Problem 14 - Thomas Calculus

Question

Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that $$ \int_{0}^{1} f(x)\left(\int_{0}^{x} g(x-y) f(y) d y\right) d x $$ $$ \begin{array}{l}{=\int_{0}^{1} f(y)\left(\int_{y}^{1} g(x-y) f(x) d x\right) d y} \\ {=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} g(|x-y|) f(x) f(y) d x d y}\end{array} $$

Step-by-Step Solution

Answer

Rearrange integrals to constant limits; integrate symmetrically with factor \(\frac{1}{2}\).

Step 1: Understand the Original Integral

We are given the integral expression \( \int_{0}^{1} f(x)\left(\int_{0}^{x} g(x-y) f(y) \, dy\right) \, dx \). This is a double integral where the inner integral has variable limits from 0 to \( x \). Our task is to change the order of integration to make the limits constant.

Step 2: Sketch the Region of Integration

Visualize or draw the region of integration for the original integral. The condition \(0 \leq y \leq x \leq 1\) describes a triangular region in the \(xy\)-plane bounded by \(x = 0\), \(y = 0\), and \(x = 1\). The line \(y = x\) forms the hypotenuse of this triangular region.

Step 3: Change the Order of Integration

To change the order of integration, we need to reverse the order of \( x \) and \( y \). For the region \(0 \leq y \leq x \leq 1\), the equivalent description of the region in terms of \(x\) is \(0 \leq x \leq 1\) with \(0 \leq y \leq x\) being rearranged to \(0 \leq y \leq 1\) and \(y \leq x \leq 1\). Hence, rewrite the integral as \( \int_{0}^{1} f(y)\left(\int_{y}^{1} g(x-y) f(x) \, dx\right) \, dy \).

Step 4: Derive the Symmetric Form using Absolute Value Function

Now, let us express this in a symmetric form. Consider the function \(g(|x-y|)\) which handles symmetry. The expression becomes: \[\frac{1}{2} \int_{0}^{1} \int_{0}^{1} g(|x-y|) f(x) f(y) \, dx \, dy.\] Here, each integration variable has constant limits from 0 to 1, symmetrically distributing the function over \(x\) and \(y\).

Step 5: Verification of Symmetric Integral

Verify through substitution if the symmetric form holds. By setting up a change of variables or symmetry argument \((x,y) \to (y,x)\), confirm the transformed integrals match within the specified region, ultimately providing a factor of \(\frac{1}{2}\) due to the symmetry over the triangular region.

Key Concepts

Integration Order

The order of integration in a double integral is crucial as it defines the sequence in which you handle the integrals. In some cases, changing the order of integration can simplify the evaluation process, especially when dealing with variable limits.

When we have a double integral like \( \int_{0}^{1} f(x)\left(\int_{0}^{x} g(x-y) f(y) \, dy\right) \, dx \), the inner integral initially has variable limits ranging from \( 0 \) to \( x \).
To change the order of integration, we reinterpret the bounds for the region of integration, often visualized in a sketch. The region described by \( 0 \leq y \leq x \leq 1 \) is a triangle in the xy-plane, bounded by \( x = 0 \), \( y = 0 \), and \( x = 1 \). By changing the order of integration, this becomes a region where \( 0 \leq y \leq 1 \) and \( y \leq x \leq 1 \), allowing us to express the integral in a new form: \( \int_{0}^{1} f(y)\left(\int_{y}^{1} g(x-y) f(x) \, dx\right) \, dy \).

Changing the order can turn variable limits into constants, and in some cases, this might make calculations easier or reveal a deeper symmetrized form of the integral.

Symmetric Integration

Symmetric integration is a technique used to simplify integrals, especially when the integrand has properties of symmetry. In this exercise, after changing the order of integration, we use symmetry to our advantage.

By introducing a symmetric expression, \( g(|x-y|) \), we can express the double integral in a more simplified and balanced form. The problem transforms into:

• \( \frac{1}{2} \int_{0}^{1} \int_{0}^{1} g(|x-y|) f(x) f(y) \, dx \, dy \).This symmetric form, where each of the integration variables \( x \) and \( y \) span from 0 to 1, mirrors the functions' values equally over the range.

Here, symmetry is reflected in the absolute difference \( |x-y| \), which treats both \( x \) and \( y \) equally. The factor \( \frac{1}{2} \) arises from counting the same paired interactions in the total region twice because of symmetry, essentially correcting for this duplication and reflecting the actual region of interest — a triangle.

Variable Limits

When dealing with double integrals, understanding and manipulating variable limits is vital for transforming integrals effectively. Variables limits are those where the bounds of integration depend on other variables.

In this specific problem, the existence of variable limits \( 0 \leq y \leq x \leq 1 \) initially defines a triangular region. Switching the order converts these into constant limits.

Initially, the integral features these variable bounds nested (the inner integral's limit depends on \( x \)), making direct evaluation complex. By changing the order, these variable limits are restructured into constant limits, which can make the process of integration clearer and, in some cases, analytically simpler.
This transformation into constant limits after changing the integration order involves:

• \( x \) running from \( 0 \) to \( 1 \)
• \( y \) also going from \( 0 \) to \( 1 \) within the described symmetric expression

By achieving constant limits, we greatly simplify working through the integration process.