Improper multiple integrals

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 $n\ge N$ 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

$$I=\underset{n\to \mathrm{\infty }}{lim}{\iint }_{K_n}f(x,y)dxdy$$

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) $\{{\iint }_{K_n}f(x,y)dxdy\}$ 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 $f(x,y)\ge 0$ 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 $I=\underset{n\to \mathrm{\infty }}{lim}{I}_n$ exists. Then, for any approximating sequence $\{K_n^{\prime }\}$ of $D$ on which $f(x,y)$ is integrable, the limit $I^{\prime }=\underset{n\to \mathrm{\infty }}{lim}{\iint }_{K_n^{\prime }}f(x,y)dxdy$ exists and $I^{\prime }=I$.

Proof. Let us define

$$\begin{array}{rcl}{I}_n& =& {\iint }_{K_n}f(x,y)dxdy,\\ I_n^{\prime }& =& {\iint }_{K_n^{\prime }}f(x,y)dxdy.\end{array}$$

Since $f(x,y)$ is non-negative and $\{K_n\}$ and $\{K_n^{\prime }\}$ are increasing sequences, we have

$$\begin{array}{r}{I}_1\le I_2\le \cdots \le I_n\le I_{n+1}\le \cdots \\ I_1^{\prime }\le I_2^{\prime }\le \cdots \le I_n^{\prime }\le I_{n+1}^{\prime }\le \cdots \end{array}$$

That is, the sequences $\{I_n\}$ and $\{I_n^{\prime }\}$ are monotone increasing. Since $\{I_n\}$ converges to $I$, we have $I=\underset{n}{sup}{I}_n$ in particular.

For any $n\in \mathbb{N}$, we can choose (why?) a sufficiently large $m\in \mathbb{N}$ such that $K_n^{\prime }\subset K_m$. Thus, $I_n^{\prime }\le I_m\le I$, which implies that the sequence $\{I_n^{\prime }\}$ is bounded above. Since $\{I_n^{\prime }\}$ is also monotone increasing, the limit $I^{\prime }=\underset{n\to \mathrm{\infty }}{lim}{I}_n^{\prime }$ exists (Remember the monotone convergence theorem?). Since $I^{\prime }=\underset{n}{sup}{I}_n^{\prime }$, we have

$$\begin{array}{}\text{(Eq:I'leqI)}& I^{\prime }\le I.\end{array}$$

For any $n\in \mathbb{N}$, we can choose a sufficiently large $m\in \mathbb{N}$ such that $K_n\subset K_m^{\prime }$. Thus, $I_n\le I_m^{\prime }\le I^{\prime }$. Therefore we have

$$\begin{array}{}\text{(eq:IleqI')}& I\le I^{\prime }.\end{array}$$

By (Eq:I'leqI) and (Eq:IleqI'), we have $I=I^{\prime }$. 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 $f(x,y)\le 0$ 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 $I=\underset{n\to \mathrm{\infty }}{lim}{I}_n$ exists. Then, for any approximating sequence $\{K_n^{\prime }\}$ of $D$ on which $f(x,y)$ is integrable, the limit $I^{\prime }=\underset{n\to \mathrm{\infty }}{lim}{\iint }_{K_n^{\prime }}f(x,y)dxdy$ exists and $I^{\prime }=I$.

Proof. Let $g(x,y)=-f(x,y)$. Then we can apply the above Lemma. ■

Example. Given $D=\{(x,y)\mid x\ge 0,y\ge 0\}$, let us find the improper multiple integral

$${\iint }_D\frac{dxdy}{(1+x^2)(1+y^2)}.$$

First, let us define $K_n=\{(x,y)\mid 0\le x\le n,0\le y\le n\}$ for each $n\in \mathbb{N}$. Then, $\{K_n\}$ is an approximating sequence of $D$.

$$\begin{array}{rcl}{I}_n& =& {\iint }_{K_n}\frac{dxdy}{(1+x^2)(1+y^2)}\\ & =& {\int }_0^n\frac{dx}{1+x^2}{\int }_0^n\frac{dy}{1+y^2}\\ & =& {[\arctan x]}_0^n{[\arctan y]}_0^n\\ & =& (\arctan n{)}^2.\end{array}$$

Noting that $\underset{n\to \mathrm{\infty }}{lim}\arctan n=\frac{\pi }{2}$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{I}_n={\left(\frac{\pi }{2}\right)}^2=\frac{{\pi }^2}{4}.$$

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.

