Leibniz integral rule

In Calculus, Leibniz's rule for differentiation under the integral sign, named after Gottfried Leibniz, tells us that if we have an integral of the form

$\int _{y_{0}}^{y_{1}}f(x,y)\,dy$

then for $x\in (x_{0},x_{1})$ the derivative of this integral is thus expressible

${d \over dx}\,\int _{y_{0}}^{y_{1}}f(x,y)\,dy=\int _{y_{0}}^{y_{1}}{\partial \over \partial x}f(x,y)\,dy$

provided that $f$ and $\partial f/\partial x$ are both continuous over a region in the form

$[x_{0},x_{1}]\times [y_{0},y_{1}]$

Thus under certain conditions, one may interchange the integral and partial differential operators. This important result 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 one may interchange integral and differential operators is essentially a question about the interchange of limits.

Formal Statement

Let $f(x,\theta )$ be a function such that ${\frac {\partial }{\partial \theta }}f(x,\theta )$ exists, and is continuous. Then,

$\ {\frac {\partial }{\partial \theta }}\int _{a(\theta )}^{b(\theta )}f(x,\theta )\,dx=\int _{a(\theta )}^{b(\theta )}{\frac {\partial }{\partial \theta }}\,f(x,\theta )\,dx+\,f(b(\theta ),\theta )b'(\theta )-\,f(a(\theta ),\theta )a'(\theta )$  

where the partial derivative of $f$ indicates that inside the integral only the variation of $f(x,\theta )$ with $\theta$ is considered in taking the derivative.

Three-dimensional, time-dependent case

See also: Differentiation under the integral sign in higher dimensions

Figure 1: A vector field F ( r, t ) defined throughout space, and a surface Σ bounded by curve ∂Σ moving with velocity v over which the field is integrated.

A Leibniz integral rule for three dimensions is:

 ${\frac {d}{dt}}\iint _{\Sigma (t)}\mathbf {F} (\mathbf {r} ,\ t)\cdot d\mathbf {A}$

  $=\iint _{\Sigma (t)}\left[{\frac {\partial }{\partial t}}\mathbf {F} (\mathbf {r} ,\ t)+\left(\mathrm {\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 moving surface in three-space bounded by the closed curve ∂Σ

d A is a vector element of the surface Σ

d s 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.

Proofs

Proof of basic form

Let:

$u(x)=\int _{y_{0}}^{y_{1}}f(x,y)\,dy\quad (1)$

So that, using difference quotients

$u'(x)=\lim _{h\rightarrow 0}{\frac {1}{h}}\left[{u(x+h)-u(x)}\right]\quad (2)$

Substitute equation (1) into equation (2), and combine the integrals (since the difference of two integrals equals the integral of the difference).

$u'(x)=\lim _{h\rightarrow 0}{\frac {1}{h}}\left[{\int _{y_{0}}^{y_{1}}f(x+h,y)\,dy-\int _{y_{0}}^{y_{1}}f(x,y)\,dy}\right]=\lim _{h\rightarrow 0}{\frac {1}{h}}{\int _{y_{0}}^{y_{1}}\left[f(x+h,y)-f(x,y)\right]\,dy}$

The ${\frac {1}{h}}$ term is constant, so can be included in the integrand.

$u'(x)=\lim _{h\rightarrow 0}\int _{y_{0}}^{y_{1}}{\frac {1}{h}}\left[f(x+h,y)-f(x,y)\right]\,dy$

Provided that the limit can be passed under the integral sign, we obtain

${\begin{aligned}u'(x)&=\int _{y_{0}}^{y_{1}}{\partial \over \partial x}f(x,y)\,dy\\\Rightarrow {d \over dx}u(x)&=\int _{y_{0}}^{y_{1}}{\partial \over \partial x}f(x,y)\,dy\end{aligned}}$

We claim that the passage of the limit under the integral sign is valid. Indeed, the bounded convergence theorem (a corollary of the dominated convergence theorem) of real analysis 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. To complete the proof, we show that these hypotheses are satisfied by the family of difference quotients $f_{n}(y)={\frac {f(x+1/n,y)-f(x,y)}{1/n}}$. Continuity of $f_{x}(x,y)$ and compactness implies that $f_{x}(x,y)$ is uniformly bounded. Uniform boundedness of the difference quotients follows from uniform boundedness of $f_{x}(x,y)$ and the Mean Value Theorem, since for all y and n, there exists z in the interval $[x,x+1/n]$ such that $f_{x}(z,y)={\frac {f(x+1/n,y)-f(x,y)}{1/n}}$. The difference quotients converge pointwise to $f_{x}(x,y)$ since $f_{x}(x,y)$ exists. This completes the proof.

