]> Beyond Riemann

Chapter 2: Integration; Section 2: Definite Integrals; page 3

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Beyond Riemann

Improper integral of the first kind

We already know that according to Riemann: if a function  f(x)  is defined in an interval of  x = [a,b] , bounded and continuous except at a countable number of points, the definite integral of the function for this interval exists.

We are now going to discuss the generalization of this definition beyond Riemann, called the improper integral of the first kind, in the following cases:

1. extending the interval of  x  up to  [a,$\infty$)
2. or to  (-$\infty$,b]
3. and even to  (-$\infty$,$\infty$)
by keeping all the other requirements of Riemann.

The first case is defined by

 $\begin{array}{l}\text{if}\text{\hspace{0.28em}}\text{for}\text{\hspace{0.28em}}\text{any}\text{ }v>a\text{ }\text{exists}\\ I\left(v\right)=\underset{a}{\overset{v}{\int }}f\left(x\right)\text{d}x\text{ }\text{and}\\ \underset{v\to \infty }{\mathrm{lim}}I\left(v\right)\text{ }\text{converges,}\text{\hspace{0.28em}}\text{then}\\ \underset{a}{\overset{\infty }{\int }}f\left(x\right)\text{d}x=\underset{v\to \infty }{\mathrm{lim}}\underset{a}{\overset{v}{\int }}f\left(x\right)\text{d}x\end{array}\right\}$ (2.2.3.1)

which without all the semantics, we simply write

 $\underset{a}{\overset{\infty }{\int }}f\left(x\right)\text{d}x=\underset{v\to \infty }{\mathrm{lim}}\underset{a}{\overset{v}{\int }}f\left(x\right)\text{d}x$ (2.2.3.2)

The second case is very similar. Again without semantics:

 $\underset{-\infty }{\overset{b}{\int }}f\left(x\right)\text{d}x=\underset{u\to -\infty }{\mathrm{lim}}\underset{u}{\overset{b}{\int }}f\left(x\right)\text{d}x$ (2.2.3.3)

As an example of the first case we'll take

 $\underset{0}{\overset{\infty }{\int }}\mathrm{exp}\left(-x\right)\text{d}x=\underset{v\to \infty }{\mathrm{lim}}\left[\mathrm{exp}\left(0\right)-\mathrm{exp}\left(-v\right)\right]=1$ (2.2.3.4)

but

 $\underset{0}{\overset{\infty }{\int }}\mathrm{exp}\left(x\right)\text{d}x=\underset{v\to \infty }{\mathrm{lim}}\left[\mathrm{exp}\left(v\right)-\mathrm{exp}\left(0\right)\right]\text{ }\text{diverges}$ (2.2.3.5)

For case 2 a very similar example will be

 $\underset{-\infty }{\overset{0}{\int }}\mathrm{exp}\left(x\right)\text{d}x=\underset{u\to -\infty }{\mathrm{lim}}\left[\mathrm{exp}\left(0\right)-\mathrm{exp}\left(u\right)\right]=1$ (2.2.3.6)

In case 3 an integral will be convergent, if it fulfills the requirements of the two cases together according to

 $\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)\text{d}x=\underset{u\to -\infty }{\mathrm{lim}}\underset{u}{\overset{{x}_{0}}{\int }}f\left(x\right)\text{d}x+\underset{v\to \infty }{\mathrm{lim}}\underset{{x}_{0}}{\overset{v}{\int }}f\left(x\right)\text{d}x$ (2.2.3.7)

where  xo  is an arbitrary finite value of  x .

