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Incalculus, theLeibniz integral rule for differentiation under the integral sign, named afterGottfried Wilhelm Leibniz, states that for anintegral of the form$${\displaystyle {\int }_{a(x)}^{b(x)}f(x,t)dt,}$$where and the integrands arefunctions dependent on${\displaystyle x,}$ the derivative of this integral is expressible aswhere thepartial derivative${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }x}}$ 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:$${\displaystyle \frac{d}{dx}\left({\int }_a^{b}f(x,t)dt\right)={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }x}f(x,t)dt.}$$

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

This important result may, under certain conditions, be used to interchange the integral and partial differentialoperators, and is particularly useful in the differentiation ofintegral transforms. An example of such is themoment generating function inprobability theory, a variation of theLaplace transform, which can be differentiated to generate themoments of arandom variable. Whether Leibniz's integral rule applies is essentially a question about the interchange oflimits.

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

Stronger versions of the theorem only require that the partial derivative existalmost everywhere, and not that it be continuous.^[2] This formula is the general form of the Leibniz integral rule and can be derived using thefundamental 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 anoperator equation:$${\displaystyle {\mathcal{I}}_t{\mathrm{\partial }}_x={\mathrm{\partial }}_x{\mathcal{I}}_t}$$where${\displaystyle {\mathrm{\partial }}_x}$ is thepartial derivative with respect to${\displaystyle x}$ and${\displaystyle {\mathcal{I}}_t}$ is the integral operator with respect to${\displaystyle t}$ over a fixedinterval. That is, it is related to thesymmetry 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 theinterchange 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 atwo dimensional surface moving in three dimensional space is^[3][4]

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

where:

• F(r,t) is a vector field at the spatial positionr at timet,
• Σ is a surface bounded by the closed curve∂Σ,
• dA is a vector element of the surfaceΣ,
• ds is a vector element of the curve∂Σ,
• v is the velocity of movement of the regionΣ,
• ∇⋅ is the vectordivergence,
• × is thevector cross product,
• The double integrals aresurface integrals over the surfaceΣ, and theline 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 offluid dynamics as theReynolds transport theorem:$${\displaystyle \frac{d}{dt}{\int }_{D(t)}F(\mathbf{x},t)dV={\int }_{D(t)}\frac{\mathrm{\partial }}{\mathrm{\partial }t}F(\mathbf{x},t)dV+{\int }_{\mathrm{\partial }D(t)}F(\mathbf{x},t){\mathbf{v}}_b\cdot d\mathbf{\Sigma },}$$

where${\displaystyle F(\mathbf{x},t)}$ is a scalar function,D(t) and∂D(t) denote a time-varying connected region ofR³ and its boundary, respectively,${\displaystyle {\mathbf{v}}_b}$ is the Eulerian velocity of the boundary (seeLagrangian and Eulerian coordinates) anddΣ =ndS is the unit normal component of thesurfaceelement.

The general statement of the Leibniz integral rule requires concepts fromdifferential geometry, specificallydifferential forms,exterior derivatives,wedge products andinterior products. With those tools, the Leibniz integral rule inn dimensions is^[4]$${\displaystyle \frac{d}{dt}{\int }_{\mathrm{\Omega }(t)}\omega ={\int }_{\mathrm{\Omega }(t)}{i}_{\mathbf{v}}(d_x\omega )+{\int }_{\mathrm{\partial }\mathrm{\Omega }(t)}{i}_{\mathbf{v}}\omega +{\int }_{\mathrm{\Omega }(t)}\dot{\omega },}$$whereΩ(t) is a time-varying domain of integration,ω is ap-form,${\displaystyle \mathbf{v}=\frac{\mathrm{\partial }\mathbf{x}}{\mathrm{\partial }t}}$ is the vector field of the velocity,${\displaystyle i_{\mathbf{v}}}$ denotes theinterior product with${\displaystyle \mathbf{v}}$,d_xω is theexterior derivative ofω with respect to the space variables only and${\displaystyle \dot{\omega }}$ is the time derivative ofω.

