Leibniz integral rule

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In mathematics, 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

$[x_0,x_1]\times[y_0,y_1].\,$

1 Limits that are variable
2 Three-dimensional, time-dependent case
3 Proofs
4 References and notes
5 See also

[edit] Limits that are variable

A more general result, applicable when the limits of integration a and b and the integrand f ( x, α ) all are functions of the parameter α is:

$\ \frac{d}{d\alpha}\int_a^b f(x,\alpha)dx = \int_a^b\frac{\partial}{\partial \alpha}\,f(x,\alpha)\,dx+f(b,\alpha)\frac{d b}{d \alpha}-f(a,\alpha)\frac{d a}{d \alpha}\,$

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

[edit] Three-dimensional, time-dependent case

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:^[1]

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

$= \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.

[edit] Proofs

[edit] Basic form

Let us first make the assignment

$u(x) = \int_{y_0}^{y_1} f(x, y) \,dy.$

Then

${d\over dx} u(x) = \lim_{h\rightarrow 0} {u(x+h)-u(x) \over h}.$

Substituting back

$= \lim_{h\rightarrow 0} {\int_{y_0}^{y_1}f(x+h,y)\,dy-\int_{y_0}^{y_1}f(x,y)\,dy \over h}.$

Since integration is linear, we can write the two integrals as one:

$= \lim_{h\rightarrow 0} {\int_{y_0}^{y_1}(f(x+h,y)-f(x,y))\,dy\over h}.$

And we can take the constant inside, with the integrand

$= \lim_{h\rightarrow 0} \int_{y_0}^{y_1} {f(x+h,y)-f(x,y)\over h}\,dy.$

And now, since the integrand is in the form of a difference quotient:

$= \int_{y_0}^{y_1} {\partial \over \partial x} f(x,y)\,dy$

which can be justified by uniform continuity, and therefore

${d\over dx} u(x) = \int_{y_0}^{y_1} {\partial \over \partial x} f(x, y) \,dy.$

[edit] 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.

[edit] 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,

$\Delta\phi=\phi(\alpha+\Delta\alpha)-\phi(\alpha)=\int_{a+\Delta a}^{b+\Delta b}f(x,\alpha+\Delta\alpha)dx\,-\int_a^b f(x,\alpha)dx\,$

$=\int_{a+\Delta a}^af(x,\alpha+\Delta\alpha)dx+\int_a^bf(x,\alpha+\Delta\alpha)dx+\int_b^{b+\Delta b}f(x,\alpha+\Delta\alpha)dx\,-\int_a^b f(x,\alpha)dx\,$

$=-\int_a^{a+\Delta a}\,f(x,\alpha+\Delta\alpha)dx+\int_a^b[f(x,\alpha+\Delta\alpha)-f(x,\alpha)]dx+\int_b^{b+\Delta b}\,f(x,\alpha+\Delta\alpha)dx\,$ .

A form of the mean value theorem, $\int_a^bf(x)dx=(b-a)f(\xi)\,$ , 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 $\Delta\alpha\rarr0\,$ , and noticing $\xi_1\rarr a\,$ and $\xi_2\rarr b\,$ and using the result

$\frac{d\phi}{d\alpha} = \int_a^b \frac{\partial} {\partial \alpha} \,f(x,\alpha)\,dx$

yields the general form of the Leibniz integral rule below:

$\ \frac{d\phi}{d\alpha} = \int_a^b\frac{\partial}{\partial \alpha}\,f(x,\alpha)\,dx+f(b,\alpha)\frac{\partial b}{\partial \alpha}-f(a,\alpha)\frac{\partial a}{\partial \alpha}\ .$

[edit] Three-dimensional, time-dependent form

See also: Differentiation under the integral sign#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,

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

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