Let's take an example of this case:

 $\begin{array}{l}\underset{-\infty }{\overset{\infty }{\int }}\frac{\text{d}x}{1+{x}^{2}}=\underset{u\to -\infty }{\mathrm{lim}}\underset{u}{\overset{0}{\int }}\frac{\text{d}x}{1+{x}^{2}}+\underset{v\to \infty }{\mathrm{lim}}\underset{0}{\overset{v}{\int }}\frac{\text{d}x}{1+{x}^{2}}=\\ =\underset{u\to -\infty }{\mathrm{lim}}\text{atan}x\underset{u}{\overset{0}{|}}+\underset{v\to \infty }{\mathrm{lim}}\text{atan}x\underset{0}{\overset{v}{|}}=\\ =0-\underset{u\to -\infty }{\mathrm{lim}}\text{atan}u+\underset{v\to \infty }{\mathrm{lim}}\text{atan}v-0=\\ =-\left(-\frac{\pi }{2}\right)+\frac{\pi }{2}=\pi \end{array}\right\}$ (2.2.3.8)

The two examples from (2.2.3.4) and (2.2.3.8) are also shown graphically in Fig. Improper integral - first kind.

Since the integrand of (2.2.3.8) is an even function, if the improper integral of case 1 or 2 converges, then it also converges for case 3. In this example we can obtain the same result in a different way:

 $\begin{array}{l}\underset{-\infty }{\overset{\infty }{\int }}\frac{\text{d}x}{1+{x}^{2}}=\underset{u\to \infty }{\mathrm{lim}}\left(\underset{-u}{\overset{0}{\int }}\frac{\text{d}x}{1+{x}^{2}}+\underset{0}{\overset{u}{\int }}\frac{\text{d}x}{1+{x}^{2}}\right)=\\ =\underset{u\to \infty }{\mathrm{lim}}\underset{-u}{\overset{u}{\int }}\frac{\text{d}x}{1+{x}^{2}}=\underset{u\to \infty }{\mathrm{lim}}\left[\text{atan}u-\text{atan}\left(-u\right)\right]=\\ =\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)=\pi \end{array}\right\}$ (2.2.3.9)

but this is correct only if the limits of the two integrals exist separately.

The following example shows why. Let's take any odd function integrable for any closed interval of  x , e.g.  sinx . We obtain that

 $\underset{u\to \infty }{\mathrm{lim}}\left(\underset{-u}{\overset{0}{\int }}\mathrm{sin}x\text{d}x+\underset{0}{\overset{u}{\int }}\mathrm{sin}x\text{d}x\right)\equiv 0$ (2.2.3.10)

meaning a value which is identically zero for any value of  u . On the other hand the separate limits of each one of the terms diverges.

In case that an integral exists for any closed interval, but cannot be expressed by elementary functions, one has to apply tests, in order to check if the appropriate improper integral of the first kind exists. We are only going to see one of them: the comparison test, which is presented here, and is only for positive functions and case 1, but with appropriate modifications can be used also in different circumstances.

Comparison test for convergence

 $\begin{array}{l}\text{if}\text{ }0\le f\left(x\right)\le g\left(x\right)\text{ }\text{for}\text{ }x\ge a\\ \text{and}\text{ }\underset{a}{\overset{\infty }{\int }}g\left(x\right)\text{d}x\text{ }\text{converges}\\ \text{then}\text{ }\underset{a}{\overset{\infty }{\int }}f\left(x\right)\text{d}x\text{ }\text{also}\text{\hspace{0.28em}}\text{converges}\end{array}\right\}$ (2.2.3.11)

Let's take the following example of the Gaussian, mentioned in the previous section (exercise 2), which the indefinite integral cannot be expressed in terms of the elementary functions.

 $\begin{array}{l}0\le \mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\le \mathrm{exp}\left(-x\right)\text{ }\text{for}\text{ }x\ge 2\\ \text{and}\text{ }\underset{2}{\overset{\infty }{\int }}\mathrm{exp}\left(-x\right)\text{d}x\text{ }\text{converges}\\ \text{then}\text{ }\underset{2}{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}}{2}\right)\text{d}x\text{ }\text{also}\text{\hspace{0.28em}}\text{converges}\end{array}\right\}$ (2.2.3.12)

Graphical presentation of (2.2.3.12) is shown in Fig. Comparison test.

