Improper multiple integrals

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Consider integrating a function \(f(x,y)\) over a region \(D\) which may not be bounded or closed. In the case of a univariate function, this corresponds to the improper integral where we took the limits of the endpoints of a closed interval. In the case of multiple integrals, we adopt the notion of a "sequence of regions."



Consider a sequence of regions \(\{K_n\}\) where each \(K_n\) is a subset of \(\mathbb{R}^2\) that satisfies the following conditions:

  • (a) \(K_1 \subset K_2\)\(\subset \cdots \subset\) \(K_n \subset K_{n+1} \subset \cdots\).
  • (b) For all \(n\in \mathbb{N}\), \(K_n \subset D\).
  • (c) For all \(n \in\mathbb{N}\), \(K_n\) is bounded and closed.
  • (d) For any bounded closed set \(F\) that is included in \(D\) (i.e., \(F \subset D\)), if \(n\) is sufficiently large, then \(F \subset K_n\). 
    • In other words: for all bounded closed \(F \subset D\), there exists some \(N\in \mathbb{N}\) such that, for all \(n\in \mathbb{N}\), if \(n \geq N\) then \(F \subset K_n\).
Such a sequence \(\{K_n\}\) is called an approximating sequence of \(D\). Let \(f(x,y)\) be a function on \(D\). Furthermore, suppose the approximating sequence \(\{K_n\}\) of \(D\) satisfies the following
  • (e) For all \(n\in\mathbb{N}\), \(\iint_{K_n}f(x,y)dxdy\) exists.
Then, \(\{K_n\}\) is called an approximating sequence of \(D\) on which \(f(x,y)\) is integrable.

Definition (Improper multiple integral)

\(f(x,y)\) is said to be improper (Riemann-)integrable if there exists at least one approximating sequence \(\{K_n\}\) of \(D\) on which \(f(x,y)\) is integrable such that the limit
\[I = \lim_{n\to\infty}\iint_{K_n}f(x,y)dxdy\]
exists, and the limit \(I\) does not depend on the way the approximating sequence \(\{K_n\}\) on which \(f(x,y)\) is integrable is constructed. In this case, we say the improper multiple integral
\[I = \iint_{D}f(x,y)dxdy\]
converges.
Remark. Basically, this definition means that the sequence (of real numbers) \(\left\{\iint_{K_n}f(x,y)dxdy\right\}\) converges to the same value whatever the approximating sequence \(\{K_n\}\) is (as long as \(f(x,y)\) is integrable on all \(K_n\)). □
However, there are infinitely many possibilities for the approximating sequence of \(D\), so checking the convergence of the integral for all of them is impossible in practice. What should we do, then?

Lemma (Improper integral of a non-negative function)

Let \(f(x,y)\) be a function such that \(f(x,y) \geq 0\) for all \((x,y) \in D\). Suppose there exists an approximating sequence \(\{K_n\}\) of \(D\) on which \(f(x,y)\) is integrable. Let \(I_n = \iint_{K_n}f(x,y)dxdy\) and assume that \(I = \lim_{n\to\infty}I_n\) exists. Then, for any approximating sequence \(\{K_n'\}\) of \(D\) on which \(f(x,y)\) is integrable, the limit \(I' = \lim_{n\to\infty}\iint_{K_n'}f(x,y)dxdy\) exists and \(I' = I\).
Proof. Let us define
\[\begin{eqnarray*} I_n &=& \iint_{K_n}f(x,y)dxdy,\\ I_n' &=& \iint_{K_n'}f(x,y)dxdy. \end{eqnarray*}\]
Since \(f(x,y)\) is non-negative and \(\{K_n\}\) and \(\{K_n'\}\) are increasing sequences, we have
\[\begin{eqnarray*} I_1 \leq I_2 \leq \cdots \leq I_n \leq I_{n+1} \leq \cdots\\ I_1' \leq I_2' \leq \cdots \leq I_n' \leq I_{n+1}' \leq \cdots \end{eqnarray*}\]
That is, the sequences \(\{I_n\}\) and \(\{I_n'\}\) are monotone increasing. Since \(\{I_n\}\) converges to \(I\), we have \(I = \sup_{n}I_n\) in particular.
For any \(n\in\mathbb{N}\), we can choose (why?) a sufficiently large \(m\in\mathbb{N}\) such that \(K_n' \subset K_m\). Thus, \(I_n' \leq I_m \leq I\), which implies that the sequence \(\{I_n'\}\) is bounded above. Since \(\{I_n'\}\) is also monotone increasing, the limit \(I' = \lim_{n\to\infty}I_n'\) exists (Remember the monotone convergence theorem?). Since \(I' = \sup_{n}I_n'\), we have
\[I' \leq I. \tag{Eq:I'leqI}\]
For any \(n\in\mathbb{N}\), we can choose a sufficiently large \(m\in\mathbb{N}\) such that \(K_n \subset K_m'\). Thus, \(I_n \leq I_m' \leq I'\). Therefore we have
\[I \leq I'. \tag{eq:IleqI'}\]
By (Eq:I'leqI) and (Eq:IleqI'), we have \(I = I'\). In other words, the value of the limit \(I\) does not depend on the choice of the approximating sequence \(\{K_n\}\). ■

