Part of a series of articles about
Calculus
${\displaystyle f(a)-f(b)=\int _{b}^{a}f'(t)\,dt}$
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 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
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
Miscellaneous 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 states that for an integral of the form

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

${\displaystyle -\infty <a(x),b(x)<\infty }$

${\displaystyle x,}$

$${\displaystyle {\begin{aligned}&{\frac {d}{dx))\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)\\&=f{\big (}x,b(x){\big )}\cdot {\frac {d}{dx))b(x)-f{\big (}x,a(x){\big )}\cdot {\frac {d}{dx))a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x))f(x,t)\,dt\end{aligned))}$$

${\displaystyle {\tfrac {\partial }{\partial x))}$

${\displaystyle f(x,t)}$

${\displaystyle x}$

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.

General form: differentiation under the integral sign

Theorem — Let ${\displaystyle f(x,t)}$ be a function such that both ${\displaystyle f(x,t)}$ and its partial derivative ${\displaystyle f_{x}(x,t)}$ are continuous in ${\displaystyle t}$ and ${\displaystyle x}$ in some region of the ${\displaystyle xt}$-plane, including ${\displaystyle a(x)\leq t\leq b(x),}$ ${\displaystyle x_{0}\leq x\leq x_{1}.}$ Also suppose that the functions ${\displaystyle a(x)}$ and ${\displaystyle b(x)}$ are both continuous and both have continuous derivatives for ${\displaystyle x_{0}\leq x\leq x_{1}.}$ Then, for ${\displaystyle x_{0}\leq x\leq x_{1},}$

$${\displaystyle {\frac {d}{dx))\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right)=f{\big (}x,b(x){\big )}\cdot {\frac {d}{dx))b(x)-f{\big (}x,a(x){\big )}\cdot {\frac {d}{dx))a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x))f(x,t)\,dt.}$$

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:

$${\displaystyle {\mathcal {I))_{t}\partial _{x}=\partial _{x}{\mathcal {I))_{t))$$

${\displaystyle \partial _{x))$

${\displaystyle x}$

${\displaystyle {\mathcal {I))_{t))$

${\displaystyle t}$

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).

Three-dimensional, time-dependent case

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.

Higher dimensions

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]

$${\displaystyle {\frac {d}{dt))\int _{\Omega (t)}\omega =\int _{\Omega (t)}i_{\mathbf {v} }(d_{x}\omega )+\int _{\partial \Omega (t)}i_{\mathbf {v} }\omega +\int _{\Omega (t)}{\dot {\omega )),}$$

${\displaystyle \mathbf {v} ={\frac {\partial \mathbf {x} }{\partial t))}$

${\displaystyle i_{\mathbf {v} ))$

${\displaystyle \mathbf {v} }$

${\displaystyle {\dot {\omega ))}$

However, all of these identities can be derived from a most general statement about Lie derivatives:

$${\displaystyle \left.{\frac {d}{dt))\right|_{t=0}\int _{\operatorname {im} _{\psi _{t))(\Omega )}\omega =\int _{\Omega }{\mathcal {L))_{\Psi }\omega ,}$$

Here, the ambient manifold on which the differential form ${\displaystyle \omega }$ lives includes both space and time.

• ${\displaystyle \Omega }$ is the region of integration (a submanifold) at a given instant (it does not depend on ${\displaystyle t}$, since its parametrization as a submanifold defines its position in time),
• ${\displaystyle {\mathcal {L))}$ is the Lie derivative,
• ${\displaystyle \Psi }$ is the spacetime vector field obtained from adding the unitary vector field in the direction of time to the purely spatial vector field ${\displaystyle \mathbf {v} }$ from the previous formulas (i.e, ${\displaystyle \Psi }$ is the spacetime velocity of ${\displaystyle \Omega }$),
• ${\displaystyle \psi _{t))$ is a diffeomorphism from the one-parameter group generated by the flow of ${\displaystyle \Psi }$, and
• ${\displaystyle {\text{im))_{\psi _{t))(\Omega )}$ is the image of ${\displaystyle \Omega }$ under such diffeomorphism.

Something remarkable about this form, is that it can account for the case when ${\displaystyle \Omega }$ changes its shape and size over time, since such deformations are fully determined by ${\displaystyle \Psi }$.

Measure theory statement

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).