For completness, here is the comparison test for divergence

 $\begin{array}{l}\text{if}\text{ }f\left(x\right)\ge g\left(x\right)\ge 0\text{ }\text{for}\text{ }x\ge a\\ \text{and}\text{ }\underset{a}{\overset{\infty }{\int }}g\left(x\right)\text{d}x\text{ }\text{diverges}\\ \text{then}\text{ }\underset{a}{\overset{\infty }{\int }}f\left(x\right)\text{d}x\text{ }\text{also}\text{\hspace{0.28em}}\text{diverges}\end{array}\right\}$ (2.2.3.13)

Improper integral of the second kind

A point where a function is unbounded is called a singular point. The improper integral of the second kind extends the definition of definite integrals to the case, where singular points are included in the interval of integration.

We are now going to discuss the improper integral in an interval  $a\le x\le b\text{\hspace{0.28em}}\left(a with only one singular point, but with all the other Riemann's requirements satisfied in the following cases:

1. a  is singular.
2. b  is singular.
3. x0  satisfying  $a<{x}_{0} , is singular.

For the definitions of these cases we will use the notation  $\epsilon \to {0}_{+}$ , meaning a positive  ε  converging to zero.

The definition of the improper integral of case 1, with a singular point of the function at  x = a  is

 $\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x=\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\underset{a+\epsilon }{\overset{b}{\int }}f\left(x\right)\text{d}x$ (2.2.3.14)

if the limit of (2.2.3.14) exists.

For example we can use

 $\begin{array}{l}f\left(x\right)={x}^{-\frac{1}{2}}\text{ }\text{for}\text{ }0\le x\le b\\ \underset{0}{\overset{b}{\int }}{x}^{-\frac{1}{2}}\text{d}x=\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\underset{0+\epsilon }{\overset{b}{\int }}{x}^{-\frac{1}{2}}\text{d}x=\\ =\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\left(2\sqrt{x}\right)\underset{0+\epsilon }{\overset{b}{|}}=2\sqrt{b}\end{array}\right\}$ (2.2.3.15)

Notice that the square-root used in (2.2.3.15) is single valued (positive), in order to perform the integration.

Another example on the other hand

 $\begin{array}{l}f\left(x\right)={x}^{-1}\text{ }\text{for}\text{ }0\le x\le b\\ \underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\underset{0+\epsilon }{\overset{b}{\int }}{x}^{-1}\text{d}x=\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\left(\mathrm{ln}x\right)\underset{0+\epsilon }{\overset{b}{|}}\end{array}\right\}$ (2.2.3.16)

diverges to $\infty$.

The definition of the improper integral of case 2, with a singular point of the function at  x = b  is

 $\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x=\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\underset{a}{\overset{b-\epsilon }{\int }}f\left(x\right)\text{d}x$ (2.2.3.17)

We will take an example that the integral diverges and the improper integral does not exist:

 $\begin{array}{l}f\left(x\right)=\mathrm{tan}x\text{ }\text{for}\text{ }0\le x\le \frac{\pi }{2}\\ \underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\underset{0}{\overset{\frac{\pi }{2}-\epsilon }{\int }}\mathrm{tan}x\text{d}x=-\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\underset{0}{\overset{\frac{\pi }{2}-\epsilon }{\int }}\frac{\text{d}\left(\mathrm{cos}x\right)}{\mathrm{cos}x}=\\ =-\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\left[\mathrm{ln}\left(\mathrm{cos}x\right)\right]\underset{0}{\overset{\frac{\pi }{2}-\epsilon }{|}}\to \infty \end{array}\right\}$ (2.2.3.18)

The definition of the improper integral of case 3, with a singular point of the function at  x0  with  $a<{x}_{0}  is

 $\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x=\underset{{\epsilon }_{1}\to {0}_{+}}{\mathrm{lim}}\underset{a}{\overset{{x}_{0}-{\epsilon }_{1}}{\int }}f\left(x\right)\text{d}x+\underset{{\epsilon }_{2}\to {0}_{+}}{\mathrm{lim}}\underset{{x}_{0}+{\epsilon }_{2}}{\overset{b}{\int }}f\left(x\right)\text{d}x$ (2.2.3.19)

One should notice that the definition of case 3, as it is expressed by (2.2.3.19), requires two independent convergences for the same singular point.

