omega \in \Omega }$.

Then, for all ${\displaystyle x\in X}$, $${\displaystyle {\frac {d}{dx}}\int _{\Omega }f(x,\omega )\,d\omega =\int _{\Omega }f_{x}(x,\omega )\,d\omega .}$$

The proof relies on the dominated convergence theorem and the mean value theorem (details below).

We first prove the case of constant limits of integration a and b.

We use Fubini's theorem to change the order of integration. For every x and h, such that h > 0 and both x and x +h are within [x₀,x₁], we have: $${\displaystyle {\begin{aligned}\int _{x}^{x+h}\int _{a}^{b}f_{x}(x,t)\,dt\,dx&=\int _{a}^{b}\int _{x}^{x+h}f_{x}(x,t)\,dx\,dt\\[2ex]&=\int _{a}^{b}\left(f(x+h,t)-f(x,t)\right)\,dt\\[2ex]&=\int _{a}^{b}f(x+h,t)\,dt-\int _{a}^{b}f(x,t)\,dt\end{aligned}}}$$

Note that the integrals at hand are well defined since ${\displaystyle f_{x}(x,t)}$ is continuous at the closed rectangle ${\displaystyle [x_{0},x_{1}]\times [a,b]}$ and thus also uniformly continuous there; thus its integrals by either dt or dx are continuous in the other variable and also integrable by it (essentially this is because for uniformly continuous functions, one may pass the limit through the integration sign, as elaborated below).

Therefore: $${\displaystyle {\begin{aligned}{\frac {\int _{a}^{b}f(x+h,t)\,dt-\int _{a}^{b}f(x,t)\,dt}{h}}&={\frac {1}{h}}\int _{x}^{x+h}\int _{a}^{b}f_{x}(x,t)\,dt\,dx\\[2ex]&={\frac {F(x+h)-F(x)}{h}}\end{aligned}}}$$

Where we have defined: $${\displaystyle F(u):=\int _{x_{0}}^{u}\int _{a}^{b}f_{x}(x,t)\,dt\,dx}$$ (we may replace x₀ here by any other point between x₀ and x)

F is differentiable with derivative ${\textstyle \int _{a}^{b}f_{x}(x,t)\,dt}$, so we can take the limit where h approaches zero. For the left hand side this limit is: $${\displaystyle {\frac {d}{dx}}\int _{a}^{b}f(x,t)\,dt}$$

For the right hand side, we get: $${\displaystyle F'(x)=\int _{a}^{b}f_{x}(x,t)\,dt}$$ And we thus prove the desired result: $${\displaystyle {\frac {d}{dx}}\int _{a}^{b}f(x,t)\,dt=\int _{a}^{b}f_{x}(x,t)\,dt}$$

If the integrals at hand are Lebesgue integrals, we may use the bounded convergence theorem (valid for these integrals, but not for Riemann integrals) in order to show that the limit can be passed through the integral sign.

Note that this proof is weaker in the sense that it only shows that f_x(x,t) is Lebesgue integrable, but not that it is Riemann integrable. In the former (stronger) proof, if f(x,t) is Riemann integrable, then so is f_x(x,t) (and thus is obviously also Lebesgue integrable).

Let

$${\displaystyle u(x)=\int _{a}^{b}f(x,t)\,dt.}$$

(1)

By the definition of the derivative,

$${\displaystyle u'(x)=\lim _{h\to 0}{\frac {u(x+h)-u(x)}{h}}.}$$

(2)

Substitute equation (1) into equation (2). The difference of two integrals equals the integral of the difference, and 1/h is a constant, so $${\displaystyle {\begin{aligned}u'(x)&=\lim _{h\to 0}{\frac {\int _{a}^{b}f(x+h,t)\,dt-\int _{a}^{b}f(x,t)\,dt}{h}}\\&=\lim _{h\to 0}{\frac {\int _{a}^{b}\left(f(x+h,t)-f(x,t)\right)\,dt}{h}}\\&=\lim _{h\to 0}\int _{a}^{b}{\frac {f(x+h,t)-f(x,t)}{h}}\,dt.\end{aligned}}}$$

We now show that the limit can be passed through the integral sign.

