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Leibniz Integral Rule 📂Analysis

Leibniz Integral Rule

Theorem

Let’s assume that f(x,t)f(x,t) and ∂f∂x(x,t)\dfrac{\partial f}{\partial x}(x,t) are consecutive. Then, the following equation holds.

ddx∫abf(x,t)dt=∫ab∂f∂x(x,t)dt \frac{d}{dx} \int_{a}^b f(x,t)dt = \int_{a}^b\frac{\partial f}{\partial x}(x,t)dt

Description

Being able to interchange the order of differentiation and integration is undoubtedly useful.

Besides, there are many theorems or formulas related to differentiation and integration named after Leibniz.

Proof

Since if continuous, then integrable, let’s assume uu as follows.

u(x):=∫abf(x,t)dt u(x):=\int_{a}^b f(x,t)dt

Then, the following is true.

u(x+h)−u(x)h=∫abf(x+h,t)dt−∫abf(x,t)dth=∫ab[f(x+h,t)−f(x,t)]dth=∫abf(x+h,t)−f(x,t)hdt \begin{equation} \begin{aligned} \frac{ u(x+h)-u(x)}{h} &= \frac{\int_{a}^{b} f(x+h,t)dt -\int_{a}^{b}f(x,t)dt}{h} \\ &= \frac{ \int_{a}^{b} \big[f(x+h,t)-f(x,t) \big] dt}{h} \\ &= \int_{a}^{b} \frac{f(x+h,t)-f(x,t)}{h}dt \end{aligned} \end{equation}

Moreover, for a fixed yy, applying the Mean Value Theorem to f(x,y)f(x,y), there exists c∈[x,x+h]c\in[x,x+h] that satisfies the following equation.

f(x+h,t)−f(x,t)h=∂f∂x(c,t) \frac{f(x+h,t)-f(x,t)}{h}=\frac{\partial f}{\partial x}(c,t)

Integrating both sides with respect to tt, by (1)(1), the following is true.

u(x+h)−u(x)h=∫ab∂f∂x(c,t)dt \frac{u(x+h)-u(x)}{h}=\int_{a}^{b}\frac{\partial f}{\partial x}(c,t) dt

Now, for any given ϵ>0\epsilon >0, let’s say ϵ0=ϵb−a\epsilon_{0}=\dfrac{\epsilon}{b-a}. Since [x,x+h]×[a,b][x,x+h]\times [a,b] is compact and by assumption ∂f∂x\dfrac{\partial f}{\partial x} is uniformly continuous on a compact interval, then for sufficiently small hh, the following is true.

∣∂f∂x(x+h,t)−∂f∂x(x,t)∣<ϵ0 \left| \frac{\partial f}{\partial x}(x+h,t)-\frac{\partial f}{\partial x} (x,t)\right| < \epsilon_{0}

Also, since c∈[x,x+h]c\in[x,x+h], by the definition of uniform continuity, the following is true.

∣∂f∂x(c,t)−∂f∂x(x,t)∣<ϵ0 \left| \frac{\partial f}{\partial x}(c,t)-\frac{\partial f}{\partial x} (x,t)\right| < \epsilon_{0}

Now, calculating we obtain the following.

∣lim⁡h→0u(x+h,t)−u(x,t)h−∫ab∂f∂x(x,t)dt∣= ∣lim⁡h→0∫ab∂f∂x(c,t)dt−∫ab∂f∂x(x,t)dt∣= ∣lim⁡h→0∫ab[∂f∂x(c,t)−∂f∂x(x,t)]dt∣= lim⁡h→0∣∫ab[∂f∂x(c,t)−∂f∂x(x,t)]dt∣≤lim⁡h→0∫ab∣∂f∂x(c,t)−∂f∂x(x,t)∣dt≤lim⁡h→0∫abϵ0dt= lim⁡h→0(b−a)ϵ0= ϵ \begin{align*} & \left| \lim \limits_{h\rightarrow 0}\frac{u(x+h,t)-u(x,t)}{h} - \int_{a}^{b} \frac{\partial f}{\partial x}(x,t)dt\right| \\ =&\ \left| \lim \limits_{h\rightarrow 0}\int_{a}^{b}\frac{\partial f}{\partial x}(c,t) dt - \int_{a}^{b} \frac{\partial f}{\partial x}(x,t)dt\right| \\ =&\ \left| \lim \limits_{h \rightarrow 0} \int_{a}^{b}\left[ \frac{\partial f}{\partial x}(c,t)-\frac{\partial f}{\partial x}(x,t) \right] dt\right| \\ =&\ \lim \limits_{h \rightarrow 0}\left| \int_{a}^{b}\left[ \frac{\partial f}{\partial x}(c,t)-\frac{\partial f}{\partial x}(x,t) \right] dt\right| \\ \le& \lim \limits_{h \rightarrow 0} \int_{a}^{b}\left| \frac{\partial f}{\partial x}(c,t)-\frac{\partial f}{\partial x}(x,t) \right| dt \\ \le& \lim \limits_{h \rightarrow 0} \int_{a}^{b} \epsilon_{0}dt \\ =&\ \lim \limits_{h \rightarrow 0} (b-a)\epsilon_{0} \\ =&\ \epsilon \end{align*}

This equation holds for any given ϵ>0\epsilon>0, hence the following is obtained.

lim⁡h→0u(x+h,t)−u(x,t)h=∫ab∂f∂x(x,t)dt \lim \limits_{h\rightarrow 0} \frac{u(x+h,t)-u(x,t)}{h}=\int_{a}^{b}\frac{\partial f}{\partial x}(x,t)dt

Also, since u(x):=∫abf(x,t)dt\displaystyle u(x):=\int_{a}^b f(x,t)dt, the following is true.

ddx∫abf(x,t)dt=∫ab∂f∂x(x,t)dt \frac{d}{dx} \int_{a}^b f(x,t)dt = \int_{a}^b\frac{\partial f}{\partial x}(x,t)dt

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