The above formula can be deduced directly from the fact that theLie derivative interacts nicely with integration of differential forms$${\displaystyle \frac{d}{dt}{\int }_{\mathrm{\Omega }(t)}\omega ={\int }_{\mathrm{\Omega }(t)}{\mathcal{L}}_{\mathrm{\Psi }}\omega ,}$$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${\displaystyle \mathrm{\Omega }(t)}$ has spacetime velocity field${\displaystyle \mathrm{\Psi }=\frac{\mathrm{\partial }}{\mathrm{\partial }t}+\mathbf{v}}$.Since${\displaystyle \omega }$ has only spatial components, the Lie derivative can be simplified usingCartan's magic formula, to$${\displaystyle {\mathcal{L}}_{\mathrm{\Psi }}\omega ={\mathcal{L}}_{\mathbf{v}}\omega +{\mathcal{L}}_{\frac{\mathrm{\partial }}{\mathrm{\partial }t}}\omega =i_{\mathbf{v}}d\omega +d{i}_{\mathbf{v}}\omega +i_{\frac{\mathrm{\partial }}{\mathrm{\partial }t}}d\omega =i_{\mathbf{v}}{d}_x\omega +d{i}_{\mathbf{v}}\omega +\dot{\omega }}$$which, after integrating over${\displaystyle \mathrm{\Omega }(t)}$ and usinggeneralized Stokes' theorem on the second term, reduces to the three desired terms.

Let${\displaystyle X}$ be an open subset of${\displaystyle \mathbf{R}}$, and${\displaystyle \mathrm{\Omega }}$ be ameasure space. Suppose${\displaystyle f:X\times \mathrm{\Omega }\to \mathbf{R}}$ 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. Foralmost all${\displaystyle \omega \in \mathrm{\Omega }}$ , the partial derivative${\displaystyle f_x}$ exists for all${\displaystyle x\in X}$.
3. There is an integrable function${\displaystyle \theta :\mathrm{\Omega }\to \mathbf{R}}$ such that${\displaystyle |f_x(x,\omega )|\le \theta (\omega )}$ for all${\displaystyle x\in X}$ and almost every${\displaystyle \omega \in \mathrm{\Omega }}$.

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

The proof relies on thedominated convergence theorem and themean value theorem (details below).

We first prove the case of constant limits of integrationa andb.

We useFubini's theorem to change the order of integration. For everyx andh, such thath > 0 and bothx andx +h are within[x₀,x₁], we have:

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 eitherdt ordx 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:

Where we have defined:$${\displaystyle F(u):={\int }_{x_0}^u{\int }_a^b{f}_x(x,t)dtdx}$$(we may replacex₀ here by any other point betweenx₀ andx)

F is differentiable with derivative${\int }_a^b{f}_x(x,t)dt$, so we can take the limit whereh 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:$${\displaystyle F^{\prime }(x)={\int }_a^b{f}_x(x,t)dt}$$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 areLebesgue integrals, we may use thebounded convergence theorem (valid for these integrals, but not forRiemann 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 thatf_x(x,t) is Lebesgue integrable, but not that it is Riemann integrable. In the former (stronger) proof, iff(x,t) is Riemann integrable, then so isf_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,

$${\displaystyle u^{\prime }(x)=\underset{h\to 0}{lim}\frac{u(x+h)-u(x)}{h}.}$$

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

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 thedominated convergence theorem). For eachδ > 0, consider thedifference quotient$${\displaystyle f_{\delta }(x,t)=\frac{f(x+\delta ,t)-f(x,t)}{\delta }.}$$Fort fixed, themean value theorem implies there existsz in the interval [x,x +δ] such that$${\displaystyle f_{\delta }(x,t)=f_x(z,t).}$$Continuity off_x(x,t) and compactness of the domain together imply thatf_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 derivativef_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 tof_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 asδ → 0 may be passed through the integral sign.

