Problem Pit: The Deceptive Double Integral

The Problem

Evaluate the double integral: $$\iint_R \frac{x-y}{(x+y)^3} \, dx \, dy$$ where $R$ is the region bounded by $x = 0$, $y = 0$, $x + y = 1$, and $x + y = 2$. This looked like a standard double integration problem. The region seemed clear, the integrand manageable. What could go wrong? As it turns out, this problem would take me through coordinate transformations and some surprising insights about symmetry.

Setting Up the Region

First, let me understand the region $R$: - $x \geq 0$ (right of y-axis) - $y \geq 0$ (above x-axis) - $x + y \geq 1$ (above the line $x + y = 1$) - $x + y \leq 2$ (below the line $x + y = 2$) The vertices of this trapezoidal region are: - $(1,0)$ where $x + y = 1$ meets $y = 0$ - $(0,1)$ where $x + y = 1$ meets $x = 0$ - $(2,0)$ where $x + y = 2$ meets $y = 0$ - $(0,2)$ where $x + y = 2$ meets $x = 0$ So $R$ is a trapezoid with these four vertices.

First Approach: Direct Integration

I'll integrate by setting up bounds for $y$ first, then $x$. For the region: - If $0 \leq y \leq 1$: then $x$ ranges from $1-y$ to $2-y$ - If $1 \leq y \leq 2$: then $x$ ranges from $0$ to $2-y$ The integral becomes: $$\int_0^1 \int_{1-y}^{2-y} \frac{x-y}{(x+y)^3} \, dx \, dy + \int_1^2 \int_0^{2-y} \frac{x-y}{(x+y)^3} \, dx \, dy$$ For the inner integral, I need: $$\int \frac{x-y}{(x+y)^3} \, dx$$ Using substitution $u = x + y$, so $x = u - y$ and $dx = du$: $$\int \frac{(u-y)-y}{u^3} \, du = \int \frac{u-2y}{u^3} \, du = \int \left(\frac{1}{u^2} - \frac{2y}{u^3}\right) du$$ $$= -\frac{1}{u} + \frac{y}{u^2} + C = -\frac{1}{x+y} + \frac{y}{(x+y)^2} + C$$

Computing the First Integral

For $0 \leq y \leq 1$: $$\int_{1-y}^{2-y} \frac{x-y}{(x+y)^3} \, dx = \left[-\frac{1}{x+y} + \frac{y}{(x+y)^2}\right]_{x=1-y}^{x=2-y}$$ When $x = 2-y$: $x + y = 2$, giving $-\frac{1}{2} + \frac{y}{4}$ When $x = 1-y$: $x + y = 1$, giving $-1 + y$ Therefore: $$\int_{1-y}^{2-y} \frac{x-y}{(x+y)^3} \, dx = \left(-\frac{1}{2} + \frac{y}{4}\right) - (-1 + y) = \frac{1}{2} - \frac{3y}{4}$$ The first part becomes: $$\int_0^1 \left(\frac{1}{2} - \frac{3y}{4}\right) dy = \left[\frac{y}{2} - \frac{3y^2}{8}\right]_0^1 = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}$$

Computing the Second Integral

For $1 \leq y \leq 2$: $$\int_0^{2-y} \frac{x-y}{(x+y)^3} \, dx = \left[-\frac{1}{x+y} + \frac{y}{(x+y)^2}\right]_{x=0}^{x=2-y}$$ When $x = 2-y$: gives $-\frac{1}{2} + \frac{y}{4}$ When $x = 0$: gives $-\frac{1}{y} + \frac{1}{y} = 0$ So: $$\int_0^{2-y} \frac{x-y}{(x+y)^3} \, dx = -\frac{1}{2} + \frac{y}{4}$$ The second part becomes: $$\int_1^2 \left(-\frac{1}{2} + \frac{y}{4}\right) dy = \left[-\frac{y}{2} + \frac{y^2}{8}\right]_1^2 = \left(-1 + \frac{1}{2}\right) - \left(-\frac{1}{2} + \frac{1}{8}\right) = -\frac{1}{8}$$ Total result: $\frac{1}{8} + (-\frac{1}{8}) = 0$. But I felt like I was missing something. The answer being exactly zero seemed too neat.

The Symmetry Observation

Looking at the integrand $\frac{x-y}{(x+y)^3}$, I noticed something important. Under the transformation $(x,y) \mapsto (y,x)$: $$\frac{x-y}{(x+y)^3} \mapsto \frac{y-x}{(y+x)^3} = -\frac{x-y}{(x+y)^3}$$ The integrand is antisymmetric! However, the region $R$ isn't symmetric under this swap, so this observation doesn't immediately help.

