6.11 Improper double integrals and iterated limits
In the last lecture we introduced integrals of functions over the whole plane \(\mathbb{R}^2\).
We obtained the definition by taking the idea from the one-dimensional case:
Definition
Let \(f: \mathbb{R}^2 \rightarrow \mathbb{R}\) be a function. Assume that \(\forall\) bounded set \(S \subseteq \mathbb{R}^2\) with area, \(f\) is integrable on \(S\).
We say that \(f\) is integrable on \(\mathbb{R}^2\) if there is \(I \in \mathbb{R}\) such that \(\forall \varepsilon > 0\), \(\exists r > 0\) s.t. \(\forall\) bounded set \(S\) with area with \([-r,r] \times [-r,r] \subseteq S\), we have \(\displaystyle \left| \int_{S} f - I \right| < \varepsilon\)
Example
We have already did an example like this when computing the integral of \(f: \mathbb{R}^2 \to \mathbb{R}\) given by \(f(x, y) = e^{-x^2-y^2}\)
Let us show that \(\displaystyle\int_{\mathbb{R}^2}f = \pi\)
We know \(\int_{D_a}e^{-x^2-y^2}dxdy=\pi-e^{-a^2}\) for disk
Take \(\varepsilon>0\), since \(e^{-a^2}\to 0\) when \(a\to\infty\), there exists \(r>0\), \(\forall a \geq r\) such that \(|e^{-a^2}|=e^{-a^2}<\varepsilon\)
For a bounded set \(S\), there exists \(a>0\) s.t. \(S\subset D_a\). If \([-r,r]\times[-r,r]\subseteq S\), then \(a>r\)
Since \(D_r\subset S\subset D_a\), then \(\displaystyle\int_{D_r}e^{-x^2-y^2}dxdy\leq \int_{S}e^{-x^2-y^2}dxdy\leq \int_{D_a} e^{-x^2-y^2}dxdy=\pi-e^{-a^2}\)
Then \(\displaystyle\pi-\int_{S}e^{-x^2-y^2}dxdy\leq \pi-\int_{D_r}e^{-x^2-y^2}dxdy=\pi-(\pi-e^{-r^2} )=e^{-r^2}<\varepsilon\)
Now we turn our attention to integrals of functions that are unbounded in some sets of points in a bounded region.
We focus on non-negative functions, because these are easy to handle in the two-dimensional case.
Moreover, we have a remarkable fact about them:
If the iterated integral exists, then the function is integrable!
This follows from a version of Fubini's theorem for improper integrals.
The following definition is an extension of the definition of integrable function on \(\mathbb{R}^2\).
Definition
Let \(D \subseteq \mathbb{R}^2\) be an open set and \(f: D \longrightarrow \mathbb{R}\) be a non-negative function.
Assume that \(f\) is integrable in every bounded subset \(S \subseteq D\) with area on which \(f\) is integrable.
The integral of \(f\) on \(D\) is \(I = \sup \left\{ \displaystyle\int_{S}f \mid S \subseteq D, S \text{ bounded with area and } f \text{ bounded on }S \right\}\)
In case an arbitrary function \(f: D \longrightarrow \mathbb{R}\) is not non-negative, we define:
\(f_+(x,y) = \begin{cases} f(x,y) & \text{if } f(x,y) \ge 0 \\ 0 & \text{otherwise} \end{cases}\) and \(f_-(x,y) = \begin{cases} -f(x,y) & \text{if } f(x,y) \le 0 \\ 0 & \text{otherwise} \end{cases}\)
Then \(f = f_+ - f_-\) with \(f_+\) and \(f_-\) non-negative functions on \(D\).
We say that \(f\) is absolutely integrable if both \(f_+\) and \(f_-\) are integrable.
In such a case, we define the integral as the difference of the integrals of \(f_+\) and \(f_-\): \(\displaystyle \int_{D}f = \int_{D}f_{+}- \int_{D}f_{-}\)
Recall that, under certain hypothesis, one can use iterated integrals on a set \(S\) with area to compute the Riemann integral.
We are going to use this result to make explicit calculations.