Following is an extension of the example (2.2.3.15) to the negative values of x, in order to have the singular point at  x = 0, inside the interval of integration:

 $\begin{array}{l}f\left(x\right)={|x|}^{-\frac{1}{2}}\text{ }\text{for}\text{ }a<0 (2.2.3.20)

The result of (2.2.3.20) was obtained by following the procedure of (2.2.3.19). We could obtain the same result by converging the two expressions by a single limit:

 $\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x=\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\left[\underset{a}{\overset{{x}_{0}-\epsilon }{\int }}f\left(x\right)\text{d}x+\underset{{x}_{0}+\epsilon }{\overset{b}{\int }}f\left(x\right)\text{d}x\right]$ (2.2.3.21)

but this can only be done, after we know that the improper integral exists.

A comparison test for convergence (divergence) of the second kind improper integrals is defined similarly as it is for the first kind.

In a more general case, there could be more than one singular point in an interval of integration. In such a case, one has to apply the criterium for the improper integral to each singular point separetely.

In the case of an improper integral of the first kind (unbounded interval), and at the same time of the second kind (unbounded function inside the interval), the integral is called improper of the third kind.

Cauchy principal value

If we return to (2.2.3.21), could it happen that the limit exists, while the improper integral does not? As matter of fact, yes! In such a case the limit is called the Cauchy principal value of the integral appearing on the left of (2.2.3.21).

One such example is the function  $f\left(x\right)={x}^{-1}$  used for (2.2.3.16). The singular point is at  x = 0  and the function is defined for negative, as well as positive, values of x. Let's calculate the Cauchy principal value for the integral of an interval including  x = 0 .

 $\begin{array}{l}f\left(x\right)={x}^{-1}\text{ }\text{for}\text{ }a<0\text{ }\text{end}\text{ }b>0\\ \text{the Cauchy principal value of}\\ \underset{a}{\overset{b}{\int }}{x}^{-1}\text{d}x=\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\left[\underset{a}{\overset{-\epsilon }{\int }}\frac{\text{d}x}{x}+\underset{\epsilon }{\overset{b}{\int }}\frac{\text{d}x}{x}\right]=\\ =\underset{\epsilon \to {0}_{+}}{\mathrm{lim}}\left[\mathrm{ln}\epsilon -\mathrm{ln}|a|+\mathrm{ln}b-\mathrm{ln}\epsilon \right]=\\ =\mathrm{ln}b-\mathrm{ln}|a|\end{array}\right\}$ (2.2.3.22)

The following examples of improper integrals of the second kind:

• $f\left(x\right)={x}^{-\frac{1}{2}}$  with relation to (2.2.3.15)
• $f\left(x\right)={x}^{-1}$  with relation to (2.2.3.16) and (2.2.3.22)
are presented graphically in Fig. Improper integral - second kind.

Mean value and improper integrals

The definition of the mean value can also be extended to improper integrals. In order to see this, we are going to discuss an example from physics, which will also show some manipulations with improper integrals. The example used is radioactive decay.

Radioactive decay was introduced in the previous section, and can be summarized in (2.2.3.23)

 $\left\{\begin{array}{l}\text{differential}\text{\hspace{0.28em}}\text{equation}\text{ }\frac{\text{d}N}{\text{d}t}=-\frac{N}{\tau }\\ \text{solution}\text{ }\text{ }N={N}_{0}\mathrm{exp}\left(-\frac{t}{\tau }\right)\end{array}\right\}$ (2.2.3.23)

where

• N  is the number of atoms that did not decay till the time  t
• N0  is the corresponding number for time zero
• τ  is a positive constant of the decay, with dimensions of time

According to the definition of the mean value

 $〈T〉=\frac{\underset{a}{\overset{b}{\int }}T\text{d}x}{\underset{a}{\overset{b}{\int }}\text{d}x}$ (2.2.3.24)

where

• <T>  is the mean value of the variable  T
• with respect to the independent variable  x
• in the interval  [a,b]

It is legitimate to ask for the mean value of  N  with respect to time for a time interval  [0,b] .