We claim that the passage of the limit under the integral sign is valid by the bounded convergence theorem (a corollary of the dominated convergence theorem). For each δ > 0, consider the difference quotient $${\displaystyle f_{\delta }(x,t)={\frac {f(x+\delta ,t)-f(x,t)}{\delta }}.}$$ For t fixed, the mean value theorem implies there exists z in the interval [x, x + δ] such that $${\displaystyle f_{\delta }(x,t)=f_{x}(z,t).}$$ Continuity of f_x(x, t) and compactness of the domain together imply that f_x(x, t) is bounded. The above application of the mean value theorem therefore gives a uniform (independent of ${\displaystyle t}$) bound on ${\displaystyle f_{\delta }(x,t)}$. The difference quotients converge pointwise to the partial derivative f_x by the assumption that the partial derivative exists.

The above argument shows that for every sequence {δ_n} → 0, the sequence ${\displaystyle \{f_{\delta _{n}}(x,t)\}}$ is uniformly bounded and converges pointwise to f_x. The bounded convergence theorem states that if a sequence of functions on a set of finite measure is uniformly bounded and converges pointwise, then passage of the limit under the integral is valid. In particular, the limit and integral may be exchanged for every sequence {δ_n} → 0. Therefore, the limit as δ → 0 may be passed through the integral sign.

If instead we only know that there is an integrable function ${\displaystyle \theta \colon \Omega \to \mathbf {R} }$ such that ${\displaystyle |f_{x}(x,\omega )|\leq \theta (\omega )}$, then ${\displaystyle |f_{\delta }(x,t)|=|f_{x}(z,t)|\leq \theta (\omega )}$ and the dominated convergence theorem allows us to move the limit inside of the integral.

For a continuous real valued function g of one real variable, and real valued differentiable functions ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ of one real variable, $${\displaystyle {\frac {d}{dx}}\left(\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt\right)=g\left(f_{2}(x)\right){f_{2}'(x)}-g\left(f_{1}(x)\right){f_{1}'(x)}.}$$

This follows from the chain rule and the First Fundamental Theorem of Calculus. Define $${\displaystyle G(x)=\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt,}$$ and $${\displaystyle \Gamma (x)=\int _{0}^{x}g(t)\,dt.}$$ (The lower limit just has to be some number in the domain of ${\displaystyle g}$)

Then, ${\displaystyle G(x)}$ can be written as a composition: ${\displaystyle G(x)=(\Gamma \circ f_{2})(x)-(\Gamma \circ f_{1})(x)}$. The Chain Rule then implies that $${\displaystyle G'(x)=\Gamma '\left(f_{2}(x)\right)f_{2}'(x)-\Gamma '\left(f_{1}(x)\right)f_{1}'(x).}$$ By the First Fundamental Theorem of Calculus, ${\displaystyle \Gamma '(x)=g(x)}$. Therefore, substituting this result above, we get the desired equation: $${\displaystyle G'(x)=g\left(f_{2}(x)\right){f_{2}'(x)}-g\left(f_{1}(x)\right){f_{1}'(x)}.}$$

Note: This form can be particularly useful if the expression to be differentiated is of the form: $${\displaystyle \int _{f_{1}(x)}^{f_{2}(x)}h(x)\,g(t)\,dt}$$ Because ${\displaystyle h(x)}$ does not depend on the limits of integration, it may be moved out from under the integral sign, and the above form may be used with the Product rule, i.e., $${\displaystyle {\begin{aligned}{\frac {d}{dx}}\left(\int _{f_{1}(x)}^{f_{2}(x)}h(x)g(t)\,dt\right)&={\frac {d}{dx}}\left(h(x)\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt\right)\\&=h'(x)\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt+h(x){\frac {d}{dx}}\left(\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt\right)\end{aligned}}}$$

