Leibniz Rule with Moving Endpoints

Why Moving Endpoints Add Terms

When a parameter changes, an integral over a moving interval changes for two reasons: the integrand may change, and the interval itself may gain or lose pieces at its endpoints. The Leibniz rule records these effects separately. Its boundary terms have opposite signs because increasing the lower endpoint removes a slice, while increasing the upper endpoint adds one.

We work with Lebesgue measure $m$ on $\mathbb{R}$, and real-valued integrands. All integrals below are over finite intervals and are Lebesgue integrals. For $c>d$, use the oriented-integral convention $\int_c^d f(x)\,dx=-\int_d^c f(x)\,dx$; the theorem itself will assume the endpoints are ordered. A function is integrable here if its absolute value has finite Lebesgue integral.

Theorem (Leibniz Rule with Moving Endpoints): Let $t_0$ lie in an open interval $T$, and let $K=[A,B]$ be a finite interval. Suppose $a,b:T\to K$ are differentiable, with $a(t)\le b(t)$ for $t\in T$. Suppose $f:T\times K\to\mathbb{R}$ is continuous, its partial derivative $\partial_t f(t,x)$ exists for every $(t,x)\in T\times K$, and $x\mapsto\partial_t f(t,x)$ is measurable for each $t$. Assume there is a nonnegative $g\in L^1(K,m)$ such that $|\partial_t f(t,x)|\le g(x)$ for all $t\in T$ and $x\in K$. Define $F(t)=\int_{a(t)}^{b(t)} f(t,x)\,dx$. Then $F$ is differentiable at $t_0$, and $$ F'(t_0)=f(t_0,b(t_0))b'(t_0)-f(t_0,a(t_0))a'(t_0) +\int_{a(t_0)}^{b(t_0)}\partial_t f(t_0,x)\,dx. $$ The same conclusion holds at each $t_0\in T$.

Proof: Separate the Two Sources of Change

Fix $t=t_0$, and write $a=a(t)$, $b=b(t)$, $a_h=a(t+h)$, and $b_h=b(t+h)$. For sufficiently small $h$, $t+h\in T$. Add and subtract the integral of $f(t,\cdot)$ over the new interval. This gives the exact decomposition

\frac{F(t+h)-F(t)}{h} = \int_K \chi_{[a_h,b_h]}(x) \frac{f(t+h,x)-f(t,x)}{h}\,dx + \frac{1}{h}\left(\int_{a_h}^{b_h}f(t,x)\,dx-\int_a^b f(t,x)\,dx\right).

For the first term, take any sequence of nonzero $h_n\to0$ with $t+h_n\in T$. At each $x\in K$, the difference quotient tends to $\partial_t f(t,x)$. By the one-variable mean value theorem applied with $x$ fixed, its absolute value is at most $g(x)$. The indicators $\chi_{[a_{h_n},b_{h_n}]}$ converge to $\chi_{[a,b]}$ except possibly at $a$ and $b$; those two points have Lebesgue measure zero. Thus the integrands converge $m$-a.e. on $K$, and their absolute values are bounded by the integrable function $g$. The Dominated Convergence Theorem therefore applies and gives the limit $\int_a^b\partial_t f(t,x)\,dx$. Since this holds for every such sequence $h_n$, it is the limit as $h\to0$.

For the second term, additivity of the oriented integral yields

\int_{a_h}^{b_h}f(t,x)\,dx-\int_a^b f(t,x)\,dx = \int_b^{b_h}f(t,x)\,dx-\int_a^{a_h}f(t,x)\,dx.

Continuity of $x\mapsto f(t,x)$ implies that, as $\delta\to0$, the average of $f(t,\cdot)$ over the oriented segment from $c$ to $c+\delta$ tends to $f(t,c)$. In particular, if $\delta\ne0$, then $\delta^{-1}\int_c^{c+\delta}f(t,x)\,dx\to f(t,c)$; if $\delta=0$, the integral is zero. Differentiability of the endpoints gives $(b_h-b)/h\to b'(t)$ and $(a_h-a)/h\to a'(t)$. Consequently, the second term tends to $f(t,b)b'(t)-f(t,a)a'(t)$. Combining the two limits proves the formula. The argument also covers $a=b$, since the interval integral is then zero and the two endpoint limits are still valid. $\square$

Worked Applications

Worked Example: A Polynomial Integrand and Two Moving Endpoints

Let $F(t)=\int_t^{1+2t}(tx+x^2)\,dx$ for $t$ in a neighborhood of $0$ small enough that $t\le1+2t$. Find $F'(0)$ by the Leibniz rule and verify it directly.

Here $a(t)=t$, $b(t)=1+2t$, and $f(t,x)=tx+x^2$. These functions are continuous and differentiable on any finite rectangle containing the endpoints for $t$ near $0$. The derivative $\partial_t f(t,x)=x$ is bounded on that rectangle, so an integrable constant function is a dominator. At $t=0$, the endpoint values are $f(0,1)=1$ and $f(0,0)=0$, while $b'(0)=2$ and $a'(0)=1$. The integrated derivative is $\int_0^1x\,dx=1/2$. Therefore

F'(0)=1\cdot2-0\cdot1+\frac12=\frac52.

