User:JLKrause/sandbox

How do I integrate $$ \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 $$? Tharindu Ranathunga

What if I told you that:

I=$$\int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2+x_3\color{#66f}{-}x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2\color{#66f}{-}x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 $$

[math] \displaystyle\ =\int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1\color{#66f}{-}x_2+x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 = \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{\color{#66f}{-}x_1+x_2+x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 [/math]

These are all equivalent because it does not matter which numerator variable has the minus sign (the position of the negative sign does not matter).

Adding all four of these gives :

[math]\displaystyle\ 4I= \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{2x_1+2x_2+2x_3+2x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 [/math]

[math]\displaystyle\ 4I = 2 \int_1^2\int_1^2 \int_1^2 \int_1^2 \frac{x_1+x_2+x_3+x_4}{x_1+x_2+x_3+x_4}dx_1dx_2dx_3dx_4 [/math]

[math]\displaystyle\ 4I=2 \int_1^2\int_1^2 \int_1^2 \int_1^2 1 dx_1dx_2dx_3dx_4[/math]

[math]\displaystyle\ 4I=2 \implies I=\frac{1}{2}[/math]