Question

Leibniz formula for differentiation of integration

Answer 1

Answer:

Theorem. Let f(x, t) be a function such that both f(x, t) and its partial derivative fx(x, t) are continuous in t and x in some region of the (x, t)-plane, including a(x) ≤ t ≤ b(x), x0 ≤ x ≤ x1. Also suppose that the functions a(x) and b(x) are both continuous and both have continuous derivatives for x0 ≤ x ≤ x1. Then, for x0 ≤ x ≤ x1,

This formula is the general form of the Leibniz integral rule and can be derived using the fundamental theorem of calculus. The (first) fundamental theorem of calculus is just the particular case of the above formula where a(x) = a, a constant, b(x) = x, and f(x, t) = f(t).

If both upper and lower limits are taken as constants, then the formula takes the shape of an operator equation:

where is the partial derivative with respect to and is the integral operator with respect to over a fixed interval. That is, it is related to the symmetry of second derivatives, but involving integrals as well as derivatives. This case is also known as the Leibniz integral rule.

The following three basic theorems on the interchange of limits are essentially equivalent:

• the interchange of a derivative and an integral (differentiation under the integral sign; i.e., Leibniz integral rule);

• the change of order of partial derivatives;

• the change of order of integration (integration under the integral sign; i.e., Fubini's theorem).

Answer 2

Answer:

In its simplest form, called the Leibniz integral rule, differentiation under the integral sign makes the following equation valid under light assumptions on f: \frac{d}{dx} \int_{a}^{b} f(x,t) \,dt = \int_{a}^{b} \frac{\partial}{\partial x} f(x,t) \, dt .