\begin{align} \frac{\partial f}{\partial x} &= \frac{ - e^{-(x+y)} \left( (x-y)^2 + a^2 \right) - e^{-(x+y)} \cdot 2(x-y) }{ \left( (x-y)^2 + a^2 \right)^2 } \\ &= \frac{ - e^{-(x+y)} \left( (x-y)^2 + a^2 + 2(x-y) \right)} { \left( (x-y)^2 + a^2 \right)^2 } \end{align}
\begin{align} x = \frac{1}{2} (u+v), \ \ y = \frac{1}{2} (-u+v) \end{align} なので、 \begin{align} \frac{\partial (x,y)}{\partial (u,v)} &= \frac{1}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \left( - \frac{1}{2} \right) \\ &= \frac{1}{2} \end{align} である。
\begin{align} I &= \iint_D f(x,y) dx dy \\ &= \iint_E \frac{e^{-v}}{u^2 + a^2} \left| \frac{\partial (x,y)}{\partial (u,v)} \right| du dv \\ &= \frac{1}{2} \iint_E \frac{e^{-v}}{u^2 + a^2} du dv \\ &= \frac{1}{2} \int_0^a e^{-v} dv \int_0^a \frac{du}{u^2 + a^2} \\ &= \frac{1}{2} \left[ -e^{-v} \right]_0^a \cdot \frac{1}{a} \int_0^\frac{\pi}{4} d \theta \ \ \ \ \ \ \ \ (u = a \tan \theta) \\ &= \frac{\pi}{8a} \left( 1 - e^{-a} \right) \end{align}
\begin{align} \lim_{a \to +0} I &= \frac{\pi}{8} \lim_{a \to +0} \frac{ 1 - e^{-a} }{a} \\ &= \frac{\pi}{8} \lim_{a \to +0} \frac{ e^{-a} }{1} \\ &= \frac{\pi}{8} \end{align}