For simplicity of the exposition, we restrict ourselves to \(y\)-simple regions \(D\)
We assume that there are continuous functions \(\varphi_1, \varphi_2: [a,b] \to \mathbb{R}\) such that \(D=\{(x,y)\in\mathbb{R}^{2}\mid x\in[a,b],\varphi_{1}(x)\leq y\leq\varphi_{2}(x)\}\)
Example
When trying to compute the integral of the function \(f(x, y) = \frac{1}{\sqrt{1 - x^2 - y^2}}\) on the unit disc \(D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \leq 1\}\)
We see that the function is unbounded on the set of points \(\{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1\},\) which are infinitely many!
Thus, we need an strategy to compute the improper integral defined above. For this we introduce the notion of exhausting regions
Exhausting regions
For \(D\) as above, consider a function \(f: D \to \mathbb{R}\) which is bounded and integrable on every bounded subset \(S \subseteq D\) with area.
We integrate \(f\) on regions \(D^*\) where we know that it is bounded and integrable and then try to take the limit when \(D^* \to D\)
We restrict more ourselves to some kind of regions \(D^*\):
Let \(\eta > 0\) be small enough such that \(a + \eta < b - \eta\) and \(\delta > 0\) be small enough such that \(\varphi_1(x) + \delta < \varphi_2(x) - \delta\)
Note that if \(\varphi_1(x) = \varphi_2(x)\) for some \(x \in [a, b]\), this \(\delta\) does not exists.
One may solve this issue by modifying a little bit the argument we describe below.
The region \(D_{\eta, \delta} = \{(x,y) \in \mathbb{R}^2 \ | \ a+\eta < x < b-\eta, \quad \varphi_1(x) + \delta \leq y \leq \varphi_2(x) - \delta\}\) is a subset of \(D\) and \(D_{\eta, \delta}\) tends to \(D\) when \((\eta, \delta) \to (0,0)\).
Because \(f\) is integrable on \(D_{\eta, \delta}\), the integral \(\int_{D_{\eta, \delta}} f\) exists and under certain hypothesis it can be computed with iterated integrals (for example, when \(f\) is continuous).
Moreover, \(f\) is integrable on \(D\) if and only if the limit \(\lim_{(\eta, \delta) \to (0,0)} \int_{D_{\eta, \delta}} f\) exists and equals \(I\). (Exercise)
Remark
Note that here we are computing an integral using a limit of a function of two variables \(\displaystyle F(\eta, \delta) = \int_{a+\eta}^{b-\eta}\int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta} f(x,y) \,dydx\)
If this limit exists, \(\lim_{(\eta, \delta) \to (0,0)} F(\eta, \delta) = I\) it follows that the iterated limit also exists and equals \(I\): \(I = \lim_{(\eta, \delta) \to (0,0)} F(\eta, \delta) = \lim_{\eta \to 0} \left( \lim_{\delta \to 0} F(\eta, \delta) \right)\)
In terms of integrals, this reads \(\displaystyle \int_{D}f = \lim_{(\eta, \delta) \to (0,0)}\int_{D_{\eta, \delta}}f = \lim_{(\eta, \delta) \to (0,0)}\int_{a+\eta}^{b-\eta}\int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta} f(x,y) \,dydx\)
\(\displaystyle = \lim_{\eta \to 0}\left( \lim_{\delta \to 0}\int_{a+\eta}^{b-\eta} \int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta}f(x,y) \,dydx \right)\)
We know that the converse for limits does not hold in general: the iterated limits may exists, but the limit may not.
Example
\(D=[0,1]\times [0,1]\), compute \(\displaystyle\int_D\frac{1}{\sqrt[3]{xy}}dxdy\)
\(\displaystyle\int_{D}\frac{1}{\sqrt[3]{xy}}dxdy=\int_{\delta}^{1-\delta}\int_{\eta}^{1-\eta}\frac{1}{\sqrt[3]{x}}\frac{1}{\sqrt[3]{y}}dxdy=\int_{\delta}^{1-\delta}\frac{1}{\sqrt[3]{y}}\int_{\eta}^{1-\eta}\frac{1}{\sqrt[3]{x}}dxdy=\int_{\delta}^{1-\delta}\frac{1}{\sqrt[3]{y}}dy\cdot\int_{\eta}^{1-\eta}\frac{1}{\sqrt[3]{x}}dx=\frac{9}{4}\left(\sqrt[3]{(1-\eta)^2}-\sqrt[3]{\eta^2}\right)\left(\sqrt[3]{(1-\delta)^2}-\sqrt[3]{\delta^2}\right)=F(\eta,\delta)\)
Then \(\lim_{(\eta,\delta)\to (0,0)}F(\eta,\delta)=\frac{9}{4}=I\)
The quite remarkable fact is that for functions \(F(\eta, \delta)\) that are integrals, if the iterated limit exists, then the limit exists!