Set $${\displaystyle \varphi (\alpha )=\int _{a}^{b}f(x,\alpha )\,dx,}$$ where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα. Then, $${\displaystyle {\begin{aligned}\Delta \varphi &=\varphi (\alpha +\Delta \alpha )-\varphi (\alpha )\\[4pt]&=\int _{a+\Delta a}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[4pt]&=\int _{a+\Delta a}^{a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}f(x,\alpha +\Delta \alpha )\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[4pt]&=-\int _{a}^{a+\Delta a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx.\end{aligned}}}$$

A form of the mean value theorem, ${\textstyle \int _{a}^{b}f(x)\,dx=(b-a)f(\xi )}$, where a < ξ < b, may be applied to the first and last integrals of the formula for Δφ above, resulting in $${\displaystyle \Delta \varphi =-\Delta af(\xi _{1},\alpha +\Delta \alpha )+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\Delta bf(\xi _{2},\alpha +\Delta \alpha ).}$$

Divide by Δα and let Δα → 0. Notice ξ₁ → a and ξ₂ → b. We may pass the limit through the integral sign: $${\displaystyle \lim _{\Delta \alpha \to 0}\int _{a}^{b}{\frac {f(x,\alpha +\Delta \alpha )-f(x,\alpha )}{\Delta \alpha }}\,dx=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}f(x,\alpha )\,dx,}$$ again by the bounded convergence theorem. This yields the general form of the Leibniz integral rule, $${\displaystyle {\frac {d\varphi }{d\alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}f(x,\alpha )\,dx+f(b,\alpha ){\frac {db}{d\alpha }}-f(a,\alpha ){\frac {da}{d\alpha }}.}$$

The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of the basic form of Leibniz's Integral Rule, the multivariable chain rule, and the first fundamental theorem of calculus. Suppose ${\displaystyle f}$ is defined in a rectangle in the ${\displaystyle x-t}$ plane, for ${\displaystyle x\in [x_{1},x_{2}]}$ and ${\displaystyle t\in [t_{1},t_{2}]}$. Also, assume ${\displaystyle f}$ and the partial derivative ${\textstyle {\frac {\partial f}{\partial x}}}$ are both continuous functions on this rectangle. Suppose ${\displaystyle a,b}$ are differentiable real valued functions defined on ${\displaystyle [x_{1},x_{2}]}$, with values in ${\displaystyle [t_{1},t_{2}]}$ (i.e. for every ${\displaystyle x\in [x_{1},x_{2}],a(x),b(x)\in [t_{1},t_{2}]}$). Now, set $${\displaystyle F(x,y)=\int _{t_{1}}^{y}f(x,t)\,dt,\qquad {\text{for}}~x\in [x_{1},x_{2}]~{\text{and}}~y\in [t_{1},t_{2}]}$$ and $${\displaystyle G(x)=\int _{a(x)}^{b(x)}f(x,t)\,dt,\quad {\text{for}}~x\in [x_{1},x_{2}]}$$

Then, by properties of definite Integrals, we can write $${\displaystyle G(x)=\int _{t_{1}}^{b(x)}f(x,t)\,dt-\int _{t_{1}}^{a(x)}f(x,t)\,dt=F(x,b(x))-F(x,a(x))}$$

