177–182. Infinite integrals

177. Infinite Integrals.

The Integral Test of § 174 shows that, if $ϕ (x)$ is a positive and decreasing function of $x$ , then the series $\sum ϕ (n)$ is convergent or divergent according as the integral function $Φ (x)$ does or does not tend to a limit as $x \to \infty$ . Let us suppose that it does tend to a limit, and that $lim_{x \to \infty} \int_{1}^{x} ϕ (t) d t = l .$ Then we shall say that the integral $\int_{1}^{\infty} ϕ (t) d t$ is convergent, and has the value $l$ ; and we shall call the integral an infinite integral.

So far we have supposed $ϕ (t)$ positive and decreasing. But it is natural to extend our definition to other cases. Nor is there any special point in supposing the lower limit to be unity. We are accordingly led to formulate the following definition:

If $ϕ (t)$ is a function of $t$ continuous when $t \geq a$ , and $lim_{x \to \infty} \int_{a}^{x} ϕ (t) d t = l,$ then we shall say that the infinite integral $\begin{matrix} (1) & \int_{a}^{\infty} ϕ (t) d t \end{matrix}$ is convergent and has the value $l$ .

The ordinary integral between limits $a$ and $A$ , as defined in Ch. VII, we shall sometimes call in contrast a finite integral.

On the other hand, when $\int_{a}^{x} ϕ (t) d t \to \infty,$ we shall say that the integral diverges to $\infty$ , and we can give a similar definition of divergence to $- \infty$ . Finally, when none of these alternatives occur, we shall say that the integral oscillates, finitely or infinitely, as $x \to \infty$ .

These definitions suggest the following remarks.

(i) If we write $\int_{a}^{x} ϕ (t) d t = Φ (x),$ then the integral converges, diverges, or oscillates according as $Φ (x)$ tends to a limit, tends to $\infty$ (or to $- \infty$ ), or oscillates, as $x \to \infty$ . If $Φ (x)$ tends to a limit, which we may denote by $Φ (\infty)$ , then the value of the integral is $Φ (\infty)$ . More generally, if $Φ (x)$ is any integral function of $ϕ (x)$ , then the value of the integral is $Φ (\infty) - Φ (a)$ .

(ii) In the special case in which $ϕ (t)$ is always positive it is clear that $Φ (x)$ is an increasing function of $x$ . Hence the only alternatives are convergence and divergence to $\infty$ .

(iii) The integral (1) of course depends on $a$ , but is quite independent of $t$ , and is in no way altered by the substitution of any other letter for $t$ (cf. § 157).

(iv) Of course the reader will not be puzzled by the use of the term infinite integral to denote something which has a definite value such as $2$ or $\frac{1}{2} π$ . The distinction between an infinite integral and a finite integral is similar to that between an infinite series and a finite series: no one supposes that an infinite series is necessarily divergent.

(v) The integral $\int_{a}^{x} ϕ (t) d t$ was defined in § 156 and § 157 as a simple limit, i.e. the limit of a certain finite sum. The infinite integral is therefore the limit of a limit, or what is known as a repeated limit. The notion of the infinite integral is in fact essentially more complex than that of the finite integral, of which it is a development.

(vi) The Integral Test of § 174 may now be stated in the form:

if $ϕ (x)$ is positive and steadily decreases as $x$ increases, then the infinite series $\sum ϕ (n)$ and the infinite integral $\int_{1}^{\infty} ϕ (x) d x$ converge or diverge together.

(vii) The reader will find no difficulty in formulating and proving theorems for infinite integrals analogous to those stated in (1)–(6) of § 77. Thus the result analogous to is that

if $\int_{a}^{\infty} ϕ (x) d x$ is convergent, and $b > a$ , then $\int_{b}^{\infty} ϕ (x) d x$ is convergent and $\int_{a}^{\infty} ϕ (x) d x = \int_{a}^{b} ϕ (x) d x + \int_{b}^{\infty} ϕ (x) d x .$

178. The case in which $ϕ (x)$ is positive.

It is natural to consider what are the general theorems, concerning the convergence or divergence of the infinite integral (1) of § 177, analogous to theorems A–D of § 167. That A is true of integrals as well as of series we have already seen in § 177, (ii). Corresponding to B we have the theorem that

the necessary and sufficient condition for the convergence of the integral (1) is that it should be possible to find a constant $K$ such that $\int_{a}^{x} ϕ (t) d t < K$ for all values of $x$ greater than $a$ .