Note that integral \(\displaystyle\int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta}f(x, y) d y\) might be an improper integral. So, to compute it we might need to consider a limit.
Indeed, assume that for any \(x \in [a, b]\) the limit \(\displaystyle\lim_{\delta \to 0}\int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta}f(x, y) d y\) exists.
Denote if by \(\displaystyle g(x) = \lim_{\delta \to 0}\int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta}f(x, y ) d y \quad \forall x \in [a, b].\)
Suppose further that the limit \(\displaystyle \lim_{\eta \to 0}\int_{a+\eta}^{b-\eta}g(x) dx = \lim_{\eta \to 0}\left( \int_{a+\eta} ^{b-\eta}\left( \lim_{\delta \to 0}\int_{\varphi_1(x)+\delta}^{\varphi_2(x)-\delta} f(x, y) dy \right) dx \right)\) also exists and denote it by \(\displaystyle \int_{a}^{b} \int_{\varphi_1(x)}^{\varphi_2(x)}f(x, y) dy dx\)If all limit exist, then it must coincide with the one-dimensional improper integrals.
Thus, if f is integrable and the iterated improper integral exists, then we must have that \(\displaystyle\int_{D} f = \int_{a}^{b} \int_{\varphi_1(x)}^{\varphi_2(x)}f(x, y) dy dx\)
Following this arguments, one proved a half of the following theorem. The remarkable point here is that the second part holds. Here one has to use the continuity hypothesis.
Theorem
Let \(D\) be an elementary region in the plane and let \(f: D \longrightarrow \mathbb{R}\) be a non-negative function which is continuous except in the boundary \(\partial D\) of \(D\).
If either of the following integrals \(\displaystyle \iint_{D}f, \quad \int_{a}^{b}\int_{\varphi_{1}(x)}^{\varphi_{2}(x)}f(x, y) d y d x \quad \text{ or }\quad \int_{c}^{d}\int_{\psi_{1}(y)}^{\psi_{2}(y)}f(x, y) d x d y\) exist as improper integral, then \(f\) is integrable and they are equal.
Example
Compute \(\displaystyle\int_{D_1}\frac{1}{\sqrt{1-x^2-y^2}}dxdy\) where \(D_{1}=\{(x,y)\in \R^{2}:x^{2}+y^{2}\leq 1\}\)
Thus this is \(y\) simple and \(\varphi_1(x)=-\sqrt{1-x^2}\) and \(\varphi_{2}(x)=\sqrt{1-x^{2}}\) where \(-1<x<1\)
Take \(\eta>0\) and \(\delta>0\) and \(D_{\eta,\delta}=\{(x,y)\in \R^2:-1+\eta<x<1+\eta,\varphi_1(x)+\delta<y<\varphi_2(x)-\delta\}\)
Then\(\displaystyle\int_{D_{\eta,\delta}}f=\int^{1-\eta}_{-1+\eta}\int^{\sqrt{1-x^2}+\delta}_{-\sqrt{1-x^2}+\delta}\frac{1}{\sqrt{1-x^2-y^2}}dydx\)
We know \(\displaystyle\int_{a}^{b}\frac{1}{\sqrt{1-y^{2}}}dy=\int_{\sin^{-1}(a)}^{\sin^{-1}(b)} \frac{\cos(t)}{\sqrt{\cos^{2}(t)}}dt=\int_{\sin^{-1}(a)}^{\sin^{-1}(b)}1dt= \sin ^{-1}(b) - \sin^{-1}(a)\) because let \(y = \sin(t)\), \(dy = \cos(t) dt\)