 $〈N〉=\frac{\underset{0}{\overset{b}{\int }}N\text{d}t}{\underset{0}{\overset{b}{\int }}\text{d}t}$ (2.2.3.25)

But if the interval is unlimited  ($b\to \infty$)  the numerator of (2.2.3.25) remains finite, on the other hand the denominator diverges to infinity, and the mean value becomes zero. This does not mean that the mean value always vanishes for an infinite inteval. If for example  N  remains constant (does not decay), then the numerator also diverges to infinity and the limit of the ratio is finite  (<N> = N) .

A more interesting question is what is the mean lifetime of an radioactive atom. If we divide the time in equal intervals  Δt , then the number of decaying particles in an interval, at time  t  is equal to the corresponding  ΔN . Therefore, the required relation for the mean lifetime is

 $〈t〉=\frac{\underset{0}{\overset{{N}_{0}}{\int }}t\text{d}N}{\underset{0}{\overset{{N}_{0}}{\int }}\text{d}N}$ (2.2.3.26)

By using  N  from (2.2.3.23) one obtains

 $\begin{array}{l}\underset{0}{\overset{{N}_{0}}{\int }}t\text{d}N=-\frac{{N}_{0}}{\tau }\underset{\infty }{\overset{0}{\int }}t\mathrm{exp}\left(-\frac{t}{\tau }\right)\text{d}t\\ \text{after}\text{\hspace{0.28em}}\text{substitution}\text{\hspace{0.28em}}\text{of}\text{ }u=\frac{t}{\tau }\\ \underset{0}{\overset{{N}_{0}}{\int }}t\text{d}N={N}_{0}\tau \underset{0}{\overset{\infty }{\int }}u\mathrm{exp}\left(-u\right)\text{d}u=\\ =-{N}_{0}\tau \left[\left(u+1\right)\mathrm{exp}\left(-u\right)\right]\underset{0}{\overset{\infty }{|}}=\\ =-{N}_{0}\tau \left[0-1\right]={N}_{0}\tau \end{array}\right\}$ (2.2.3.27)

where L'Hôpital rule was used for evaluating the the improper integral of the first kind. Finally from (2.2.3.26) and (2.2.3.27), we obtain

 $〈t〉=\tau$ (2.2.3.28)

We could solve (2.2.3.27) by substituting  N  as a variable of the integration. This is not new, but in this case the improper integral will become of the second kind.

 $\begin{array}{l}\text{from}\text{\hspace{0.28em}}\left(2.2.3.23\right)\text{ }t=-\tau \mathrm{ln}\left(\frac{N}{{N}_{0}}\right)\\ \text{by substituting}\text{ }v=\left(\frac{N}{{N}_{0}}\right)\\ \underset{0}{\overset{{N}_{0}}{\int }}t\text{d}N=-\tau \underset{0}{\overset{{N}_{0}}{\int }}\mathrm{ln}\left(\frac{N}{{N}_{0}}\right)\text{d}N=\\ =-\tau {N}_{0}\underset{0}{\overset{1}{\int }}\mathrm{ln}v\text{d}v=-\tau {N}_{0}\left(v\mathrm{ln}v-v\right)\underset{0}{\overset{1}{|}}=\\ =-\tau {N}_{0}\left[-1-\underset{v\to 0}{\mathrm{lim}}\left(v\mathrm{ln}v\right)\right]=\tau {N}_{0}\end{array}\right\}$ (2.2.3.29)

and we conclude with the same result (2.2.3.28).

In case (2.2.3.27), the integration was performed on the unbounded independent variable  t , and therefore we had an improper integral of the first kind. In case (2.2.3.29 ) the integration is done on the bounded independent variable  N , but the dependent variable is unbounded, and therefore we have improper integral of the second kind.

