Question

Use your CAS to compute iterated integrals. \(\int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dydx} } \) and . \(\int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dxdy} } \). Explain with the help of Fubini’s theorem.

Short answer

Fubini’s theorem dies not apply here.

Step-by-step solution

Using maple to find exact value of integrals.

\(\begin{array}{l}evalf\left( {{\mathop{\rm int}} \left( {{\mathop{\rm int}} \left( {\frac{{x - y}}{{{{(x + y)}^3}}}} \right)x = 0....1,y = 0....1} \right)} \right);\\ > evalf\left( {{\mathop{\rm int}} \frac{{(x - y)}}{{{{(x + y)}^3}}},y = 0....1,x = 0....1} \right);\end{array}\)

Maple result:

\(evalf\left( {{\mathop{\rm int}} \left( {{\mathop{\rm int}} \left( {\frac{{x - y}}{{{{(x + y)}^3}}},x = 0....1} \right),y = 0....1} \right)} \right);\)

\( - 0.50000000\)

\(evalf\left( {{\mathop{\rm int}} \left( {{\mathop{\rm int}} \left( {\frac{{x - y}}{{{{(x + y)}^3}}},y = 0....1} \right),x = 0....1} \right)} \right);\)

\(0.50000000\)

So,\(\int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dydx} } = 0.5\)

\(\int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dxdy} } = - 0.5\)

Therefore, \(\int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dydx} } \ne \int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dxdy} } \)

Sketching the graph of the function  

\(f(x,y) = \frac{{x - y}}{{{{(x + y)}^3}}}\)on \(R = (0,1) \times (0,1)\).

According to Fubini’s theorem, if the integral is continuous on\(R = (0,1) \times (0,1)\), then only we can write\(\int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dydx} } = \int\limits_0^1 {\int\limits_0^1 {\frac{{x - y}}{{{{(x + y)}^3}}}dxdy} } \).

But, from graph, the function is not continuous.

So, Fubini’s theorem does not apply here