Corollary (Improper integral of a non-positive function)

Let \(f(x,y)\) be a function such that \(f(x,y) \leq 0\) for all \((x,y) \in D\). Suppose there exists an approximating sequence \(\{K_n\}\) of \(D\) on which \(f(x,y)\) is integrable. Let \(I_n = \iint_{K_n}f(x,y)dxdy\) and assume that \(I = \lim_{n\to\infty}I_n\) exists. Then, for any approximating sequence \(\{K_n'\}\) of \(D\) on which \(f(x,y)\) is integrable, the limit \(I' = \lim_{n\to\infty}\iint_{K_n'}f(x,y)dxdy\) exists and \(I' = I\).
Proof. Let \(g(x,y) = -f(x,y)\). Then we can apply the above Lemma. ■

Example. Given \(D = \{(x,y) \mid x \geq 0, ~ y \geq 0\}\), let us find the improper multiple integral
\[\iint_D\frac{dxdy}{(1 + x^2)(1 + y^2)}.\]
First, let us define \(K_n = \{(x,y) \mid 0 \leq x \leq n, ~ 0 \leq y \leq n\}\) for each \(n \in \mathbb{N}\). Then, \(\{K_n\}\) is an approximating sequence of \(D\).
\[\begin{eqnarray*} I_n &=& \iint_{K_n}\frac{dxdy}{(1 + x^2)(1 + y^2)}\\ &=&\int_0^n\frac{dx}{1+x^2}\int_0^n\frac{dy}{1 + y^2}\\ &=& \left[\arctan x\right]_0^n\left[\arctan y\right]_0^n\\ &=& (\arctan n)^2. \end{eqnarray*}\]
Noting that \(\lim_{n\to\infty}\arctan n = \frac{\pi}{2}\), we have
\[\lim_{n\to\infty}I_n = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}.\]
Therefore,
\[\iint_D\frac{dxdy}{(1 + x^2)(1 + y^2)}  = \frac{\pi^2}{4}.\]
□

Example (Gaussian integral). The following integral is called the Gaussian integral and is of great practical importance.
\[\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}.\]
Let's prove this equality.
Since \(e^{-x^2}\) is an even function, \(\int_{-\infty}^{\infty}e^{-x^2}dx = 2\int_{0}^{\infty}e^{-x^2}dx\). To calculate the latter integral, we consider the following double integral
\[\iint_De^{-x^2 - y^2}dxdy \text{ where } D = \{(x,y) \mid x\geq 0, ~ y \geq 0\}.\tag{Eq:gaussian}\]
Let \(K_n = \{(x,y) \mid x\geq 0, ~ y \geq 0, ~ x^2 + y^2 \leq n^2\}\) be an approximating sequence of \(D\). Using the polar coordinates, we have
\[\begin{eqnarray*} I_n &=& \iint_{K_n}e^{-x^2 - y^2}dxdy\\ &=& \int_0^{n}re^{-r^2}dr\int_0^{\frac{\pi}{2}}d\theta\\ &=&\frac{\pi}{2}\left[-\frac{1}{2}e^{-r^2}\right]_0^n\\ &=& \frac{\pi}{4}(1 - e^{-n^2}). \end{eqnarray*}\]
Since \(\lim_{n\to\infty}I_n = \frac{\pi}{4}\), the integral in (Eq:gaussian) exists and is equal to \(\frac{\pi}{4}\).

On the other hand, using another approximating sequence of \(D\)
\[D_n = \{(x,y)\mid 0\leq x \leq n,  ~ 0\leq y \leq n\},\]

the same integral can be calculated as
\[\begin{eqnarray*} J_n &=& \iint_{D_n}e^{-x^2 - y^2}dxdy = \iint_{D_n}e^{-x^2}e^{-y^2}dxdy\\ &=&\int_0^ne^{-x^2}dx\int_0^ne^{-y^2}dy = \left(\int_0^ne^{-x^2}dx\right)^2. \end{eqnarray*}\]
Noting \(e^{-x^2 - y^2} > 0\), by the above Lemma, \(J_n\) converges to the same value as \(I_n\) does, that is, \(\frac{\pi}{4}\). Since \(e^{-x^2}\) is positive, we have
\[\int_0^\infty e^{-x^2}dx = \frac{\sqrt{\pi}}{2},\]
and hence,
\[\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.\]
□

Example (Gamma function). The gamma function \(\Gamma(s)\) is defined as
\[\Gamma(s) = \int_0^\infty e^{-x}x^{s-1}dx.\]
Thus,
\[\Gamma\left(\frac{1}{2}\right) = \int_0^\infty e^{-x}x^{-\frac{1}{2}}dx.\]
Let \(x = t^2\), then
\[\Gamma\left(\frac{1}{2}\right) = \int_0^\infty e^{-x}x^{-\frac{1}{2}}dx = 2\int_0^{\infty}e^{-t^2}dt = \sqrt{\pi}.\]
□

Theorem (\(\Gamma\left(n - \frac{1}{2}\right)\))

For any \(n\in\mathbb{N}\),
\[\Gamma\left(n - \frac{1}{2}\right) = \frac{(2n-3)!!}{2^{n-1}}\sqrt{\pi}.\]
Here, \(n!! = n(n-2)(n-4)\cdots 1\) and \((-1)!! = 1\) (These are called double factorials).
Proof. Exercise. ■

The beta function is defined as
\[B(p, q) = \int_0^1x^{p-1}(1-x)^{q-1}dx.\]
The following theorem connects the gamma function with the beta function.