Since the functions ${\displaystyle F,a,b}$ are all differentiable (see the remark at the end of the proof), by the multivariable chain rule, it follows that ${\displaystyle G}$ is differentiable, and its derivative is given by the formula: $${\displaystyle G'(x)=\left({\frac {\partial F}{\partial x}}(x,b(x))+{\frac {\partial F}{\partial y}}(x,b(x))b'(x)\right)-\left({\frac {\partial F}{\partial x}}(x,a(x))+{\frac {\partial F}{\partial y}}(x,a(x))a'(x)\right)}$$Now, note that for every ${\displaystyle x\in [x_{1},x_{2}]}$, and for every ${\displaystyle y\in [t_{1},t_{2}]}$, we have that ${\textstyle {\frac {\partial F}{\partial x}}(x,y)=\int _{t_{1}}^{y}{\frac {\partial f}{\partial x}}(x,t)\,dt}$, because when taking the partial derivative with respect to ${\displaystyle x}$ of ${\displaystyle F}$, we are keeping ${\displaystyle y}$ fixed in the expression ${\textstyle \int _{t_{1}}^{y}f(x,t)\,dt}$; thus the basic form of Leibniz's Integral Rule with constant limits of integration applies. Next, by the first fundamental theorem of calculus, we have that ${\textstyle {\frac {\partial F}{\partial y}}(x,y)=f(x,y)}$; because when taking the partial derivative with respect to ${\displaystyle y}$ of ${\displaystyle F}$, the first variable ${\displaystyle x}$ is fixed, so the fundamental theorem can indeed be applied.

Substituting these results into the equation for ${\displaystyle G'(x)}$ above gives: $${\displaystyle {\begin{aligned}G'(x)&=\left(\int _{t_{1}}^{b(x)}{\frac {\partial f}{\partial x}}(x,t)\,dt+f(x,b(x))b'(x)\right)-\left(\int _{t_{1}}^{a(x)}{\dfrac {\partial f}{\partial x}}(x,t)\,dt+f(x,a(x))a'(x)\right)\\[2pt]&=f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial f}{\partial x}}(x,t)\,dt,\end{aligned}}}$$ as desired.

There is a technical point in the proof above which is worth noting: applying the Chain Rule to ${\displaystyle G}$ requires that ${\displaystyle F}$ already be differentiable. This is where we use our assumptions about ${\displaystyle f}$. As mentioned above, the partial derivatives of ${\displaystyle F}$ are given by the formulas ${\textstyle {\frac {\partial F}{\partial x}}(x,y)=\int _{t_{1}}^{y}{\frac {\partial f}{\partial x}}(x,t)\,dt}$ and ${\textstyle {\frac {\partial F}{\partial y}}(x,y)=f(x,y)}$. Since ${\textstyle {\dfrac {\partial f}{\partial x}}}$ is continuous, its integral is also a continuous function,^[7] and since ${\displaystyle f}$ is also continuous, these two results show that both the partial derivatives of ${\displaystyle F}$ are continuous. Since continuity of partial derivatives implies differentiability of the function,^[8] ${\displaystyle F}$ is indeed differentiable.

At time t the surface Σ in Figure 1 contains a set of points arranged about a centroid ${\displaystyle \mathbf {C} (t)}$. The function ${\displaystyle \mathbf {F} (\mathbf {r} ,t)}$ can be written as $${\displaystyle \mathbf {F} (\mathbf {C} (t)+\mathbf {r} -\mathbf {C} (t),t)=\mathbf {F} (\mathbf {C} (t)+\mathbf {I} ,t),}$$ with ${\displaystyle \mathbf {I} }$ independent of time. Variables are shifted to a new frame of reference attached to the moving surface, with origin at ${\displaystyle \mathbf {C} (t)}$. For a rigidly translating surface, the limits of integration are then independent of time, so: $${\displaystyle {\frac {d}{dt}}\left(\iint _{\Sigma (t)}d\mathbf {A} _{\mathbf {r} }\cdot \mathbf {F} (\mathbf {r} ,t)\right)=\iint _{\Sigma }d\mathbf {A} _{\mathbf {I} }\cdot {\frac {d}{dt}}\mathbf {F} (\mathbf {C} (t)+\mathbf {I} ,t),}$$ where the limits of integration confining the integral to the region Σ no longer are time dependent so differentiation passes through the integration to act on the integrand only: $${\displaystyle {\frac {d}{dt}}\mathbf {F} (\mathbf {C} (t)+\mathbf {I} ,t)=\mathbf {F} _{t}(\mathbf {C} (t)+\mathbf {I} ,t)+\mathbf {v\cdot \nabla F} (\mathbf {C} (t)+\mathbf {I} ,t)=\mathbf {F} _{t}(\mathbf {r} ,t)+\mathbf {v} \cdot \nabla \mathbf {F} (\mathbf {r} ,t),}$$ with the velocity of motion of the surface defined by $${\displaystyle \mathbf {v} ={\frac {d}{dt}}\mathbf {C} (t).}$$