Here is another interesting fact. In this particular case the following relation holds

 $\underset{0}{\overset{\infty }{\int }}N\text{d}t=\underset{0}{\overset{{N}_{0}}{\int }}t\text{d}N$ (2.2.3.30)

Indeed

 $\begin{array}{l}\underset{0}{\overset{\infty }{\int }}N\text{d}t={N}_{0}\underset{0}{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{t}{\tau }\right)\text{d}t\\ \text{after}\text{\hspace{0.28em}}\text{substitution}\text{\hspace{0.28em}}\text{of}\text{ }u=\frac{t}{\tau }\\ \underset{0}{\overset{\infty }{\int }}N\text{d}t={N}_{0}\tau \underset{0}{\overset{\infty }{\int }}\mathrm{exp}\left(-u\right)\text{d}u=\\ ={N}_{0}\tau \left[-\mathrm{exp}\left(-u\right)\right]\underset{0}{\overset{\infty }{|}}={N}_{0}\tau \end{array}\right\}$ (2.2.3.31)

gives the same result as (2.2.3.29). This fact is used in some textbooks dealing with random decays, as this one.

The graphical presentation related to the mean-time of radioactive decay is shown in Fig. Radioactive decay.

Polar coordinates

So far the Riemann's integral was used with Cartesian coordinates. Without introducing new concepts, we are going to extend its meaning for use with polar coordinates.

The polar coordinates were introduced and used in the previous chapter for expressing mathematically some of the geometrical curves. Their relations with the Cartesian coordinates (1.2.5.29-30) are given here as (2.2.3.32-33) for reference.

 $\left\{\begin{array}{l}x=r\mathrm{cos}\phi \\ y=r\mathrm{sin}\phi \end{array}\right\}$ (2.2.3.32)
 $\left\{\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\\ \mathrm{cos}\phi =\frac{x}{r}\\ \mathrm{sin}\phi =\frac{y}{r}\end{array}\right\}$ (2.2.3.33)

A curve given in polar coordinates is usually expressed as

 $r=r\left(\phi \right)$ (2.2.3.34)

We want to obtain the area between a segment of the curve limited by the two end end points of the angle and between two straight lines from the origin with lengths  r1  and  r2 , corresponding to these end points. The origin and the end points define a triangle with area

 $A=\frac{{r}_{1}{r}_{2}}{2}\mathrm{sin}\left({\phi }_{2}-{\phi }_{1}\right)$ (2.2.3.35)

which is graphically explained in Fig. Area of a triangle.

Let's divide the angular interval in small intervals of  Δ , and add up the corresponding areas of the triangles, in the limit Δ→0 , this sum should give the required area. For each small  Δ  the corresponding product   ${r}_{1}{r}_{2}\to {r}^{2}$   and  sin(Δ)→Δ , since   $\underset{\Delta \phi \to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\Delta \phi \right)}{\Delta \phi }=1$   as we already know. Finally the required area is

 $\text{area}\text{\hspace{0.17em}}=\frac{1}{2}\underset{{\phi }_{1}}{\overset{{\phi }_{2}}{\int }}{r}^{2}\left(\phi \right)\text{d}\phi$ (2.2.3.36)

Let's take an example. We will use the parabola

 $2y=1-{x}^{2}$ (2.2.3.37)

which after the application of (2.2.3.32) becomes

 $\begin{array}{l}2r\mathrm{sin}\phi =1-{\left(r\mathrm{cos}\phi \right)}^{2}\\ \text{yielding}\text{\hspace{0.28em}}\text{for}\text{ }r\ge 0\\ r=\frac{1}{1+\mathrm{sin}\phi }\end{array}\right\}$ (2.2.3.38)

The following integrations using different coordinate systems, give the same result as expected.

 $\begin{array}{l}\underset{0}{\overset{1}{\int }}y\text{d}x=\frac{1}{2}\underset{0}{\overset{1}{\int }}\left(1-{x}^{2}\right)\text{d}x=\frac{1}{3}\\ \frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}{r}^{2}\text{d}\phi =\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\text{d}\phi }{{\left(1+\mathrm{sin}\phi \right)}^{2}}=\frac{1}{3}\end{array}\right\}$ (2.2.3.39)

The calculations of (2.2.3.38-39) are not given here in detail, but left as simple exercises for the user.

A graphical illustration of the integration with polar coordinates, based on this example, is presented in Fig. Area with polar coordinates.

From (2.2.3.36) it follows that the integration yields a positive sign if it is done from lower to higher  φ , meaning in a counter-clockwise sense. In the case of an area including the origin and enclosed by a continuous loop, one can simply do the integration, starting at any  φ  and finishing with  φ+2π .