Theorem

For all \(p, q > 0\),
\[B(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.\tag{Eq:BG}\]
Proof. For each \(n\in \mathbb{N}\), let us define the following sequences of regions:
\[\begin{eqnarray*} K_n &=& \{(x,y) \mid x\geq 0, ~ y \geq 0, ~ x^2 + y^2 \leq n^2\},\\ D_n &=& \{(x,y) \mid 0 \leq x\leq n, ~ 0 \leq y \leq n\} \end{eqnarray*}\]
as approximating sequences of the first quadrant
\[D = \{(x,y) \mid x \geq 0, ~ y \geq 0\}.\]
We consider integrating the function
\[f(x,y) = 4e^{-x^2 -y^2}x^{2p-1}y^{2q-1}\]
on \(D\) using \(\{K_n\}\) and \(\{D_n\}\).

First, using \(\{K_n\}\) and polar coordinates,
\[\begin{eqnarray*} \iint_{K_n}f(x,y)dxdy &=& 4\int_0^{n}dr\int_0^{\frac{\pi}{2}}re^{-r^2}(r\cos\theta)^{2p-1}(r\sin\theta)^{2q-1}d\theta\\ &=&\left(2\int_0^ne^{-r^2}r^{2p+2q-1}dr\right)\left(2\int_0^{\frac{\pi}{2}}(\cos\theta)^{2p-1}(\sin\theta)^{2q-1}\,d\theta\right)\\ &=&\left(\int_0^ne^{-t}t^{p+q-1}dt\right)\left(2\int_0^{\frac{\pi}{2}}(\cos^{2}\theta)^{p-1}(\sin^{2}\theta)^{q-1}\cos\theta\sin\theta d\theta\right)\\ &\to&\Gamma(p+q)B(p,q). \end{eqnarray*}\]
Next, using \(\{D_n\}\),
\[\begin{eqnarray*} \iint_{D_n}f(x,y)dxdy &=& \left(2\int_0^ne^{-x^2}x^{2p-1}dx\right)\left(2\int_0^ne^{-y^2}y^{2q-1}dy\right)\\ &\to&\Gamma(p)\Gamma(q). \end{eqnarray*}\]
Thus, 
\[\Gamma(p+q)B(p,q) = \Gamma(p)\Gamma(q).\]
Dividing both sides by \(\Gamma(p+q) (\neq 0)\), we have the desired equality (Eq:BG). ■

Exercise. Prove
\[\int_0^{\frac{\pi}{2}}\sin^a\theta \cos^b\theta d\theta = \frac{\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{b+1}{2}\right)}{2\Gamma\left(\frac{a+b+2}{2}\right)}\]
for \(a > -1\) and \(b > -1\). □

Example. 
\[\begin{eqnarray*} \int_0^{\frac{\pi}{2}}\sin^5\theta \cos^6\theta d\theta &=& \frac{\Gamma\left(3\right)\Gamma\left(\frac{7}{2}\right)}{2\Gamma\left(\frac{13}{2}\right)}\\ &=&\frac{2!\cdot\frac{5!!}{2^3}\sqrt{\pi}}{\frac{11!!}{2^5}\sqrt{\pi}}\\ &=&\frac{2^3\cdot 5!!}{11!!}\\ &=& \frac{8}{11\cdot 9\cdot 7} = \frac{8}{693}. \end{eqnarray*}\]
□

Example. Let us express the following integral in terms of the gamma function:
\[\int_0^{\infty}\frac{x^{b-1}}{1 + x^a}dx\]
where \(a > 0\) and \(b > 0\).

Let \(t = \frac{1}{1 + x^a}\). Then \(x = \left(\frac{1 - t}{t}\right)^{\frac{1}{a}}\), \(dx = -\frac{1}{a}\left(\frac{1-t}{t}\right)^{\frac{1}{a} - 1}\frac{dt}{t^2}\).
\[\begin{eqnarray*} \int_0^{\infty}\frac{x^{b-1}}{1 + x^a}dx &=& -\frac{1}{a}\int_1^0t^{-\frac{b}{a}}(1 - t)^{\frac{b}{a}-1}dt\\ &=&\frac{1}{a}B\left(1 - \frac{b}{a}, \frac{b}{a}\right) = \frac{1}{a}\frac{\Gamma\left(1 - \frac{b}{a}\right)\Gamma\left(\frac{b}{a}\right)}{\Gamma(1)}\\ &=&\frac{1}{a}{\Gamma\left(1 - \frac{b}{a}\right)\Gamma\left(\frac{b}{a}\right)}. \end{eqnarray*}\]
□

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