Dead End: Integration by Parts

I wondered if integration by parts might simplify things. Setting $u = x-y$ and $dv = \frac{dx}{(x+y)^3}$: To find $v$: $\int (x+y)^{-3} dx = -\frac{1}{2(x+y)^2}$ By integration by parts: $$\int \frac{x-y}{(x+y)^3} \, dx = -\frac{x-y}{2(x+y)^2} - \int \left(-\frac{1}{2(x+y)^2}\right) dx$$ $$= -\frac{x-y}{2(x+y)^2} - \frac{1}{2(x+y)} + C$$ This wasn't simpler than my original antiderivative, so I abandoned this approach.

The Breakthrough: Change of Variables

The real insight came when I realized the natural coordinates for this problem are: - $u = x + y$ - $v = x - y$ This transformation directly addresses the structure of both the integrand and region boundaries. Solving for $x$ and $y$: - $x = \frac{u + v}{2}$ - $y = \frac{u - v}{2}$ The Jacobian is: $$J = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}$$ So $|J| = \frac{1}{2}$. The integrand transforms beautifully: $$\frac{x-y}{(x+y)^3} = \frac{v}{u^3}$$

Transforming the Region

The original constraints become: 1. $x \geq 0 \Rightarrow u + v \geq 0$ 2. $y \geq 0 \Rightarrow u - v \geq 0$ 3. $x + y \geq 1 \Rightarrow u \geq 1$ 4. $x + y \leq 2 \Rightarrow u \leq 2$ From constraints 1 and 2: $v \geq -u$ and $v \leq u$, so $-u \leq v \leq u$. From constraints 3 and 4: $1 \leq u \leq 2$. In $(u,v)$ coordinates, the region is: $$R': \{(u,v) : 1 \leq u \leq 2, -u \leq v \leq u\}$$ This is much cleaner! For each $u$ from 1 to 2, $v$ ranges symmetrically from $-u$ to $u$.

The Final Integration

Our integral becomes: $$\iint_R \frac{x-y}{(x+y)^3} \, dx \, dy = \iint_{R'} \frac{v}{u^3} \cdot \frac{1}{2} \, du \, dv = \frac{1}{2} \int_1^2 \int_{-u}^u \frac{v}{u^3} \, dv \, du$$ The inner integral: $$\int_{-u}^u \frac{v}{u^3} \, dv = \frac{1}{u^3} \int_{-u}^u v \, dv = \frac{1}{u^3} \left[\frac{v^2}{2}\right]_{-u}^u = \frac{1}{u^3} \left(\frac{u^2}{2} - \frac{u^2}{2}\right) = 0$$ The inner integral is zero because we're integrating an odd function ($v$) over the symmetric interval $[-u, u]$! Therefore: $$\iint_R \frac{x-y}{(x+y)^3} \, dx \, dy = \frac{1}{2} \int_1^2 0 \, du = 0$$

Why This Makes Perfect Sense

The coordinate transformation revealed the true structure of the problem. In $(u,v)$ coordinates: - $u = x + y$ captures the "sum" part - $v = x - y$ captures the "difference" part The integrand $\frac{v}{u^3}$ separates these components cleanly. We're integrating the antisymmetric part ($v$) over a region that's symmetric in the $v$-direction for each fixed $u$. By symmetry, this integral must be zero. My original calculation gave the same result, but the coordinate transformation shows why it must be zero, rather than just verifying it numerically.

Lessons Learned

This problem taught me several valuable lessons: 1. When expressions involve sums and differences of variables, consider coordinates that separate these components. 2. Symmetry arguments often provide deeper insight than brute-force calculation. 3. The structure of both the integrand and the region should guide coordinate choice. 4. A "messy" calculation yielding a simple answer is often a hint that there's a more elegant approach. 5. Understanding why an answer is what it is is more valuable than just computing it. The coordinate transformation $(u,v) = (x+y, x-y)$ was the key insight that converted a computational slog into an elegant symmetry argument. This technique of separating symmetric and antisymmetric components appears throughout mathematics and physics.

Conclusion

The answer is 0, but more importantly, we now understand why it must be zero. The integrand has an antisymmetric component that vanishes when integrated over the symmetric region in the natural coordinates. This insight provides a general framework for tackling similar problems involving symmetric and antisymmetric structures.