# Show that $I(y)=int_0^1 frac{x-y}{(x+y)^3}dx$ exists

Mathematics Asked on December 26, 2020

Show that the Riemann Integral

$$I(y)=int_0^1 frac{x-y}{(x+y)^3}dx$$

exists for $$yin(0,1]$$.

$$A=int_0^1(int_0^1 frac{x-y}{(x+y)^3}dx)dy$$, $$B=int_0^1(int_0^1 frac{x-y}{(x+y)^3}dy)dx$$

Show that $$A$$ and $$B$$ exist and $$Aneq B$$.

Showing $$Aneq B$$ wasn’t that difficult, it was just integration by parts, I got $$A=-frac{1}{2}$$ and $$B=frac{1}{2}$$. But I don’t know how to show that these integrals exist. I was able to calculate them and they’re continuous but I don’t really believe that that’s enough.
I do realize that $$I(y)$$ wouldn’t exist if $$yin[0,1]$$ because you’d divide by $$0$$.

The integrand $$displaystyle x mapsto frac{x-y}{(x+y)^3}$$ is continuous on the closed interval $$[0,1]$$ for $$0 < y leqslant 1$$ and, therefore, $$I(y)$$ exists as a Riemann integral.

We also have $$displaystyle frac{x-y}{(x+y)^3}= frac{partial}{partial x}frac{-x}{(x+y)^2},$$ which gives us for $$y in (0,1]$$,

$$I(y) = int_0^1frac{x-y}{(x+y)^3} , dx = left.frac{-x}{(x+y)^2}right|_0^1 = frac{-1}{(1+y)^2}$$

Technically $$I(0)$$ does not exist since, when $$y = 0$$, the integrand $$x^{-2}$$ is not integrable on $$[0,1]$$ in any sense. However, the value of $$I$$ at $$y=0$$ does not affect the Riemann integral since $${0}$$ is a set of measure zero. We can extend $$I|_{(0,1]}$$ as a Riemann integrable function on $$[0,1]$$ so that we have existence of the iterated integral

$$int_0^1left(int_0^1 frac{x-y}{(x+y)^3},dxright),dy = int_0^1 frac{-1}{(1+y)^2}, dy = frac{1}{2}$$

By a similar argument, the iterated integral with the order switched exists.

Since the iterated integrals are not equal it follows that the function

$$(x,y) mapsto frac{x-y}{(x+y)^3},$$

is not absolutely (or Lebesgue) integrable over $$[0,1]times [0,1]$$. Otherwise, the iterated integrals would have to be equal as a consequence of Fubini's theorem.

Correct answer by RRL on December 26, 2020

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