Similarly, corresponding to C, we have the theorem:

if $\int_{a}^{\infty} ϕ (x) d x$ is convergent, and $ψ (x) \leq K ϕ (x)$ for all values of $x$ greater than $a$ , then $\int_{a}^{\infty} ψ (x) d x$ is convergent and

$\int_{a}^{\infty} ψ (x) d x \leq K \int_{a}^{\infty} ϕ (x) d x .$ We leave it to the reader to formulate the corresponding test for divergence.

We may observe that d’Alembert’s test (§ 168), depending as it does on the notion of successive terms, has no analogue for integrals; and that the analogue of Cauchy’s test is not of much importance, and in any case could only be formulated when we have investigated in greater detail the theory of the function $ϕ (x) = r^{x}$ , as we shall do in Ch. IX. The most important special tests are obtained by comparison with the integral $\int_{a}^{\infty} \frac{d x}{x^{s}} (a > 0),$ whose convergence or divergence we have investigated in § 175, and are as follows:

if $ϕ (x) < K x^{- s}$ , where $s > 1$ , when $x \geq a$ , then $\int_{a}^{\infty} ϕ (x) d x$ is convergent; and if $ϕ (x) > K x^{- s}$ , where $s \leq 1$ , when $x \geq a$ , then the integral is divergent; and in particular, if $lim x^{s} ϕ (x) = l$ , where $l > 0$ , then the integral is convergent or divergent according as $s > 1$ or $s \leq 1$ .

There is one fundamental property of a convergent infinite series in regard to which the analogy between infinite series and infinite integrals breaks down. If $\sum ϕ (n)$ is convergent then $ϕ (n) \to 0$ ; but it is not always true, even when $ϕ (x)$ is always positive, that if $\int_{a}^{\infty} ϕ (x) d x$ is convergent then $ϕ (x) \to 0$ .

Consider for example the function $ϕ (x)$ whose graph is indicated by the thick line in the figure. Here the height of the peaks corresponding to the points $x = 1$ , $2$ , $3$ , … is in each case unity, and the breadth of the peak corresponding to $x = n$ is $2 / (n + 1)^{2}$ . The area of the peak is $1 / (n + 1)^{2}$ , and it is evident that, for any value of $ξ$ , $\int_{0}^{ξ} ϕ (x) d x < \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}},$ so that $\int_{0}^{\infty} ϕ (x) d x$ is convergent; but it is not true that $ϕ (x) \to 0$

Example LXXIII

1. The integral

\int_{a}^{\infty} \frac{α x^{r} + β x^{r - 1} + \dots + λ}{A x^{s} + B x^{s - 1} + \dots + L} d x,

where

α

and

A

are positive and

a

is greater than the greatest root of the denominator, is convergent if

s > r + 1

and otherwise divergent.

2. Which of the integrals $\int_{a}^{\infty} \frac{d x}{\sqrt{x}}$ , $\int_{a}^{\infty} \frac{d x}{x^{4 / 3}}$ , $\int_{a}^{\infty} \frac{d x}{c^{2} + x^{2}}, \int_{a}^{\infty} \frac{x d x}{c^{2} + x^{2}}, \int_{a}^{\infty} \frac{x^{2} d x}{c^{2} + x^{2}}, \int_{a}^{\infty} \frac{x^{2} d x}{α + 2 β x^{2} + γ x^{4}}$ are convergent? In the first two integrals it is supposed that $a > 0$ , and in the last that $a$ is greater than the greatest root (if any) of the denominator.

3. The integrals $\int_{a}^{ξ} \cos x d x, \int_{a}^{ξ} \sin x d x, \int_{a}^{ξ} \cos (α x + β) d x$ oscillate finitely as $ξ \to \infty$ .

4. The integrals $\int_{a}^{ξ} x \cos x d x, \int_{a}^{ξ} x^{2} \sin x d x \int_{a}^{ξ} x^{n} \cos (α x + β) d x,$ where $n$ is any positive integer, oscillate infinitely as $ξ \to \infty$ .

