In calculus, the Leibniz integral rule for differentiation under the integral sign, named after Gottfried Leibniz, states that for an integral of the form
where , the derivative of this integral is expressible as
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where the partial derivative indicates that inside the integral, only the variation of with is considered in taking the derivative.^{[1]} Notice that if and are constants rather than functions of , we have the special case:
Besides, if and , which is a common situation as well (for example, in the proof of Cauchy's repeated integration formula), we have:
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 Leibniz's integral rule applies is essentially a question about the interchange of limits.
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:
where is a scalar function, D(t) and ∂D(t) denote a timevarying connected region of R^{3} and its boundary, respectively, 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^{[2]}
where Ω(t) is a timevarying domain of integration, ω is a pform, is the vector field of the velocity, denotes the interior product with , d_{x}ω is the exterior derivative of ω with respect to the space variables only and is the time derivative of ω.
However, all of these identities can be derived from a most general statement about Lie derivatives:
Here, the ambient manifold on which the differential form lives includes both space and time.
Something remarkable about this form, is that it can account for the case when changes its shape and size over time, since such deformations are fully determined by .
Let be an open subset of , and be a measure space. Suppose satisfies the following conditions:
Then, for all ,
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 [x_{0},x_{1}], we have:
Note that the integrals at hand are well defined since is continuous at the closed rectangle 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:
Where we have defined:
(we may replace x_{0} here by any other point between x_{0} and x)
F is differentiable with derivative , so we can take the limit where h approaches zero. For the left hand side this limit is:
For the right hand side, we get:
And we thus prove the desired result:
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

(1) 
By the definition of the derivative,

(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 the dominated convergence theorem). For each δ > 0, consider the difference quotient
For t fixed, the mean value theorem implies there exists z in the interval [x, x + δ] such that
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 ) bound on . 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 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.
For a continuous real valued function g of one real variable, and real valued differentiable functions and of one real variable,
This follows from the chain rule and the First Fundamental Theorem of Calculus. Define
and
Then, can be written as a composition: . The Chain Rule then implies that
By the First Fundamental Theorem of Calculus, . Therefore, substituting this result above, we get the desired equation:
Note: This form can be particularly useful if the expression to be differentiated is of the form:
Because does not depend on the limits of integration, it may be move out from under the integral sign, and the above form may be used with the Product rule, i.e.,
Set
where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα. Then,
A form of the mean value theorem, , where a < ξ < b, may be applied to the first and last integrals of the formula for Δφ above, resulting in
Divide by Δα and let Δα → 0. Notice ξ_{1} → a and ξ_{2} → b. We may pass the limit through the integral sign:
again by the bounded convergence theorem. This yields the general form of the Leibniz integral 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 is defined in a rectangle in the plane, for and . Also, assume and the partial derivative are both continuous functions on this rectangle. Suppose are differentiable real valued functions defined on , with values in (i.e. for every ). Now, set
and
Then, by properties of Definite Integrals, we can write
Since the functions are all differentiable (see the remark at the end of the proof), by the Multivariable Chain Rule, it follows that is differentiable, and its derivative is given by the formula:
Now, note that for every , and for every , we have that , because when taking the partial derivative with respect to of , we are keeping fixed in the expression ; 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 ; because when taking the partial derivative with respect to of , the first variable is fixed, so the fundamental theorem can indeed be applied.
Substituting these results into the equation for above gives:
as desired.
There is a technical point in the proof above which is worth noting: applying the Chain Rule to requires that already be differentiable. This is where we use our assumptions about . As mentioned above, the partial derivatives of are given by the formulas and . Since is continuous, its integral is also a continuous function,^{[3]} and since is also continuous, these two results show that both the partial derivatives of are continuous. Since continuity of partial derivatives implies differentiability of the function,^{[4]} is indeed differentiable.
At time t the surface Σ in Figure 1 contains a set of points arranged about a centroid . The function can be written as
with independent of time. Variables are shifted to a new frame of reference attached to the moving surface, with origin at . For a rigidly translating surface, the limits of integration are then independent of time, so:
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:
with the velocity of motion of the surface defined by
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)
and that Stokes theorem equates the surface integral of the curl over Σ with a line integral over ∂Σ:
The sign of the line integral is based on the righthand rule for the choice of direction of line element ds. To establish this sign, for example, suppose the field F points in the positive zdirection, and the surface Σ is a portion of the xyplane with perimeter ∂Σ. We adopt the normal to Σ to be in the positive zdirection. Positive traversal of ∂Σ is then counterclockwise (righthand rule with thumb along zaxis). Then the integral on the lefthand side determines a positive flux of F through Σ. Suppose Σ translates in the positive xdirection at velocity v. An element of the boundary of Σ parallel to the yaxis, 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 zdirection on the left side of ∂Σ (where ds points downward), and in the positive zdirection 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,
which is the quoted result. This proof does not consider the possibility of the surface deforming as it moves.