Proofs

Proof of basic form

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:

$${\displaystyle \int _{x}^{x+h}\int _{a}^{b}f_{x}(x,t)\,dt\,dx=\int _{a}^{b}\int _{x}^{x+h}f_{x}(x,t)\,dx\,dt=\int _{a}^{b}\left(f(x+h,t)-f(x,t)\right)\,dt=\int _{a}^{b}f(x+h,t)\,dt-\int _{a}^{b}f(x,t)\,dt}$$

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:

$${\displaystyle {\frac {\int _{a}^{b}f(x+h,t)\,dt-\int _{a}^{b}f(x,t)\,dt}{h))={\frac {1}{h))\int _{x}^{x+h}\int _{a}^{b}f_{x}(x,t)\,dt\,dx={\frac {F(x+h)-F(x)}{h))}$$

Where we have defined:

$${\displaystyle F(u)\equiv \int _{x_{0))^{u}\int _{a}^{b}f_{x}(x,t)\,dt\,dx}$$

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:

$${\displaystyle F'(x)=\int _{a}^{b}f_{x}(x,t)\,dt}$$

$${\displaystyle {\frac {d}{dx))\int _{a}^{b}f(x,t)\,dt=\int _{a}^{b}f_{x}(x,t)\,dt}$$

Another proof using the bounded convergence theorem

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

$${\displaystyle f_{\delta }(x,t)={\frac {f(x+\delta ,t)-f(x,t)}{\delta )).}$$

$${\displaystyle f_{\delta }(x,t)=f_{x}(z,t).}$$

${\displaystyle t}$

${\displaystyle f_{\delta }(x,t)}$

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.

Variable limits form

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

$${\displaystyle G(x)=\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt,}$$

$${\displaystyle \Gamma (x)=\int _{0}^{x}g(t)\,dt.}$$

${\displaystyle g}$

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

$${\displaystyle G'(x)=\Gamma '\left(f_{2}(x)\right)f_{2}'(x)-\Gamma '\left(f_{1}(x)\right)f_{1}'(x).}$$

${\displaystyle \Gamma '(x)=g(x)}$

$${\displaystyle G'(x)=g\left(f_{2}(x)\right){f_{2}'(x)}-g\left(f_{1}(x)\right){f_{1}'(x)}.}$$

Note: This form can be particularly useful if the expression to be differentiated is of the form:

$${\displaystyle \int _{f_{1}(x)}^{f_{2}(x)}h(x)g(t)\,dt}$$

${\displaystyle h(x)}$

$${\displaystyle {\frac {d}{dx))\left(\int _{f_{1}(x)}^{f_{2}(x)}h(x)g(t)\,dt\right)={\frac {d}{dx))\left(h(x)\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt\right)=h'(x)\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt+h(x){\frac {d}{dx))\left(\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt\right)}$$

General form with variable limits

Set

$${\displaystyle \varphi (\alpha )=\int _{a}^{b}f(x,\alpha )\,dx,}$$

$${\displaystyle {\begin{aligned}\Delta \varphi &=\varphi (\alpha +\Delta \alpha )-\varphi (\alpha )\\[4pt]&=\int _{a+\Delta a}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[4pt]&=\int _{a+\Delta a}^{a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}f(x,\alpha +\Delta \alpha )\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[4pt]&=-\int _{a}^{a+\Delta a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx.\end{aligned))}$$

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:

$${\displaystyle \lim _{\Delta \alpha \to 0}\int _{a}^{b}{\frac {f(x,\alpha +\Delta \alpha )-f(x,\alpha )}{\Delta \alpha ))\,dx=\int _{a}^{b}{\frac {\partial }{\partial \alpha ))f(x,\alpha )\,dx,}$$

$${\displaystyle {\frac {d\varphi }{d\alpha ))=\int _{a}^{b}{\frac {\partial }{\partial \alpha ))f(x,\alpha )\,dx+f(b,\alpha ){\frac {db}{d\alpha ))-f(a,\alpha ){\frac {da}{d\alpha )).}$$

Alternative proof of the general form with variable limits, using the chain rule

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

$${\displaystyle F(x,y)=\int _{t_{1))^{y}f(x,t)\,dt,\qquad {\text{for))~x\in [x_{1},x_{2}]~{\text{and))~y\in [t_{1},t_{2}]}$$

$${\displaystyle G(x)=\int _{a(x)}^{b(x)}f(x,t)\,dt,\quad {\text{for))~x\in [x_{1},x_{2}]}$$

Then, by properties of Definite Integrals, we can write

$${\displaystyle {\begin{aligned}G(x)&=\int _{t_{1))^{b(x)}f(x,t)\,dt-\int _{t_{1))^{a(x)}f(x,t)\,dt\\[4pt]&=F(x,b(x))-F(x,a(x))\end{aligned))}$$

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:

$${\displaystyle {\begin{aligned}G'(x)&=\left(\int _{t_{1))^{b(x)}{\frac {\partial f}{\partial x))(x,t)\,dt+f(x,b(x))b'(x)\right)-\left(\int _{t_{1))^{a(x)}{\dfrac {\partial f}{\partial x))(x,t)\,dt+f(x,a(x))a'(x)\right)\\[2pt]&=f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial f}{\partial x))(x,t)\,dt,\end{aligned))}$$

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.

