double integral getting different results I am trying to calculate the double integral $$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx$$ If you plug this into wolfram, you get $-\frac{1}{2}$ and if you plug it into symbolab you get $\frac{1}{2}$ I will show you my steps, I just want to make sure I got the right answer.
$$\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx=\lim_{b \to 0^+} \int_{b}^1 \int_b^1 \frac{y+x}{(y+x)^3}-\frac{2x}{(y+x)^3}dydx$$
$$=\lim_{b \to 0^+} \int_{b}^1 \frac{-1}{(1+x)^2}dx=\lim_{b \to 0^+} \frac{1}{1+x}\Big|_b^1=\frac{-1}{2}$$ I just wanted to verify because these two different websites are giving me different answers.
 A: Both are incorrect.  The integral is zero.
To understand why, you can see that the integrand is antisymmetric in $x$ and $y$; specifically, if $$f(x,y) = \frac{y-x}{(x+y)^3},$$ then $$f(y,x) = -f(x,y).$$  So on a square region $[b, 1]^2$, the integral is always zero.  Taking the limit as $b \to 0^+$ does not change this fact.
Here is how the integral should be evaluated in Mathematica:
Integrate[(y - x)/(y + x)^3, {x, b, 1}, {y, b, 1},  Assumptions -> 0 < b < 1]
The answer given is 0.  If you instead entered
Integrate[(y - x)/(y + x)^3, {x, 0, 1}, {y, 0, 1}]
You will get -1/2, which is incorrect, but I should stress here that it is wrong not because Mathematica made a computational error, but because this expression is not the same as what you are actually trying to evaluate!  That is to say, $$\int_{x=0}^1 \int_{y=0}^1 \frac{y-x}{(x+y)^3} \, dy \, dx \ne \lim_{b \to 0^+} \int_{x=b}^1 \int_{y=b}^1 \frac{y-x}{(x+y)^3} \, dy \, dx.$$  To give you an sense of why this is the case, try evaluating $$\int_{y=0}^1 \int_{x=0}^1 \frac{y-x}{(x+y)^3} \, dx \, dy.$$  If you do this in Mathematica, the result is 1/2.  The integrand does not satisfy Fubini's theorem.
A: I may be wrong but I think the correct result is $0$.
Where I think you got it wrong is in the second equality, where you exchange limit and the integral for one term.
Here I sketched some more detailed computations:
\begin{align}
  \lim_{b \to 0}
    \int_b^1 \int_b^1 \frac{y-x}{(y+x)^3}dydx
{}={} &
  \lim_{b \to 0}
     \int_b^1 \int_b^1
        \left(\frac{y+x}{(y+x)^3}-\frac{2x}{(y+x)^3}\right)dydx
\\
{}={} &
  \lim_{b \to 0}
     \int_b^1 
        \left(
          -\frac{1}{y+x}\Bigg|_{y=b}^{y=1}
          +
          \frac{2x}{2(y+x)^2}\Bigg|_{y=b}^{y=1}
         \right)dx
\\
{}={} &
  \lim_{b \to 0}
     \int_b^1 
        \left(
          \frac{1}{b+x}
          -\frac{1}{1+x}
          +
          \frac{x}{(1+x)^2}
          -\frac{x}{(b+x)^2}
         \right)dx
\\
{}={} &
  \lim_{b \to 0}
     \int_b^1 
        \left(
          \frac{-1}{(1+x)^2}
          +\frac{b}{(b+x)^2}
         \right)dx
\\
{}={} &
  \lim_{b\to0}
  \left(
    \frac{1}{1+x}
    -\frac{b}{b+x}
    \Bigg|_{x=b}^{x=1}
  \right)
\\
{}={} &
  \lim_{b\to0}
  \left(
    \frac{1}{2}
    -\frac{1}{1+b}
    +\frac{b}{2b}
    -\frac{b}{b+1}
  \right)
\\
{}={} &
   0
\end{align}