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

For acontinuousreal valued functiong of onereal variable, and real valueddifferentiable functions${\displaystyle f_1}$ and${\displaystyle f_2}$ of one real variable,$${\displaystyle \frac{d}{dx}\left({\int }_{f_1(x)}^{f_2(x)}g(t)dt\right)=g\left(f_2(x)\right)f_2^{\prime }(x)-g\left(f_1(x)\right)f_1^{\prime }(x).}$$

This follows from thechain rule and theFirst Fundamental Theorem of Calculus. Define$${\displaystyle G(x)={\int }_{f_1(x)}^{f_2(x)}g(t)dt,}$$and$${\displaystyle \mathrm{\Gamma }(x)={\int }_0^{x}g(t)dt.}$$(The lower limit just has to be some number in the domain of${\displaystyle g}$)

Then,${\displaystyle G(x)}$ can be written as acomposition:${\displaystyle G(x)=(\mathrm{\Gamma }\circ f_2)(x)-(\mathrm{\Gamma }\circ f_1)(x)}$. TheChain Rule then implies that$${\displaystyle G^{\prime }(x)={\mathrm{\Gamma }}^{\prime }\left(f_2(x)\right)f_2^{\prime }(x)-{\mathrm{\Gamma }}^{\prime }\left(f_1(x)\right)f_1^{\prime }(x).}$$By theFirst Fundamental Theorem of Calculus,${\displaystyle {\mathrm{\Gamma }}^{\prime }(x)=g(x)}$. Therefore, substituting this result above, we get the desired equation:$${\displaystyle G^{\prime }(x)=g\left(f_2(x)\right)f_2^{\prime }(x)-g\left(f_1(x)\right)f_1^{\prime }(x).}$$

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 theProduct rule, i.e.,

Set$${\displaystyle \phi (\alpha )={\int }_a^{b}f(x,\alpha )dx,}$$wherea andb are functions ofα that exhibit increments Δa and Δb, respectively, whenα is increased by Δα. Then,

A form of themean value theorem,${\int }_a^{b}f(x)dx=(b-a)f(\xi )$, wherea <ξ <b, may be applied to the first and last integrals of the formula for Δφ above, resulting in$${\displaystyle \mathrm{\Delta }\phi =-\mathrm{\Delta }af({\xi }_1,\alpha +\mathrm{\Delta }\alpha )+{\int }_a^b[f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )]dx+\mathrm{\Delta }bf({\xi }_2,\alpha +\mathrm{\Delta }\alpha ).}$$

Divide by Δα and let Δα → 0. Noticeξ₁ →a andξ₂ →b. We may pass the limit through the integral sign:$${\displaystyle \underset{\mathrm{\Delta }\alpha \to 0}{lim}{\int }_a^b\frac{f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )}{\mathrm{\Delta }\alpha }dx={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx,}$$again by the bounded convergence theorem. This yields the general form of the Leibniz integral rule,$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx+f(b,\alpha )\frac{db}{d\alpha }-f(a,\alpha )\frac{da}{d\alpha }.}$$

The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of thebasic form of Leibniz's Integral Rule, themultivariable chain rule, and thefirst 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$\frac{\mathrm{\partial }f}{\mathrm{\partial }x}$ are both continuous functions on this rectangle. Suppose${\displaystyle a,b}$ aredifferentiable 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$${\displaystyle F(x,y)={\int }_{t_1}^{y}f(x,t)dt,\text{for}x\in [x_1,x_2]\text{and}y\in [t_1,t_2]}$$and$${\displaystyle G(x)={\int }_{a(x)}^{b(x)}f(x,t)dt,\text{for}x\in [x_1,x_2]}$$