Three-dimensional, time-dependent form

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

$${\displaystyle \mathbf {F} (\mathbf {C} (t)+\mathbf {r} -\mathbf {C} (t),t)=\mathbf {F} (\mathbf {C} (t)+\mathbf {I} ,t),}$$

${\displaystyle \mathbf {I} }$

${\displaystyle \mathbf {C} (t)}$

$${\displaystyle {\frac {d}{dt))\left(\iint _{\Sigma (t)}d\mathbf {A} _{\mathbf {r} }\cdot \mathbf {F} (\mathbf {r} ,t)\right)=\iint _{\Sigma }d\mathbf {A} _{\mathbf {I} }\cdot {\frac {d}{dt))\mathbf {F} (\mathbf {C} (t)+\mathbf {I} ,t),}$$

$${\displaystyle {\frac {d}{dt))\mathbf {F} (\mathbf {C} (t)+\mathbf {I} ,t)=\mathbf {F} _{t}(\mathbf {C} (t)+\mathbf {I} ,t)+\mathbf {v\cdot \nabla F} (\mathbf {C} (t)+\mathbf {I} ,t)=\mathbf {F} _{t}(\mathbf {r} ,t)+\mathbf {v} \cdot \nabla \mathbf {F} (\mathbf {r} ,t),}$$

$${\displaystyle \mathbf {v} ={\frac {d}{dt))\mathbf {C} (t).}$$

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)

$${\displaystyle \nabla \times \left(\mathbf {v} \times \mathbf {F} \right)=(\nabla \cdot \mathbf {F} +\mathbf {F} \cdot \nabla )\mathbf {v} -(\nabla \cdot \mathbf {v} +\mathbf {v} \cdot \nabla )\mathbf {F} ,}$$

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

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,

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

Alternative derivation

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,

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

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

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:

$${\displaystyle \varphi (\alpha )=\int _{a}^{b}f(x,\alpha )\,dx,}$$

${\displaystyle \varphi }$

$${\displaystyle {\frac {d\varphi }{d\alpha ))=\int _{a}^{b}{\frac {\partial }{\partial \alpha ))f(x,\alpha )\,dx.}$$

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,

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

$${\displaystyle |R|<\int _{a}^{b}\varepsilon \,dx=\varepsilon (b-a).}$$

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

$${\displaystyle \int _{a}^{b}f(x,\alpha )\,dx=\varphi (\alpha ),}$$

$${\displaystyle {\begin{aligned}\Delta \varphi &=\varphi (\alpha +\Delta \alpha )-\varphi (\alpha )\\[6pt]&=\int _{a+\Delta a}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[6pt]&=\int _{a+\Delta a}^{a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}f(x,\alpha +\Delta \alpha )\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx-\int _{a}^{b}f(x,\alpha )\,dx\\[6pt]&=-\int _{a}^{a+\Delta a}f(x,\alpha +\Delta \alpha )\,dx+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\int _{b}^{b+\Delta b}f(x,\alpha +\Delta \alpha )\,dx.\end{aligned))}$$

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

$${\displaystyle {\frac {d\varphi }{d\alpha ))=\int _{a}^{b}{\frac {\partial }{\partial \alpha ))f(x,\alpha )\,dx}$$

$${\displaystyle {\frac {d\varphi }{d\alpha ))=\int _{a}^{b}{\frac {\partial }{\partial \alpha ))f(x,\alpha )\,dx+f(b,\alpha ){\frac {\partial b}{\partial \alpha ))-f(a,\alpha ){\frac {\partial a}{\partial \alpha )).}$$

This is the general form of the Leibniz integral rule.

Examples

Example 1: Fixed limits

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

$${\displaystyle {\frac {d}{d\alpha ))\varphi (\alpha )=\int _{0}^{1}{\frac {\partial }{\partial \alpha ))\left({\frac {\alpha }{x^{2}+\alpha ^{2))}\right)\,dx=\int _{0}^{1}{\frac {x^{2}-\alpha ^{2)){(x^{2}+\alpha ^{2})^{2))}dx=\left.-{\frac {x}{x^{2}+\alpha ^{2))}\right|_{0}^{1}=-{\frac {1}{1+\alpha ^{2))},}$$

$${\displaystyle \varphi (\alpha )={\begin{cases}0,&\alpha =0,\\-\arctan({\alpha })+{\frac {\pi }{2)),&\alpha \neq 0.\end{cases))}$$

Example 2: Variable limits

An example with variable limits:

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

Applications

Evaluating definite integrals

The formula