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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 $0$.

One Answer

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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