Then, by properties ofdefinite 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 themultivariable chain rule, it follows that${\displaystyle G}$ is differentiable, and its derivative is given by the formula:$${\displaystyle G^{\prime }(x)=\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,b(x))+\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,b(x))b^{\prime }(x)\right)-\left(\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,a(x))+\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,a(x))a^{\prime }(x)\right)}$$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$\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,y)={\int }_{t_1}^y\frac{\mathrm{\partial }f}{\mathrm{\partial }x}(x,t)dt$, because when taking the partial derivative with respect to${\displaystyle x}$ of${\displaystyle F}$, we are keeping${\displaystyle y}$ fixed in the expression${\int }_{t_1}^{y}f(x,t)dt$; thus thebasic form of Leibniz's Integral Rule with constant limits of integration applies. Next, by thefirst fundamental theorem of calculus, we have that$\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,y)=f(x,y)$; 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${\displaystyle G^{\prime }(x)}$ above gives: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 bedifferentiable. This is where we use our assumptions about${\displaystyle f}$. As mentioned above, the partial derivatives of${\displaystyle F}$ are given by the formulas$\frac{\mathrm{\partial }F}{\mathrm{\partial }x}(x,y)={\int }_{t_1}^y\frac{\mathrm{\partial }f}{\mathrm{\partial }x}(x,t)dt$ and$\frac{\mathrm{\partial }F}{\mathrm{\partial }y}(x,y)=f(x,y)$. Since${\displaystyle \frac{\mathrm{\partial }f}{\mathrm{\partial }x}}$ 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 timet the surface Σ inFigure 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:$${\displaystyle \frac{d}{dt}\left({\iint }_{\mathrm{\Sigma }(t)}d{\mathbf{A}}_{\mathbf{r}}\cdot \mathbf{F}(\mathbf{r},t)\right)={\iint }_{\mathrm{\Sigma }}d{\mathbf{A}}_{\mathbf{I}}\cdot \frac{d}{dt}\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t),}$$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:$${\displaystyle \frac{d}{dt}\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t)={\mathbf{F}}_t(\mathbf{C}(t)+\mathbf{I},t)+\mathbf{v}\cdot \mathrm{\nabla }\mathbf{F}(\mathbf{C}(t)+\mathbf{I},t)={\mathbf{F}}_t(\mathbf{r},t)+\mathbf{v}\cdot \mathrm{\nabla }\mathbf{F}(\mathbf{r},t),}$$with the velocity of motion of the surface defined by$${\displaystyle \mathbf{v}=\frac{d}{dt}\mathbf{C}(t).}$$

This equation expresses thematerial 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 (seearticle on curl)$${\displaystyle \mathrm{\nabla }\times \left(\mathbf{v}\times \mathbf{F}\right)=(\mathrm{\nabla }\cdot \mathbf{F}+\mathbf{F}\cdot \mathrm{\nabla })\mathbf{v}-(\mathrm{\nabla }\cdot \mathbf{v}+\mathbf{v}\cdot \mathrm{\nabla })\mathbf{F},}$$and thatStokes theorem equates the surface integral of the curl over Σ with a line integral over∂Σ:$${\displaystyle \frac{d}{dt}\left({\iint }_{\mathrm{\Sigma }(t)}\mathbf{F}(\mathbf{r},t)\cdot d\mathbf{A}\right)={\iint }_{\mathrm{\Sigma }(t)}({\mathbf{F}}_t(\mathbf{r},t)+\left(\mathbf{F}\cdot \mathrm{\nabla }\right)\mathbf{v}+\left(\mathrm{\nabla }\cdot \mathbf{F}\right)\mathbf{v}-(\mathrm{\nabla }\cdot \mathbf{v})\mathbf{F})\cdot d\mathbf{A}-{\oint }_{\mathrm{\partial }\mathrm{\Sigma }(t)}\left(\mathbf{v}\times \mathbf{F}\right)\cdot d\mathbf{s}.}$$

The sign of the line integral is based on theright-hand rule for the choice of direction of line elementds. To establish this sign, for example, suppose the fieldF points in the positivez-direction, and the surface Σ is a portion of thexy-plane with perimeter ∂Σ. We adopt the normal to Σ to be in the positivez-direction. Positive traversal of ∂Σ is then counterclockwise (right-hand rule with thumb alongz-axis). Then the integral on the left-hand side determines apositive flux ofF through Σ. Suppose Σ translates in the positivex-direction at velocityv. An element of the boundary of Σ parallel to they-axis, sayds, sweeps out an areavt ×ds in timet. If we integrate around the boundary ∂Σ in a counterclockwise sense,vt ×ds points in the negativez-direction on the left side of ∂Σ (whereds points downward), and in the positivez-direction on the right side of ∂Σ (whereds 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 ofF is increasing on the right of ∂Σ and decreasing on the left. However, thedot productv ×F ⋅ds = −F ×v ⋅ds = −F ⋅v ×ds. Consequently, the sign of the line integral is taken as negative.

