Problem 31 - Calculus On Manifolds: A Modern Approach To Classical Theorems Of Advanced Calculus

Question

3-31. If \(A=\left[a_{1}, b_{1}\right] \times \times\left[a_{n}, b_{n}\right]\) and \(f: A \rightarrow \mathbf{R}\) is continuous, define \(F: A \rightarrow \mathbf{R}\) by $$ F(x)=\int_{\left[a_{1}, x^{1}\right] \times \cdots \times\left[a_{n}, x^{n}\right]} f $$ What is \(D_{i} F(x)\), for \(x\) in the interior of \(A ?\)

Step-by-Step Solution

Answer

The result of differentiating \(F(x)\) with respect to \(x^i\) is \(D_i F(x) = f(x)\).

Step 1: Understand the function definition

where \(x = (x^1, \dots, x^n) \) and \(f\text{ is a continuous function defined on }A \text{.}\)

Step 2: Recall fundamentals of differentiation under the integral

Differentiation under the integral sign can be used since \(f\) is continuous. Recall that if we have an integral $$ \int_{a}^{x} f(t) \, dt $$ and if \(f\) and its derivative are continuous, then the derivative of the integral is just the integrand with respect to \(x\):\( \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) \).

Step 3: Apply differentiation to the multidimensional integral

Given the function $$\int_{[a_1, x^1] \times \cdots \times [a_n, x^n]} f$$and taking the partial derivative with respect to the variable \(x^i\), we have:\( D_i F(x) = \frac{\partial}{\partial x^i} \left( \int_{[a_1, x^1] \times \cdots \times [a_n, x^n]} f \right) \).

Step 4: Use the property of the Dirac delta function

Using the property of the Dirac delta function in partial differentiation under the integral, we get, \[ D_i F(x) = f(x) \].

Key Concepts

partial differentiation

Partial differentiation is a key concept in multivariable calculus. It involves taking the derivative of a function of multiple variables with respect to one variable, while keeping the other variables constant.
For example, consider the function \( f(x, y) \). The partial derivative of \(f\) with respect to \(x\) is denoted by \( \frac{\partial f}{\partial x} \). This means we take the derivative of \(f\) as if \(y\) is a constant.
In this exercise, partial differentiation allows us to determine how the function \(F(x)\) changes with respect to one particular variable among many. By isolating the effect of one variable, we can better understand the function's behavior in a multidimensional space.

differentiation under the integral sign

Differentiation under the integral sign is a technique that allows us to differentiate an integral whose limits are variable. This technique is particularly useful when dealing with integrals depending on a parameter.
For a simple case, consider the integral \( \int_{a}^{x} f(t) \, dt \). If \(f\) is continuous, the derivative of this integral with respect to \(x\) is given by \( \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) \).
In the given exercise, we extended this concept to a multi-dimensional integral. By applying this technique, we differentiated \( \int_{[a_1, x^1] \times \cdots \times [a_n, x^n]} f \) with respect to one of the variables, say \( x^i \), to find \( D_i F(x) \).

Dirac delta function

The Dirac delta function, \( \delta(x) \), is a special mathematical function with unique properties. It is used to model an 'infinitely sharp' spike at a particular point.
For practical purposes, the Dirac delta function helps in selecting the value of a function at a specific point within an integral. The key property of the delta function is \( \int_{-\infty}^{\infty} f(x) \delta(x - a) \, dx = f(a) \), which implies that the integral 'picks out' the value of the function \(f\) at the point \(a\).
In our exercise, when we performed differentiation under the integral sign, the Dirac delta function played a critical role. It ensured that after differentiation, the resulting expression directly gave us \( f(x) \), thereby simplifying the otherwise complex multi-dimensional differentiation problem.