This equation expresses the material derivative of the field, that is, the derivative with respect to a coordinate system attached to the moving surface. Having found the derivative, variables can be switched back to the original frame of reference. We notice that (see article on curl) $${\displaystyle \nabla \times \left(\mathbf {v} \times \mathbf {F} \right)=(\nabla \cdot \mathbf {F} +\mathbf {F} \cdot \nabla )\mathbf {v} -(\nabla \cdot \mathbf {v} +\mathbf {v} \cdot \nabla )\mathbf {F} ,}$$ and that Stokes theorem equates the surface integral of the curl over Σ with a line integral over ∂Σ: $${\displaystyle {\frac {d}{dt}}\left(\iint _{\Sigma (t)}\mathbf {F} (\mathbf {r} ,t)\cdot d\mathbf {A} \right)=\iint _{\Sigma (t)}{\big (}\mathbf {F} _{t}(\mathbf {r} ,t)+\left(\mathbf {F\cdot \nabla } \right)\mathbf {v} +\left(\nabla \cdot \mathbf {F} \right)\mathbf {v} -(\nabla \cdot \mathbf {v} )\mathbf {F} {\big )}\cdot d\mathbf {A} -\oint _{\partial \Sigma (t)}\left(\mathbf {v} \times \mathbf {F} \right)\cdot d\mathbf {s} .}$$

The sign of the line integral is based on the right-hand rule for the choice of direction of line element ds. To establish this sign, for example, suppose the field F points in the positive z-direction, and the surface Σ is a portion of the xy-plane with perimeter ∂Σ. We adopt the normal to Σ to be in the positive z-direction. Positive traversal of ∂Σ is then counterclockwise (right-hand rule with thumb along z-axis). Then the integral on the left-hand side determines a positive flux of F through Σ. Suppose Σ translates in the positive x-direction at velocity v. An element of the boundary of Σ parallel to the y-axis, say ds, sweeps out an area vt × ds in time t. If we integrate around the boundary ∂Σ in a counterclockwise sense, vt × ds points in the negative z-direction on the left side of ∂Σ (where ds points downward), and in the positive z-direction on the right side of ∂Σ (where ds points upward), which makes sense because Σ is moving to the right, adding area on the right and losing it on the left. On that basis, the flux of F is increasing on the right of ∂Σ and decreasing on the left. However, the dot product v × F ⋅ ds = −F × v ⋅ ds = −F ⋅ v × ds. Consequently, the sign of the line integral is taken as negative.

If v is a constant, $${\displaystyle {\frac {d}{dt}}\iint _{\Sigma (t)}\mathbf {F} (\mathbf {r} ,t)\cdot d\mathbf {A} =\iint _{\Sigma (t)}{\big (}\mathbf {F} _{t}(\mathbf {r} ,t)+\left(\nabla \cdot \mathbf {F} \right)\mathbf {v} {\big )}\cdot d\mathbf {A} -\oint _{\partial \Sigma (t)}\left(\mathbf {v} \times \mathbf {F} \right)\cdot \,d\mathbf {s} ,}$$ which is the quoted result. This proof does not consider the possibility of the surface deforming as it moves.