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

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

Proof. From theproof of the fundamental theorem of calculus,

and

Supposea andb are constant, and thatf(x) involves a parameterα which is constant in the integration but may vary to form different integrals. Assume thatf(x,α) is a continuous function ofx andα in the compact set {(x,α) :α₀ ≤α ≤α₁ anda ≤x ≤b}, and that the partial derivativef_α(x,α) exists and is continuous. If one defines:$${\displaystyle \phi (\alpha )={\int }_a^{b}f(x,\alpha )dx,}$$then${\displaystyle \phi }$ may be differentiated with respect toα by differentiating under the integral sign, i.e.,$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx.}$$

By theHeine–Cantor theorem it is uniformly continuous in that set. In other words, for anyε > 0 there exists Δα such that for all values ofx in [a,b],

On the other hand,

Henceφ(α) is a continuous function.

Similarly if${\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )}$ exists and is continuous, then for allε > 0 there exists Δα such that:

Therefore,$${\displaystyle \frac{\mathrm{\Delta }\phi }{\mathrm{\Delta }\alpha }={\int }_a^b\frac{f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )}{\mathrm{\Delta }\alpha }dx={\int }_a^b\frac{\mathrm{\partial }f(x,\alpha )}{\mathrm{\partial }\alpha }dx+R,}$$where

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

This is the formula we set out to prove.

Now, suppose$${\displaystyle {\int }_a^{b}f(x,\alpha )dx=\phi (\alpha ),}$$wherea andb are functions ofα which take increments Δa and Δb, respectively, whenα is increased by Δα. Then,

A form of themean value theorem,${\int }_a^{b}f(x)dx=(b-a)f(\xi ),$ wherea <ξ <b, can be applied to the first and last integrals of the formula for Δφ above, resulting in$${\displaystyle \mathrm{\Delta }\phi =-\mathrm{\Delta }af({\xi }_1,\alpha +\mathrm{\Delta }\alpha )+{\int }_a^b[f(x,\alpha +\mathrm{\Delta }\alpha )-f(x,\alpha )]dx+\mathrm{\Delta }bf({\xi }_2,\alpha +\mathrm{\Delta }\alpha ).}$$

Dividing by Δα, letting Δα → 0, noticingξ₁ →a andξ₂ →b and using the above derivation for$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx}$$yields$${\displaystyle \frac{d\phi }{d\alpha }={\int }_a^b\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }f(x,\alpha )dx+f(b,\alpha )\frac{\mathrm{\partial }b}{\mathrm{\partial }\alpha }-f(a,\alpha )\frac{\mathrm{\partial }a}{\mathrm{\partial }\alpha }.}$$

This is the general form of the Leibniz integral rule.

Consider the function

${\displaystyle \phi (\alpha )={\int }_0^1\frac{\alpha }{x^2+{\alpha }^2}dx.}$

The function under the integral sign is not continuous at the point${\displaystyle (x,\alpha )=(0,0)}$, and the function${\displaystyle \phi (\alpha )}$ has a discontinuity at${\displaystyle \alpha =0}$ because${\displaystyle \phi (\alpha )}$ approaches${\displaystyle \pm \pi /2}$ as${\displaystyle \alpha \to 0^{\pm }}$.

If we differentiate${\displaystyle \phi (\alpha )}$ with respect to${\displaystyle \alpha }$ under the integral sign, we get$${\displaystyle \frac{d}{d\alpha }\phi (\alpha )={\int }_0^1\frac{\mathrm{\partial }}{\mathrm{\partial }\alpha }\left(\frac{\alpha }{x^2+{\alpha }^2}\right)dx={\int }_0^1\frac{x^2-{\alpha }^2}{(x^2+{\alpha }^2{)}^2}dx={-\frac{x}{x^2+{\alpha }^2}|}_0^1=-\frac{1}{1+{\alpha }^2},}$$for${\displaystyle \alpha \ne 0}$. This may be integrated (with respect to${\displaystyle \alpha }$) to find