Then \(g(y)=\frac{1}{\sqrt{1-x^{2}-y^{2}}}=\frac{1}{\sqrt{c-y^{2}}}\) \(=\frac{1}{\sqrt{c\left(1-\frac{y^{2}}{c}\right)}}=\frac{1}{\sqrt{c}}\frac{1}{\sqrt{1-\left(\frac{y}{\sqrt{c}}\right)^{2}}}\)
Then \(\displaystyle\int_{a}^{b}\frac{1}{\sqrt{c-y^{2}}}dy=\frac{1}{\sqrt{c}}\int_{a}^{b}\frac{1}{\sqrt{1-\left(\frac{y}{\sqrt{c}}\right)^{2}}} dy=\int_{\frac{a}{\sqrt{c}}}^{\frac{b}{\sqrt{c}}}\frac{1}{\sqrt{1-u^{2}}}du\) since \(u = \sin(t)\) and \(du = \cos(t) dt\)
Then \(=\sin^{-1}\left(\frac{b}{\sqrt{c}}\right)-\sin^{-1}\left(\frac{a}{\sqrt{c}}\right )=\sin^{-1}\left(\frac{b}{\sqrt{1-x^{2}}}\right)-\sin^{-1}\left(\frac{a}{\sqrt{1-x^{2}}} \right)\)
Then \(\displaystyle\int_{D_{\eta, \delta}}f = \int_{-1+\eta}^{1-\eta}\left( \sin^{-1}\left (\frac{\sqrt{1-x^{2}}-\delta}{\sqrt{1-x^{2}}} \right) - \sin^{-1}\left(\frac{\sqrt{1-x^{2}}+\delta}{\sqrt{1-x^{2}}}\right) \right ) dx\)
Then since \(\displaystyle\lim_{(\eta, \delta) \to (0,0)}\int_{D_{\eta, \delta}}f = \lim_{\eta \to 0}\lim_{\delta \to 0}\int_{D_{\eta, \delta}}f = \lim_{\eta \to 0}\int_{-1+\eta}^{1-\eta}(\sin^{-1} (1) - \sin^{-1}(-1)) dx\)
\(\displaystyle= \lim_{\eta \to 0}\int_{-1+\eta}^{1-\eta}\left(\frac{\pi}{2}- \left (-\frac{\pi}{2}\right)\right) dx=\) \(\displaystyle \lim_{\eta \to 0}\int_{-1+\eta}^{1-\eta}\pi dx = \lim_{\eta \to 0}\pi((1-\eta) - (-1+\eta)) = \lim_{\eta \to 0}\pi(1-\eta+1-\eta) = \lim_{\eta \to 0}\pi(2-2\eta ) = 2\pi\)
Unbounded functions at isolated points
Finally we consider functions that are unbounded at isolated points on elementary regions \(D\).
Let \(f: D \longrightarrow \mathbb{R}\) be a non-negative function which is continuous on every point \(D\) of but it is not defined on \((x_0, y_0) \in D\).
Consider a disk \(D_\delta\) of radius \(\delta\) around the point \((x_0, y_0) \in D\). Then the set \(D - D_\delta\) is also elementary.
Definition
We say that f is integrable on D if \(\displaystyle\lim_{\delta \to 0}\int_{D-D_\delta}f\) exists.
Example
Polar coord. \(\displaystyle\int_{D} \frac{1}{\sqrt{x^{2}+y^{2}}}\, dx dy\) let \(\begin{cases} x = r \cos(t) \\ y = r \sin(t) \end{cases}\)
\(\displaystyle \int_{D_\delta}\frac{1}{\sqrt{x^{2}+y^{2}}}dx dy = \int_{\delta}^{1} \int_{0}^{2\pi} \frac{1}{\sqrt{r^{2}}}r dt dr = \int_{\delta}^{1} \int_{0}^{2\pi}1 dt dr = \int_{\delta} ^{1} t \Big|_{0}^{2\pi}dr = \int_{\delta}^{1} 2\pi dr = 2\pi(1-\delta)\)
Then \(\lim_{\delta \to 0} 2\pi(1-\delta) = 2\pi\)