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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.

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I believe symbolab is incorrect here. – Ben Longo Jan 23 at 17:23
    
When I do the integration in $y$ I get $$\int_b^1\frac{b}{(b+x)^2}-\frac{1}{(1+x)^2}\,dx$$ left to integrate. And it integrates to zero. This seems natural, since the original function is antisymmetric in the line $y=x$. – mickep Jan 23 at 17:33
    
@mickep How do you get that, when I do the integration in $y$, I get a $\frac{1}{x+b}-\frac{x}{(b+x)^2}$ that turns to 0, what am I doing wrong? Can I not put the limit into the integral and make that term 0? – Panphobia Jan 23 at 17:35
    
Your mistake seems to be in the step where you integrate $y$ and insert limits. Since you do not show all steps there, it is difficult to see what is going wrong. A primitive in $y$ is $-y/(x+y)^2$. Insert $y=1$ and $y=b$ and subtract, and you will see that you do not get $-1/(1+x)^2$ as a result. – mickep Jan 23 at 17:40
up vote 8 down vote accepted

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.

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Can you tell me which step I went wrong? I understand your logic, but I thought all my algebra was fine. Is it because in the integration on $y$ I got rid of a term instead of keeping it around? – Panphobia Jan 23 at 17:36
4  
@Panphobia Your algebra is okay but you have made an incorrect assumption that the interval of integration for $y$ is from $0$ to $1$ regardless of the value of $b$; that is to say, you cannot switch the order in which you perform the integration with the act of taking the limit of $b$. – heropup Jan 23 at 17:43

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}

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