As an example we are going to calculate the area of an ellipse with radii   $a\text{\hspace{0.28em}}\text{and}\text{\hspace{0.28em}}b\text{\hspace{0.28em}}\left(a\ge b\right)$ . We have already obtained in a previous exercise that this area is πab. In the previous chapter we found that the expression

 $\begin{array}{l}r=\frac{a\left(1-{\beta }^{2}\right)}{\left(1+\beta \mathrm{cos}\phi \right)}\\ \beta =\sqrt{1-\frac{{b}^{2}}{{a}^{2}}}\end{array}\right\}$ (2.2.3.40)

corresponds to the ellipse with a focus at the origin. The area of the ellipse calculated by the use of (2.2.3.40) is

 $\begin{array}{l}A=\frac{1}{2}\underset{0}{\overset{2\pi }{\int }}{r}^{2}\text{d}\phi =\frac{{a}^{2}{\left(1-{\beta }^{2}\right)}^{2}}{2}\underset{0}{\overset{2\pi }{\int }}\frac{\text{d}\phi }{{\left(1+\beta \mathrm{cos}\phi \right)}^{2}}=\\ =\frac{{a}^{2}{\left(1-{\beta }^{2}\right)}^{2}}{2}\frac{2\pi }{{\left(1-{\beta }^{2}\right)}^{\frac{3}{2}}}=\pi {a}^{2}\sqrt{1-{\beta }^{2}}=\pi ab\end{array}\right\}$ (2.2.3.41)

The solution of the definite integral from (2.2.3.41) can be found in mathematical tables.

Using (2.2.3.35) for the calculation of an area enclosed by a continuous loop is not restricted only for the case when the origin is in this area. One has to take care to perform the integration counter-clockwise along the loop. The parts of the integration, where the coordinate  φ  is diminishing, yields a negative value, which is essential for obtaining the correct value of the enclosed area. This is very similar to the use of Cartesian coordinates for enclosed areas, as discussed in the previous page.

Exercises

Exercise 1. Calculate the improper integral $\underset{0}{\overset{\infty }{\int }}f\left(x\right)\text{d}x$ if it exists, or show that it does not, in the following cases:

1. $f\left(x\right)=\mathrm{cosh}x-\mathrm{sinh}x$
2. $f\left(x\right)={\mathrm{cosh}}^{2}x-{\mathrm{sinh}}^{2}x$

Exercise 2. Calculate the improper integral if it exists, othewise show that it does not.

$\underset{0}{\overset{1}{\int }}\text{atanh}x\text{ }\text{d}x$

Exercise 3. The domain of the function

$f\left(x\right)={x}^{-p}\text{ }\text{where}\text{\hspace{0.28em}}\text{the constant}\text{ }p>0$

covers all non-negative values of x.

For which values of  p do the following improper integrals exist (if at all) and what are they? Examine each case separetely!

1. $\underset{a}{\overset{\infty }{\int }}f\left(x\right)\text{d}x\text{ }\text{where}\text{ }a>0$

2. $\underset{0}{\overset{a}{\int }}f\left(x\right)\text{d}x\text{ }\text{where}\text{ }a>0$

3. $\underset{0}{\overset{\infty }{\int }}f\left(x\right)\text{d}x$

Exercise 4.

1. Does the improper integral $\underset{0}{\overset{\frac{3\pi }{4}}{\int }}\mathrm{tan}x\text{ }\text{d}x$ exist? If yes, what is its value?
2. What is its Cauchy principal value?
3. Does the integral $\underset{0}{\overset{\frac{3\pi }{4}}{\int }}|\mathrm{tan}x|\text{d}x$ have a finite Cauchy principal value?

Exercise 5. The expression in polar coordinates of a closed curve is

$r=a{\mathrm{cos}}^{2}\phi +b{\mathrm{sin}}^{2}\phi$

where $a\ge b>0$ are constants.

1. The curve is similar to the canonical form of an ellipse with radii a and b. Show that this is not an ellipse (except in the trivial case of a circle)!
2. Obtain the enclosed area!

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