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }.$$

Let's prove this equality.

Since $e^{-x^2}$ is an even function, ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=2{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx$. To calculate the latter integral, we consider the following double integral

$$\begin{array}{}\text{(Eq:gaussian)}& {\iint }_D{e}^{-x^2-y^2}dxdy\text{ where }D=\{(x,y)\mid x\ge 0,y\ge 0\}.\end{array}$$

Let $K_n=\{(x,y)\mid x\ge 0,y\ge 0,x^2+y^2\le n^2\}$ be an approximating sequence of $D$. Using the polar coordinates, we have

$$\begin{array}{rcl}{I}_n& =& {\iint }_{K_n}{e}^{-x^2-y^2}dxdy\\ & =& {\int }_0^{n}r{e}^{-r^2}dr{\int }_0^{\frac{\pi }{2}}d\theta \\ & =& \frac{\pi }{2}{[-\frac{1}{2}{e}^{-r^2}]}_0^n\\ & =& \frac{\pi }{4}(1-e^{-n^2}).\end{array}$$

Since $\underset{n\to \mathrm{\infty }}{lim}{I}_n=\frac{\pi }{4}$, the integral in (Eq:gaussian) exists and is equal to $\frac{\pi }{4}$.

On the other hand, using another approximating sequence of $D$

$$D_n=\{(x,y)\mid 0\le x\le n,0\le y\le n\},$$

the same integral can be calculated as

$$\begin{array}{rcl}{J}_n& =& {\iint }_{D_n}{e}^{-x^2-y^2}dxdy={\iint }_{D_n}{e}^{-x^2}{e}^{-y^2}dxdy\\ & =& {\int }_0^n{e}^{-x^2}dx{\int }_0^n{e}^{-y^2}dy={\left({\int }_0^n{e}^{-x^2}dx\right)}^2.\end{array}$$

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

$${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2},$$

and hence,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }.$$

□

Example (Gamma function). The gamma function $\mathrm{\Gamma }(s)$ is defined as

$$\mathrm{\Gamma }(s)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{s-1}dx.$$

Thus,

$$\mathrm{\Gamma }\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-\frac{1}{2}}dx.$$

Let $x=t^2$, then

$$\mathrm{\Gamma }\left(\frac{1}{2}\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-\frac{1}{2}}dx=2{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt=\sqrt{\pi }.$$

□

Theorem ($\mathrm{\Gamma }(n-\frac{1}{2})$)

For any $n\in \mathbb{N}$,

$$\mathrm{\Gamma }(n-\frac{1}{2})=\frac{(2n-3)!!}{2^{n-1}}\sqrt{\pi }.$$

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

$$B(p,q)={\int }_0^1{x}^{p-1}(1-x{)}^{q-1}dx.$$

The following theorem connects the gamma function with the beta function.

Theorem

For all $p,q>0$,

$$\begin{array}{}\text{(Eq:BG)}& B(p,q)=\frac{\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)}{\mathrm{\Gamma }(p+q)}.\end{array}$$