Lemma. One has:
Proof. From the proof of the fundamental theorem of calculus,
and
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) : α_{0} ≤ α ≤ α_{1} and a ≤ x ≤ b}, and that the partial derivative f_{α}(x, α) exists and is continuous. If one defines:
then may be differentiated with respect to α by differentiating under the integral sign, i.e.,
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],
On the other hand,
Hence φ(α) is a continuous function.
Similarly if exists and is continuous, then for all ε > 0 there exists Δα such that:
Therefore,
where
Now, ε → 0 as Δα → 0, so
This is the formula we set out to prove.
Now, suppose
where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then,
A form of the mean value theorem, where a < ξ < b, can be applied to the first and last integrals of the formula for Δφ above, resulting in
Dividing by Δα, letting Δα → 0, noticing ξ_{1} → a and ξ_{2} → b and using the above derivation for
yields
This is the general form of the Leibniz integral rule.
Consider the function
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
which is, of course, true for all values of α except α = 0. This may be integrated (with respect to α) to find
An example with variable limits:
The formula
can be of use when evaluating certain definite integrals. When used in this context, the Leibniz integral rule for differentiating under the integral sign is also known as Feynman's trick for integration.
First we calculate:
The limits of integration being independent of , we have:
On the other hand:
Equating these two relations then yields
In a similar fashion, pursuing yields
Adding the two results then produces
which computes as desired.
This derivation may be generalized. Note that if we define
it can easily be shown that
Given , this integral reduction formula can be used to compute all of the values of for . Integrals like and may also be handled using the Weierstrass substitution.
Here, we consider the integral
Differentiating under the integral with respect to , we have
Therefore:
But by definition so and
Here, we consider the integral
We introduce a new variable φ and rewrite the integral as
When φ = 1 this equals the original integral. However, this more general integral may be differentiated with respect to :
Now, fix φ, and consider the vector field on defined by . Further, choose the positive oriented parametrization of the unit circle given by , , so that . Then the final integral above is precisely
where is the closed unit disc. Its integrand is identically 0, so is likewise identically zero. This implies that f(φ) is constant. The constant may be determined by evaluating at :
Therefore, the original integral also equals .
There are innumerable other integrals that can be solved using the technique of differentiation under the integral sign. For example, in each of the following cases, the original integral may be replaced by a similar integral having a new parameter :
The first integral, the Dirichlet integral, is absolutely convergent for positive α but only conditionally convergent when . Therefore, differentiation under the integral sign is easy to justify when , but proving that the resulting formula remains valid when requires some careful work.
The measuretheoretic version of differentiation under the integral sign also applies to summation (finite or infinite) by interpreting summation as counting measure. An example of an application is the fact that power series are differentiable in their radius of convergence.
Differentiation under the integral sign is mentioned in the late physicist Richard Feynman's bestselling memoir Surely You're Joking, Mr. Feynman! in the chapter "A Different Box of Tools". He describes learning it, while in high school, from an old text, Advanced Calculus (1926), by Frederick S. Woods (who was a professor of mathematics in the Massachusetts Institute of Technology). The technique was not often taught when Feynman later received his formal education in calculus, but using this technique, Feynman was able to solve otherwise difficult integration problems upon his arrival at graduate school at Princeton University:
One thing I never did learn was contour integration. I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. One day he told me to stay after class. "Feynman," he said, "you talk too much and you make too much noise. I know why. You're bored. So I'm going to give you a book. You go up there in the back, in the corner, and study this book, and when you know everything that's in this book, you can talk again." So every physics class, I paid no attention to what was going on with Pascal's Law, or whatever they were doing. I was up in the back with this book: "Advanced Calculus", by Woods. Bader knew I had studied "Calculus for the Practical Man" a little bit, so he gave me the real works—it was for a junior or senior course in college. It had Fourier series, Bessel functions, determinants, elliptic functions—all kinds of wonderful stuff that I didn't know anything about. That book also showed how to differentiate parameters under the integral sign—it's a certain operation. It turns out that's not taught very much in the universities; they don't emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was selftaught using that book, I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn't do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's, and they had tried all their tools on it before giving the problem to me.