5. Integrals to $- \infty$ .If $\int_{ξ}^{a} ϕ (x) d x$ tends to a limit $l$ as $ξ \to - \infty$ , then we say that $\int_{- \infty}^{a} ϕ (x) d x$ is convergent and equal to $l$ . Such integrals possess properties in every respect analogous to those of the integrals discussed in the preceding sections: the reader will find no difficulty in formulating them.

6. Integrals from $- \infty$ to $+ \infty$ . If the integrals $\int_{- \infty}^{a} ϕ (x) d x, \int_{a}^{\infty} ϕ (x) d x$ are both convergent, and have the values $k$ , $l$ respectively, then we say that $\int_{- \infty}^{\infty} ϕ (x) d x$ is convergent and has the value $k + l$ .

7. Prove that $\int_{- \infty}^{0} \frac{d x}{1 + x^{2}} = \int_{0}^{\infty} \frac{d x}{1 + x^{2}} = \frac{1}{2} \int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}} = \frac{1}{2} π .$

8. Prove generally that $\int_{- \infty}^{\infty} ϕ (x^{2}) d x = 2 \int_{0}^{\infty} ϕ (x^{2}) d x,$ provided that the integral $\int_{0}^{\infty} ϕ (x^{2}) d x$ is convergent.

9. Prove that if $\int_{0}^{\infty} x ϕ (x^{2}) d x$ is convergent then $\int_{- \infty}^{\infty} x ϕ (x^{2}) d x = 0$ .

10. Analogue of Abel’s Theorem of § 173. If $ϕ (x)$ is positive and steadily decreases, and $\int_{a}^{\infty} ϕ (x) d x$ is convergent, then $x ϕ (x) \to 0$ . Prove this (a) by means of Abel’s Theorem and the Integral Test and (b) directly, by arguments analogous to those of § 173.

11. If $a = x_{0} < x_{1} < x_{2} < \dots$ and $x_{n} \to \infty$ , and $u_{n} = \int_{x_{n}}^{x_{n + 1}} ϕ (x) d x$ , then the convergence of $\int_{a}^{\infty} ϕ (x) d x$ involves that of $\sum u_{n}$ . If $ϕ (x)$ is always positive the converse statement is also true. [That the converse is not true in general is shown by the example in which $ϕ (x) = \cos x$ , $x_{n} = n π$ .]

179. Application to infinite integrals of the rules for substitution and integration by parts.

The rules for the transformation of a definite integral which were discussed in § 161 may be extended so as to apply to infinite integrals.

(1) Transformation by substitution. Suppose that $\begin{matrix} (1) & \int_{a}^{\infty} ϕ (x) d x \end{matrix}$ is convergent. Further suppose that, for any value of $ξ$ greater than $a$ , we have, as in § 161, $\begin{matrix} (2) & \int_{a}^{ξ} ϕ (x) d x = \int_{b}^{τ} ϕ {f (t)} f^{'} (t) d t, \end{matrix}$ where $a = f (b)$ , $ξ = f (τ)$ . Finally suppose that the functional relation $x = f (t)$ is such that $x \to \infty$ as $t \to \infty$ . Then, making $τ$ and so $ξ$ tend to $\infty$ in (2), we see that the integral $\begin{matrix} (3) & \int_{b}^{\infty} ϕ {f (t)} f^{'} (t) d t \end{matrix}$ is convergent and equal to the integral .

On the other hand it may happen that $ξ \to \infty$ as $τ \to - \infty$ or as $τ \to c$ . In the first case we obtain $\begin{aligned} 2 \int_{a}^{\infty} ϕ (x) d x & = & lim_{τ \to - \infty} \int_{b}^{τ} ϕ {f (t)} f^{'} (t) d t \\ = - & lim_{τ \to - \infty} \int_{τ}^{b} ϕ {f (t)} f^{'} (t) d t = - \int_{- \infty}^{b} ϕ {f (t)} f^{'} (t) d t . \end{aligned}$ In the second case we obtain $\begin{matrix} (4) & \int_{a}^{\infty} ϕ (x) d x = lim_{τ \to c} \int_{b}^{τ} ϕ {f (t)} f^{'} (t) d t . \end{matrix}$ We shall return to this equation in § 181.

