Theorem. Leibniz Rule for Differentiating Integrals with Variable Integration Bounds
Let $ f(x, t) $ be a function such that:
- $ f(x, t) $ and its partial derivative $ \frac{\partial f}{\partial x}(x, t) $ are continuous on the region of interest.
- The functions $ a(x) $ and $ b(x) $ are differentiable with respect to $ x $, and for each $ x $, the integrals $ \int_{a(x)}^{b(x)} f(x, t) , dt $ exist.
Then the derivative of the function
$$ F(x) = \int_{a(x)}^{b(x)} f(x, t) \, dt $$with respect to $ x $ is given by the formula:
$$ F'(x) = f(x, b(x)) \cdot b'(x) - f(x, a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(x, t) \, dt $$Case 1: Constant Bounds
For constant bounds $ a $ and $ b $, we need to show:
$$ \frac{d}{dx} \left( \int_{a}^{b} f(x, t) \, dt \right) = \int_{a}^{b} \frac{\partial f}{\partial x}(x, t) \, dt $$Define $ F(x) = \int_{a}^{b} f(x, t) , dt $. By the limit definition of the derivative:
$$ F'(x) = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = \lim_{h \to 0} \frac{1}{h} \left( \int_{a}^{b} f(x+h, t) \, dt - \int_{a}^{b} f(x, t) \, dt \right) $$Since the bounds are constant, combine the integrals:
$$ F'(x) = \lim_{h \to 0} \frac{1}{h} \int_{a}^{b} \left( f(x+h, t) - f(x, t) \right) dt $$The difference quotient inside the integral:
$$ \frac{f(x+h, t) - f(x, t)}{h} \to \frac{\partial f}{\partial x}(x, t) \quad \text{as} \quad h \to 0 $$Interchanging the limit and integral (justified by the continuity of $ f(x, t) $ and its partial derivative):
$$ F'(x) = \int_{a}^{b} \lim_{h \to 0} \frac{f(x+h, t) - f(x, t)}{h} \, dt = \int_{a}^{b} \frac{\partial f}{\partial x}(x, t) \, dt $$Thus, for constant bounds, we have:
$$ \frac{d}{dx} \left( \int_{a}^{b} f(x, t) \, dt \right) = \int_{a}^{b} \frac{\partial f}{\partial x}(x, t) \, dt $$Case 2: Variable Bounds
Now consider the case where $ a(x) $ and $ b(x) $ are functions of $ x $. We want to compute:
$$ F(x) = \int_{a(x)}^{b(x)} f(x, t) \, dt $$We decompose this into two fixed-bound integrals:
$$ F(x) = \int_{c}^{b(x)} f(x, t) \, dt - \int_{c}^{a(x)} f(x, t) \, dt $$where $ c $ is a constant. Now, differentiate each integral using the chain rule and the Fundamental Theorem of Calculus.
Call the first integral $ F_{b(x)}(b(x), f(x, t)) $ and the second integral $ F_{a(x)}(a(x), f(x, t)) $. By the multivariable chain rule, we have:
$$ \frac{d}{dx} F_{b(x)}(b(x), f(x, t)) = \frac{\partial F}{\partial b} \cdot \frac{db}{dx} + \frac{\partial F}{\partial f} \cdot \frac{\partial f}{\partial x} $$and similarly for $ F_a(x) $.
To compute $ \frac{\partial F}{\partial b} $, we need to differentiate $ F_b(x) $ with respect to $ b $. Using the Fundamental Theorem of Calculus, we have:
$$ \frac{\partial F_{b(x)}}{\partial b(x)} = \frac{\partial}{b(x)} \left( \int_{c}^{b(x)} f(x, t) \, dt \right) = f(x, b(x)) \cdot b'(x) $$Notice that since $ f(x, t) $ is constant with respect to $ b(x) $, then the bounds of integration $ c $ and $ b(x) $ are constant with respect to it. Thus, the Leibniz Rule for constant bounds applies.
$$ \frac{\partial F}{\partial f} = \frac{\partial}{\partial f} \left( \int_{c}^{b(x)} f(x, t) \, dt \right) = \int_{c}^{b(x)} \frac{\partial}{\partial f} f(x, t) \, dt = \int_{c}^{b(x)} 1 \, dt $$- First integral: $ \int_{c}^{b(x)} f(x, t) , dt $
Putting it all together:
$$ \frac{d}{dx} \left( \int_{c}^{b(x)} f(x, t) \, dt \right) = f(x, b(x)) \cdot b'(x) + \int_{c}^{b(x)} \frac{\partial f}{\partial x}(x, t) \, dt $$- Second integral: $ \int_{c}^{a(x)} f(x, t) , dt $
Similarly, for the second integral:
$$ \frac{d}{dx} \left( \int_{c}^{a(x)} f(x, t) \, dt \right) = f(x, a(x)) \cdot a'(x) + \int_{c}^{a(x)} \frac{\partial f}{\partial x}(x, t) \, dt $$- Subtract the results:
Now, subtract the derivative of the second integral from the derivative of the first:
$$ F'(x) = \left[ f(x, b(x)) \cdot b'(x) + \int_{c}^{b(x)} \frac{\partial f}{\partial x}(x, t) \, dt \right] - \left[ f(x, a(x)) \cdot a'(x) + \int_{c}^{a(x)} \frac{\partial f}{\partial x}(x, t) \, dt \right] $$Simplifying:
$$ F'(x) = f(x, b(x)) \cdot b'(x) - f(x, a(x)) \cdot a'(x) + \left( \int_{c}^{b(x)} \frac{\partial f}{\partial x}(x, t) \, dt - \int_{c}^{a(x)} \frac{\partial f}{\partial x}(x, t) \, dt \right) $$The difference of the two integrals can be combined into one:
$$ F'(x) = f(x, b(x)) \cdot b'(x) - f(x, a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial f}{\partial x}(x, t) \, dt $$Clarification of Assumptions:
In both the constant bounds and variable bounds cases, we assumed that the partial derivative $ \frac{\partial f}{\partial x}(x, t) $ is continuous. This is a sufficient condition for applying the rule, as it ensures that the limit of the difference quotient converges uniformly, allowing us to interchange the limit and the integral.
However, the Leibniz Rule still holds under weaker conditions. Specifically, the partial derivative $ \frac{\partial f}{\partial x}(x, t) $ does not need to be continuous. The key requirement is that the limit of the difference quotient can be interchanged with the integral.
More generally, the interchange of the limit and the integral (in this case, taking the derivative under the integral sign) is valid if:
- $ \frac{\partial f}{\partial x}(x, t) $ exists almost everywhere and is absolutely integrable on the interval $ [a(x), b(x)] $.
- Alternatively, uniform convergence or the Dominated Convergence Theorem can be used to justify bringing the limit inside the integral if there exists a dominating function that bounds the integrand.