Leibniz integral rule

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:

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

(we may replace x₀ here by any other point between x₀ and x)

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:

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

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 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

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

For t fixed, the mean value theorem implies there exists z in the interval [x, x + δ] such that

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

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 ${\displaystyle t}$) bound on ${\displaystyle f_{\delta }(x,t)}$. 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 ${\displaystyle \{f_{\delta _{n}}(x,t)\}}$ 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.

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

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

and

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

(The lower limit just has to be some number in the domain of ${\displaystyle g}$)

Then, ${\displaystyle G(x)}$ can be written as a composition: ${\displaystyle G(x)=(\Gamma \circ f_{2})(x)-(\Gamma \circ f_{1})(x)}$. The Chain Rule then implies that

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

By the First Fundamental Theorem of Calculus, ${\displaystyle \Gamma '(x)=g(x)}$. Therefore, substituting this result above, we get the desired equation:

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

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

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

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 the Product rule, i.e.,

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

Set

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

where a and b are functions of α that exhibit increments Δa and Δb, respectively, when α is increased by Δα. Then,

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

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:

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

again by the bounded convergence theorem. This yields the general form of the Leibniz integral rule,

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

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

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

and

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

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:

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

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 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.

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

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

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:

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

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:

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

with the velocity of motion of the surface defined by

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

This equation expresses 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)

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

and that Stokes theorem equates the surface integral of the curl over Σ with a line integral over ∂Σ:

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

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,

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

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

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,

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

and

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

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:

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

then ${\displaystyle \varphi }$ may be differentiated with respect to α by differentiating under the integral sign, i.e.,

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

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,

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

where

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

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

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

where a and b are functions of α which take increments Δa and Δb, respectively, when α is increased by Δα. Then,

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

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

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

yields

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

This is the general form of the Leibniz integral rule.

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

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

which is, of course, true for all values of α except α = 0. This may be integrated (with respect to α) to find

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

An example with variable limits:

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

The formula