For a direct check, an antiderivative in $x$ is $tx^2/2+x^3/3$, so $$ F(t)=\frac{t}{2}\big((1+2t)^2-t^2\big)+\frac13\big((1+2t)^3-t^3\big). $$ The derivative at $t=0$ of the first term is $1/2$, and that of the second term is $2$. Hence direct differentiation gives $F'(0)=1/2+2=5/2$, agreeing with the rule.

Worked Example: An Exponential Integrand with a Moving Lower Limit

Set $F(t)=\int_t^{\pi/2}e^{tx}\,dx$ for $t$ near $0$. Compute $F'(0)$ using the rule and check it from an explicit formula.

On a finite rectangle around $t=0$, $f(t,x)=e^{tx}$ is continuous and $\partial_t f(t,x)=xe^{tx}$. This derivative is bounded there, so it has an integrable constant dominator on the finite interval. The upper endpoint is constant, and the lower endpoint has derivative $1$. Thus

F'(0)=0\cdot e^{0\cdot\pi/2}-1\cdot e^{0\cdot0} +\int_0^{\pi/2}xe^{0\cdot x}\,dx =-1+\left[\frac{x^2}{2}\right]_0^{\pi/2} =\frac{\pi^2}{8}-1.

For $t\ne0$, direct integration gives $F(t)=(e^{\pi t/2}-e^{t^2})/t$, and $F(0)=\pi/2$. Taylor expansion at zero gives $e^{\pi t/2}=1+\pi t/2+\pi^2t^2/8+o(t^2)$ and $e^{t^2}=1+t^2+o(t^2)$. Therefore $F(t)=\pi/2+(\pi^2/8-1)t+o(t)$, which verifies the derivative.

Worked Example: Endpoint Contributions Can Cancel

Let $F(t)=\int_t^{t+2}x\,dx$. Compute its derivative using the rule and then verify it directly.

Here $f(t,x)=x$, so $\partial_t f=0$. The endpoints satisfy $a'(t)=b'(t)=1$. The rule gives, for every $t$,

F'(t)=(t+2)\cdot1-t\cdot1+\int_t^{t+2}0\,dx=2.

Independently, evaluating the integral gives $$ F(t)=\frac{(t+2)^2-t^2}{2} =\frac{t^2+4t+4-t^2}{2}=2t+2, $$ so $F'(t)=2$, as required. Both endpoint terms matter: their difference is $2$, even though the integrand has no parameter dependence.

Common Pitfall: Freezing the Limits

A frequent error is to differentiate only the integrand and omit the endpoint terms. In the last example, that incorrect calculation would give $F'(t)=\int_t^{t+2}\partial_t f(t,x)\,dx=0$, because $\partial_t f=0$. But direct evaluation gives $F(t)=2t+2$, whose derivative is $2$. The correction is to include both boundary contributions: $(t+2)-t=2$.

The domination condition is also substantive: it justifies the limit of the difference quotients by the Dominated Convergence Theorem. For example, on $[0,1]$, define $f(0,x)=0$, define $f(t,x)=\chi_{(0,t)}(x)$ for $t>0$, and define $f(t,x)=0$ for $t<0$, with $0<t<1$. For every fixed $x$, the derivative in $t$ at $0$ is zero: if $x>0$, then $f(t,x)=0$ whenever $t$ is sufficiently close to $0$, and at $x=0$ the indicator is always zero. Yet $\int_0^1 f(t,x)\,dx=t$ for $t>0$, so the right derivative of the integral at zero is $1$, not the integral of the pointwise derivative, which is $0$. The difference quotients $t^{-1}\chi_{(0,t)}$ have no integrable dominator independent of $t$. This example does not satisfy the theorem's continuity and domination assumptions; it illustrates why unguarded differentiation under an integral is invalid.

The rule is useful whenever parameter changes affect both a density or integrand and the region of integration. Its bookkeeping is reliable: the upper endpoint contributes with a plus sign, the lower endpoint with a minus sign, and the remaining term is the integral of the parameter derivative. The theorem above is a sufficient local version for finite endpoints; it does not assert the same formula for infinite or singular endpoints without additional hypotheses.

Check Your Understanding

For each question, identify the endpoint contributions separately from the integrated derivative.

For $F(t)=\int_{0}^{t}(1+tx)\,dx$, use the Leibniz rule to compute $F'(0)$, and verify by direct integration.
For $F(t)=\int_{t}^{3t}e^x\,dx$, write the derivative formula at a parameter value where $t\le3t$.
In the proof, why can the indicators of the moving intervals converge almost everywhere even when an endpoint moves?
Which hypothesis permits dominated convergence for the difference quotients, and what fails in the indicator example when it is removed?
If $a(t)=b(t)$ at an interior parameter value and $a(t)\le b(t)$ nearby, explain how the formula is interpreted at that value.

Leibniz Rule with Moving Endpoints

What You'll Learn