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Consider the improper double integral $$ I_{\text{Riemann}} = \int_0^1 \int_0^{\sqrt x} \frac{2xy}{1-y^4} \; dy \; dx = \lim_{B \to 1^-} \int_0^B \int_0^{\sqrt x} \frac{2xy}{1-y^4} \; dy \; dx .$$ The standard "freshman calculus" solution goes by swapping the order of integration to get $\int_0^1 \int_{y^2}^1 \frac{2xy}{1-y^4} \; dx \; dy$; then the inner integral is $\frac{y}{1-y^4} \int_{y^2}^1 2x \; dx = \frac{y}{1-y^4} \left[ x^2 \right]^1_{y^2} = y$ and the answer is $\int_0^1 y \; dy = \frac12$. I'm trying to make sense of this rigorously, particularly the bit about swapping the order of integration (which seems to require some sort of Tonelli/Fubini result).

My idea is something like the following: define the double Lebesgue integral $$I_{\text{Lebesgue}} = \int_{x \in [0,1)} \int_{y \in [0,1)} \mathbf 1(x \ge y^2) \cdot \frac{2xy}{(1-y^4)} \; dy \; dx.$$ Then Tonelli's theorem lets us swap the order of summation to get $$I_{\text{Lebesgue}} = \int_{y \in [0,1)} \frac{y}{1-y^4} \int_{x \in [0,1)} \mathbf 1(x \ge y^2) \cdot 2x \; dx \; dy.$$ Thus the inner integral is the same as the Riemann one $\int_{y^2}^1 2x \; dx = 1 - y^4$, hence $$I_{\text{Lebesgue}} = \int_{y \in [0,1)} y \; dy = \frac12.$$ However, since the original Riemann integral was improper, I don't really know how to justify the first step (cue $x^{-1} \sin x$ example).

So I have the following three questions:

  1. Is there some result/theorem that lets me quickly see that $I_{\text{Riemann}} = I_{\text{Lebesgue}}$? Bonus points for not using nonnegativity of $\frac{2xy}{1-y^4}$.

  2. Is the calculation of $I_{\text{Lebesgue}}$ correct as written?

  3. Are there other ways to justify the interchange of improper integrals? I'm fine appealing to Lebesgue since Lebesgue integrals are "better-behaved" anyways, but I'm wondering if I've missed something easier.

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  • $\begingroup$ $u=y^2$ substitution simplifies it. Make it immediately. $\endgroup$ – herb steinberg Aug 26 '18 at 22:02
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    $\begingroup$ Isn’t nonnegativity the fastest way? Simply apply Tonelli’s theorem. $\endgroup$ – Szeto Aug 26 '18 at 22:42
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You don't need to turn to Lebesgue integration to add rigor. The result can be obtained rigorously within the framework of the multi-dimensional Riemann integral.

The Riemann integral is naturally defined over bounded rectangles and extended to more general (rectifiable) regions with an indicator function. Before we even begin to consider Fubini's theorem to manipulate iterated integrals, we first need to define what the inproper integral means. In this case, the restriction of $f(x,y) = 2xy/(1-y^4)$ is continuous and Riemann integrable over $S_b = \{(x,y): 0 \leqslant y \leqslant \sqrt{x} \leqslant b$ for any $b < 1$ with the integral given by

$$I_b = \int_{S_b}f = \int_{[0,b]^2}f \, \chi_{S_b}$$

There are some technicalities regarding how to precisely define the improper (or extended) Riemann integral for arbitrary regions, but it boils down in this case to $I = \lim_{b \to 1-} I_b$ when the limit exists.

At this point we apply Fubini's theorem specifically for Riemann integrals, as discussed for example in Analysis on Manifolds by Munkres or Calculus on Manifolds by Spivak. This states simply if $f$ is integrable and the iterated Riemann integrals exist --both of which are satisfied here -- we have

$$\begin{align}I_b = \int_0^b \left(\int_0^b \frac{2xy}{1 - y^4}\, \chi_{S_b} \, dy\right) \, dx &= \int_0^b \left(\int_0^b \frac{2xy}{1 - y^4}\, \chi_{S_b} \, dx\right) \, dy \\ &= \int_0^b \left(\int_{y^2}^b \frac{2xy}{1 - y^4}\, \, dx\right) \, dy \\ &= \int_0^b \frac{(b^2-y^4)y}{1 - y^4}\, dy \\ &= \int_0^1 \frac{(b^2-y^4)y}{1 - y^4} \chi_{y \leqslant b} \, dy \end{align}$$

Since, for all $y \in [0,1]$ we have

$$\left| \frac{(b^2-y^4)y}{1 - y^4} \chi_{y \leqslant b} \right| \leqslant y \leqslant 1$$

it follows by the bounded convergence theorem that

$$I = \lim_{b \to 1-}I_b = \int_0^1 \lim_{b \to 1-} \frac{(b^2-y^4)y}{1 - y^4} \chi_{y \leqslant b} \, dy = \int_0^1 y \, dy = \frac{1}{2}$$

Lebesgue and Riemann integrals coincide on bounded rectangles, so the demonstration could also be made in the context of Lebesgue integrals in an analogous fashion with the dominated convergence theorem.

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