Proof. For each $n\in \mathbb{N}$, let us define the following sequences of regions:

$$\begin{array}{rcl}{K}_n& =& \{(x,y)\mid x\ge 0,y\ge 0,x^2+y^2\le n^2\},\\ D_n& =& \{(x,y)\mid 0\le x\le n,0\le y\le n\}\end{array}$$

as approximating sequences of the first quadrant

$$D=\{(x,y)\mid x\ge 0,y\ge 0\}.$$

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,

$$\begin{array}{rcl}{\iint }_{K_n}f(x,y)dxdy& =& 4{\int }_0^{n}dr{\int }_0^{\frac{\pi }{2}}r{e}^{-r^2}(r\cos \theta {)}^{2p-1}(r\sin \theta {)}^{2q-1}d\theta \\ & =& \left(2{\int }_0^n{e}^{-r^2}{r}^{2p+2q-1}dr\right)(2{\int }_0^{\frac{\pi }{2}}(\cos \theta {)}^{2p-1}(\sin \theta {)}^{2q-1}d\theta )\\ & =& \left({\int }_0^n{e}^{-t}{t}^{p+q-1}dt\right)(2{\int }_0^{\frac{\pi }{2}}({\cos }^2\theta {)}^{p-1}({\sin }^2\theta {)}^{q-1}\cos \theta \sin \theta d\theta )\\ & \to & \mathrm{\Gamma }(p+q)B(p,q).\end{array}$$

Next, using $\{D_n\}$,

$$\begin{array}{rcl}{\iint }_{D_n}f(x,y)dxdy& =& \left(2{\int }_0^n{e}^{-x^2}{x}^{2p-1}dx\right)\left(2{\int }_0^n{e}^{-y^2}{y}^{2q-1}dy\right)\\ & \to & \mathrm{\Gamma }(p)\mathrm{\Gamma }(q).\end{array}$$

Thus, 

$$\mathrm{\Gamma }(p+q)B(p,q)=\mathrm{\Gamma }(p)\mathrm{\Gamma }(q).$$

Dividing both sides by $\mathrm{\Gamma }(p+q)(\ne 0)$, we have the desired equality (Eq:BG). ■

Exercise. Prove

$${\int }_0^{\frac{\pi }{2}}{\sin }^a\theta {\cos }^b\theta d\theta =\frac{\mathrm{\Gamma }\left(\frac{a+1}{2}\right)\mathrm{\Gamma }\left(\frac{b+1}{2}\right)}{2\mathrm{\Gamma }\left(\frac{a+b+2}{2}\right)}$$

for $a>-1$ and $b>-1$. □

Example. 

$$\begin{array}{rcl}{\int }_0^{\frac{\pi }{2}}{\sin }^5\theta {\cos }^6\theta d\theta & =& \frac{\mathrm{\Gamma }\left(3\right)\mathrm{\Gamma }\left(\frac{7}{2}\right)}{2\mathrm{\Gamma }\left(\frac{13}{2}\right)}\\ & =& \frac{2!\cdot \frac{5!!}{2^3}\sqrt{\pi }}{\frac{11!!}{2^5}\sqrt{\pi }}\\ & =& \frac{2^3\cdot 5!!}{11!!}\\ & =& \frac{8}{11\cdot 9\cdot 7}=\frac{8}{693}.\end{array}$$

□

Example. Let us express the following integral in terms of the gamma function:

$${\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{1+x^a}dx$$

where $a>0$ and $b>0$.

Let $t=\frac{1}{1+x^a}$. Then $x={\left(\frac{1-t}{t}\right)}^{\frac{1}{a}}$, $dx=-\frac{1}{a}{\left(\frac{1-t}{t}\right)}^{\frac{1}{a}-1}\frac{dt}{t^2}$.

$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x^{b-1}}{1+x^a}dx& =& -\frac{1}{a}{\int }_1^0{t}^{-\frac{b}{a}}(1-t{)}^{\frac{b}{a}-1}dt\\ & =& \frac{1}{a}B(1-\frac{b}{a},\frac{b}{a})=\frac{1}{a}\frac{\mathrm{\Gamma }(1-\frac{b}{a})\mathrm{\Gamma }\left(\frac{b}{a}\right)}{\mathrm{\Gamma }(1)}\\ & =& \frac{1}{a}\mathrm{\Gamma }(1-\frac{b}{a})\mathrm{\Gamma }\left(\frac{b}{a}\right).\end{array}$$

□