There are of course corresponding results for the integrals $\int_{- \infty}^{a} ϕ (x) d x, \int_{- \infty}^{\infty} ϕ (x) d x,$ which it is not worth while to set out in detail: the reader will be able to formulate them for himself.

Example LXXIV

1. Show, by means of the substitution

x = t^{α}

, that if

s > 1

and

α > 0

then

\int_{1}^{\infty} x^{- s} d x = α \int_{1}^{\infty} t^{α (1 - s) - 1} d t;

and verify the result by calculating the value of each integral directly.

2. If $\int_{a}^{\infty} ϕ (x) d x$ is convergent then it is equal to one or other of $α \int_{(a - β) / α}^{\infty} ϕ (α t + β) d t, - α \int_{- \infty}^{(a - β) / α} ϕ (α t + β) d t,$ according as $α$ is positive or negative.

3. If $ϕ (x)$ is a positive and steadily decreasing function of $x$ , and $α$ and $β$ are any positive numbers, then the convergence of the series $\sum ϕ (n)$ implies and is implied by that of the series $\sum ϕ (α n + β)$ .

[It follows at once, on making the substitution

x = α t + β

, that the integrals

\int_{a}^{\infty} ϕ (x) d x, \int_{(a - β) / α}^{\infty} ϕ (α t + β) d t

converge or diverge together. Now use the Integral Test.]

4. Show that $\int_{1}^{\infty} \frac{d x}{(1 + x) \sqrt{x}} = \frac{1}{2} π .$

[Put

x = t^{2}

5. Show that $\int_{0}^{\infty} \frac{\sqrt{x}}{(1 + x)^{2}} d x = \frac{1}{2} π .$

[Put

x = t^{2}

and integrate by parts.]

6. If $ϕ (x) \to h$ as $x \to \infty$ , and $ϕ (x) \to k$ as $x \to - \infty$ , then $\int_{- \infty}^{\infty} {ϕ (x - a) - ϕ (x - b)} d x = - (a - b) (h - k) .$

[For

\begin{aligned} 2 \int_{- ξ^{'}}^{ξ} {ϕ (x - a) - ϕ (x - b)} d x & = \int_{- ξ^{'}}^{ξ} ϕ (x - a) d x & - \int_{- ξ^{'}}^{ξ} ϕ (x - b) d x \\ = \int_{- ξ^{'} - a}^{ξ - a} ϕ (t) d t & - \int_{- ξ^{'} - b}^{ξ - b} ϕ (t) d t \\ = \int_{- ξ^{'} - a}^{- ξ^{'} - b} ϕ (t) d t & - \int_{ξ - a}^{ξ - b} ϕ (t) d t . \end{aligned}

The first of these two integrals may be expressed in the form

(a - b) k + \int_{- ξ^{'} - a}^{- ξ^{'} - b} ρ d t,

where

ρ \to 0

ξ^{'} \to \infty

, and the modulus of the last integral is less than or equal to

| a - b | κ

, where

κ

is the greatest value of

ρ

throughout the interval

[- ξ^{'} - a, - ξ^{'} - b]

. Hence

\int_{- ξ^{'} - a}^{- ξ^{'} - b} ϕ (t) d t \to (a - b) k .

The second integral may be discussed similarly.]

(2) Integration by parts. The formula for integration by parts (§ 161) is $\int_{a}^{ξ} f (x) ϕ^{'} (x) d x = f (ξ) ϕ (ξ) - f (a) ϕ (a) - \int_{a}^{ξ} f^{'} (x) ϕ (x) d x .$

Suppose now that $ξ \to \infty$ . Then if any two of the three terms in the above equation which involve $ξ$ tend to limits, so does the third, and we obtain the result $\int_{a}^{\infty} f (x) ϕ^{'} (x) d x = lim_{ξ \to \infty} f (ξ) ϕ (ξ) - f (a) ϕ (a) - \int_{a}^{\infty} f^{'} (x) ϕ (x) d x .$ There are of course similar results for integrals to $- \infty$ , or from $- \infty$ to $\infty$ .

Example LXXV

Show that

\int_{0}^{\infty} \frac{x}{(1 + x)^{3}} d x = \frac{1}{2} \int_{0}^{\infty} \frac{d x}{(1 + x)^{2}} = \frac{1}{2} .