Lemma. One has: $${\displaystyle {\frac {\partial }{\partial b}}\left(\int _{a}^{b}f(x)\,dx\right)=f(b),\qquad {\frac {\partial }{\partial a}}\left(\int _{a}^{b}f(x)\,dx\right)=-f(a).}$$

Proof. From the proof of the fundamental theorem of calculus,

$${\displaystyle {\begin{aligned}{\frac {\partial }{\partial b}}\left(\int _{a}^{b}f(x)\,dx\right)&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\left(\int _{a}^{b+\Delta b}f(x)\,dx-\int _{a}^{b}f(x)\,dx\right)\\[1ex]&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\left(\int _{a}^{b}f(x)\,dx+\int _{b}^{b+\Delta b}f(x)\,dx-\int _{a}^{b}f(x)\,dx\right)\\[1ex]&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\int _{b}^{b+\Delta b}f(x)\,dx\\[1ex]&=\lim _{\Delta b\to 0}{\frac {1}{\Delta b}}\left[f(b)\Delta b+O\left(\Delta b^{2}\right)\right]\\[1ex]&=f(b),\end{aligned}}}$$ and $${\displaystyle {\begin{aligned}{\frac {\partial }{\partial a}}\left(\int _{a}^{b}f(x)\,dx\right)&=\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\left[\int _{a+\Delta a}^{b}f(x)\,dx-\int _{a}^{b}f(x)\,dx\right]\\[6pt]&=\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\int _{a+\Delta a}^{a}f(x)\,dx\\[6pt]&=\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\left[-f(a)\Delta a+O\left(\Delta a^{2}\right)\right]\\[6pt]&=-f(a).\end{aligned}}}$$

Suppose a and b are constant, and that f(x) involves a parameter α which is constant in the integration but may vary to form different integrals. Assume that f(x, α) is a continuous function of x and α in the compact set {(x, α) : α₀ ≤ α ≤ α₁ and a ≤ x ≤ b}, and that the partial derivative f_α(x, α) exists and is continuous. If one defines: $${\displaystyle \varphi (\alpha )=\int _{a}^{b}f(x,\alpha )\,dx,}$$ then ${\displaystyle \varphi }$ may be differentiated with respect to α by differentiating under the integral sign, i.e., $${\displaystyle {\frac {d\varphi }{d\alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}f(x,\alpha )\,dx.}$$

By the Heine–Cantor theorem it is uniformly continuous in that set. In other words, for any ε > 0 there exists Δα such that for all values of x in [a, b], $${\displaystyle |f(x,\alpha +\Delta \alpha )-f(x,\alpha )|<\varepsilon .}$$

On the other hand, $${\displaystyle {\begin{aligned}\Delta \varphi &=\varphi (\alpha +\Delta \alpha )-\varphi (\alpha )\\[6pt]&=\int _{a}^{b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[6pt]&=\int _{a}^{b}\left(f(x,\alpha +\Delta \alpha )-f(x,\alpha )\right)\,dx\\[6pt]&\leq \varepsilon (b-a).\end{aligned}}}$$

Hence φ(α) is a continuous function.

Similarly if ${\displaystyle {\frac {\partial }{\partial \alpha }}f(x,\alpha )}$ exists and is continuous, then for all ε > 0 there exists Δα such that: $${\displaystyle \forall x\in [a,b],\quad \left|{\frac {f(x,\alpha +\Delta \alpha )-f(x,\alpha )}{\Delta \alpha }}-{\frac {\partial f}{\partial \alpha }}\right|<\varepsilon .}$$

Therefore, $${\displaystyle {\frac {\Delta \varphi }{\Delta \alpha }}=\int _{a}^{b}{\frac {f(x,\alpha +\Delta \alpha )-f(x,\alpha )}{\Delta \alpha }}\,dx=\int _{a}^{b}{\frac {\partial f(x,\alpha )}{\partial \alpha }}\,dx+R,}$$ where $${\displaystyle |R|<\int _{a}^{b}\varepsilon \,dx=\varepsilon (b-a).}$$

Now, ε → 0 as Δα → 0, so $${\displaystyle \lim _{{\Delta \alpha }\to 0}{\frac {\Delta \varphi }{\Delta \alpha }}={\frac {d\varphi }{d\alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}f(x,\alpha )\,dx.}$$

This is the formula we set out to prove.

