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Show that ∫dx from 0 to 1 ∫(x^2-y^2)/(x^2+y^2)dy from 0 to 1=∫dy from 0 to 1 ∫(x^2-y^2)/(x^2+y^2)dx from 0 to 1?

1 Answer
Apr 13, 2018

We seek to show that:

# int_0^1 int_0^1 \ (x^2-y^2)/(x^2+y^2) \ dy \ dx = int_0^1 int_0^1 (x^2-y^2)/(x^2+y^2) \ dx \ dy #

i.e., that in this particular case, the order of integration is unimportant (In general, this is not the case)

Consider the LHS integral:

# I_L = int_0^1 int_0^1 \ (x^2-y^2)/(x^2+y^2) \ dy \ dx #

# \ \ \ \ = int_0^1 [2xarctan(y/x)-y]_(y=0)^1 \ dx #

# \ \ \ \ = int_0^1 {(2xarctan(1/x)-1) - (2xarctan(0)-0)} \ dx #

# \ \ \ \ = int_0^1 2xarctan(1/x)-1 \ dx #

# \ \ \ \ = [x^2"arccot"x-arctanx]_(x=0)^1 #

# \ \ \ \ = ("arccot"1-arctan1)-(0-arctan0) #

# \ \ \ \ = pi/4-pi/4 #

# \ \ \ \ = 0 #

And the RHS:

# I_R = int_0^1 int_0^1 \ (x^2-y^2)/(x^2+y^2) \ dx \ dy #

# \ \ \ \ = int_0^1 [x-2yarctan(x/y)]_(x=0)^1 \ dy #

# \ \ \ \ = int_0^1 {(1-2yarctan(1/y)) - (0-2yarctan0)} \ dy #

# \ \ \ \ = int_0^1 1-2yarctan(1/y) \ dy #

# \ \ \ \ = [arctanx-x^2"arccot"x]_(x=0)^1 #

# \ \ \ \ = (arctan1-"arccot"1)-(arctan0-0) #

# \ \ \ \ = pi/4-pi/4 #

# \ \ \ \ = 0 #