For a simpler proof using Fubini's theorem, see the references.

Variable limits form

For a monovariant function $g$:

${d \over dx}\,\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt=g[f_{2}(x)]{f_{2}'(x)}-g[f_{1}(x)]{f_{1}'(x)}$

This follows from the chain rule.

General form with variable limits

Now, suppose $\int _{a}^{b}f(x,\alpha )dx=\phi (\alpha )\,$, where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα.

Then,

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

A form of the mean value theorem, $\int _{a}^{b}f(x)dx=(b-a)f(\xi )\,$, where $a<\xi <b\,$, may be applied to the first and last integrals of the formula for $\Delta \phi \,$ above, resulting in

$\Delta \phi =-\Delta a\,f(\xi _{1},\alpha +\Delta \alpha )+\int _{a}^{b}[f(x,\alpha +\Delta \alpha )-f(x,\alpha )]\,dx+\Delta b\,f(\xi _{2},\alpha +\Delta \alpha )$

Dividing by $\Delta \alpha \,$, and letting $\Delta \alpha \rightarrow 0\,$, and noticing $\xi _{1}\rightarrow a\,$ and $\xi _{2}\rightarrow b\,$ and using the result

$\int _{a}^{b}{\frac {1}{\Delta \alpha }}\left[f(x,\alpha +\Delta \alpha )-f(x,\alpha )\right]\,dx=\int _{a}^{b}{\frac {\partial }{\partial \alpha }}\,f(x,\alpha )\,dx$

yields the general form of the Leibniz integral rule below:

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

Three-dimensional, time-dependent form

See also: Differentiation under the integral sign in higher dimensions

At time t the surface Σ in Figure 1 contains a set of points arranged about a centroid R ( t ) and function F ( r, t) can be written as F ( R ( t ) + r − R(t), t ) = F ( R ( t ) + ρ, t ), with ρ independent of time. Variables are shifted to a new frame of reference attached to the moving surface, with origin at R ( t ). For a rigidly translating surface, the limits of integration are then independent of time, so:

${\frac {d}{dt}}\iint _{\Sigma (t)}d\mathbf {A} _{\mathbf {r} }\mathbf {\cdot F} (\mathbf {r} ,t)$ $=\iint _{\Sigma }d\mathbf {A} _{\vec {\rho }}\cdot {\frac {d}{dt}}\mathbf {F} (\mathbf {R} (t)+{\vec {\rho }},\ 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:

${\frac {d}{dt}}\mathbf {F} (\mathbf {R} (t)+{\vec {\rho }},\ t)={\frac {\partial }{\partial t}}\mathbf {F} (\mathbf {R} (t)+{\vec {\rho }},\ t)+\mathbf {v\cdot \nabla F} (\mathbf {R} (t)+{\vec {\rho }},\ t)\$

$={\frac {\partial }{\partial t}}\mathbf {F} (\mathbf {r} ,\ t)+\mathbf {v\cdot \nabla F} (\mathbf {r} ,\ t)$

with the velocity of motion of the surface defined by:

$\mathbf {v} ={\frac {d}{dt}}\mathbf {R} (t)$

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 ):

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

and that Stokes theorem allows the surface integral of the curl over Σ to be made a line integral over ∂Σ:

 ${\frac {d}{dt}}\iint _{\Sigma (t)}\mathbf {F} (\mathbf {r} ,\ t)\cdot d\mathbf {A}$

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

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,

 ${\frac {d}{dt}}\iint _{\Sigma (t)}\mathbf {F} (\mathbf {r} ,\ t)\cdot d\mathbf {A}$

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

which is the quoted result. This proof does not consider the possibility of the surface deforming as it moves.

See also

• Chain rule
• Leibniz rule (generalized product rule)
• Differentiation under the integral sign
• Reynolds transport theorem a generalization of Leibniz rule

External links

• "The Leibniz Rule" by Rob Harron