Now, suppose $${\displaystyle \int _{a}^{b}f(x,\alpha )\,dx=\varphi (\alpha ),}$$ where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then, $${\displaystyle {\begin{aligned}\Delta \varphi &=\varphi (\alpha +\Delta \alpha )-\varphi (\alpha )\\[6pt]&=\int _{a+\Delta a}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[6pt]&=\int _{a+\Delta a}^{a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}f(x,\alpha +\Delta \alpha )\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[6pt]&=-\int _{a}^{a+\Delta a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx.\end{aligned}}}$$

A form of the mean value theorem, ${\textstyle \int _{a}^{b}f(x)\,dx=(b-a)f(\xi ),}$ where a < ξ < b, can be applied to the first and last integrals of the formula for Δφ above, resulting in $${\displaystyle \Delta \varphi =-\Delta a\,f(\xi _{1},\alpha +\Delta \alpha )+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\Delta b\,f(\xi _{2},\alpha +\Delta \alpha ).}$$

Dividing by Δα, letting Δα → 0, noticing ξ₁ → a and ξ₂ → b and using the above derivation for $${\displaystyle {\frac {d\varphi }{d\alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}f(x,\alpha )\,dx}$$ yields $${\displaystyle {\frac {d\varphi }{d\alpha }}=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}f(x,\alpha )\,dx+f(b,\alpha ){\frac {\partial b}{\partial \alpha }}-f(a,\alpha ){\frac {\partial a}{\partial \alpha }}.}$$

This is the general form of the Leibniz integral rule.

Consider the function $${\displaystyle \varphi (\alpha )=\int _{0}^{1}{\frac {\alpha }{x^{2}+\alpha ^{2}}}\,dx.}$$

The function under the integral sign is not continuous at the point (x, α) = (0, 0), and the function φ(α) has a discontinuity at α = 0 because φ(α) approaches ±π/2 as α → 0^±.

If we differentiate φ(α) with respect to α under the integral sign, we get $${\displaystyle {\frac {d}{d\alpha }}\varphi (\alpha )=\int _{0}^{1}{\frac {\partial }{\partial \alpha }}\left({\frac {\alpha }{x^{2}+\alpha ^{2}}}\right)\,dx=\int _{0}^{1}{\frac {x^{2}-\alpha ^{2}}{(x^{2}+\alpha ^{2})^{2}}}dx=\left.-{\frac {x}{x^{2}+\alpha ^{2}}}\right|_{0}^{1}=-{\frac {1}{1+\alpha ^{2}}},}$$ for α≠0. This may be integrated (with respect to α) to find $${\displaystyle \varphi (\alpha )={\begin{cases}0,&\alpha =0,\\-\arctan({\alpha })+{\frac {\pi }{2}},&\alpha \neq 0.\end{cases}}}$$

An example with variable limits: $${\displaystyle {\begin{aligned}{\frac {d}{dx}}\int _{\sin x}^{\cos x}\cosh t^{2}\,dt&=\cosh \left(\cos ^{2}x\right){\frac {d}{dx}}(\cos x)-\cosh \left(\sin ^{2}x\right){\frac {d}{dx}}(\sin x)+\int _{\sin x}^{\cos x}{\frac {\partial }{\partial x}}(\cosh t^{2})\,dt\\[6pt]&=\cosh(\cos ^{2}x)(-\sin x)-\cosh(\sin ^{2}x)(\cos x)+0\\[6pt]&=-\cosh(\cos ^{2}x)\sin x-\cosh(\sin ^{2}x)\cos x.\end{aligned}}}$$