2. $\int_{0}^{\infty} \frac{x^{2}}{(1 + x)^{4}} d x = \frac{2}{3} \int_{0}^{\infty} \frac{x}{(1 + x)^{3}} d x = \frac{1}{3}$ .

3. If $m$ and $n$ are positive integers, and $I_{m, n} = \int_{0}^{\infty} \frac{x^{m} d x}{(1 + x)^{m + n}},$ then $I_{m, n} = {m / (m + n - 1)} I_{m - 1, n} .$ Hence prove that $I_{m, n} = m! (n - 2)! / (m + n - 1)!$ .

4. Show similarly that if $I_{m, n} = \int_{0}^{\infty} \frac{x^{2 m + 1} d x}{(1 + x^{2})^{m + n}}$ then $I_{m, n} = {m / (m + n - 1)} I_{m - 1, n}, 2 I_{m, n} = m! (n - 2)! / (m + n - 1)! .$ Verify the result by applying the substitution $x = t^{2}$ to the result of Ex. 3.

180. Other types of infinite integrals.

It was assumed, in the definition of the ordinary or finite integral given in Ch. VII, that (1) the range of integration is finite and (2) the subject of integration is continuous.

It is possible, however, to extend the notion of the ‘definite integral’ so as to apply to many cases in which these conditions are not satisfied. The ‘infinite’ integrals which we have discussed in the preceding sections, for example, differ from those of Ch. VII in that the range of integration is infinite. We shall now suppose that it is the second of the conditions (1), (2) that is not satisfied. It is natural to try to frame definitions applicable to some such cases at any rate. There is only one such case which we shall consider here. We shall suppose that $ϕ (x)$ is continuous throughout the range of integration $[a, A]$ except for a finite number of values of $x$ , say $x = ξ_{1}$ , $ξ_{2}$ , …, and that $ϕ (x) \to \infty$ or $ϕ (x) \to - \infty$ as $x$ tends to any of these exceptional values from either side.

It is evident that we need only consider the case in which $[a, A]$ contains one such point $ξ$ . When there is more than one such point we can divide up $[a, A]$ into a finite number of sub-intervals each of which contains only one; and, if the value of the integral over each of these sub-intervals has been defined, we can then define the integral over the whole interval as being the sum of the integrals over each sub-interval. Further, we can suppose that the one point $ξ$ in $[a, A]$ comes at one or other of the limits $a$ , $A$ . For, if it comes between $a$ and $A$ , we can then define $\int_{a}^{A} ϕ (x) d x$ as $\int_{a}^{ξ} ϕ (x) d x + \int_{ξ}^{A} ϕ (x) d x,$ assuming each of these integrals to have been satisfactorily defined. We shall suppose, then, that $ξ = a$ ; it is evident that the definitions to which we are led will apply, with trifling changes, to the case in which $ξ = A$ .

Let us then suppose $ϕ (x)$ to be continuous throughout $[a, A]$ except for $x = a$ , while $ϕ (x) \to \infty$ as $x \to a$ through values greater than $a$ . A typical example of such a function is given by $ϕ (x) = (x - a)^{- s},$ where $s > 0$ ; or, in particular, if $a = 0$ , by $ϕ (x) = x^{- s}$ . Let us therefore consider how we can define $\begin{matrix} (1) & \int_{0}^{A} \frac{d x}{x^{s}}, \end{matrix}$ when $s > 0$ .

The integral $\int_{1 / A}^{\infty} y^{s - 2} d y$ is convergent if $s < 1$ (§ 175) and means $lim_{η \to \infty} \int_{1 / A}^{η} y^{s - 2} d y$ . But if we make the substitution $y = 1 / x$ , we obtain $\int_{1 / A}^{η} y^{s - 2} d y = \int_{1 / η}^{A} x^{- s} d x .$ Thus $lim_{η \to \infty} \int_{1 / η}^{A} x^{- s} d x$ , or, what is the same thing, $lim_{ϵ \to + 0} \int_{ϵ}^{A} x^{- s} d x,$ exists provided that $s < 1$ ; and it is natural to define the value of the integral (1) as being equal to this limit. Similar considerations lead us to define $\int_{a}^{A} (x - a)^{- s} d x$ by the equation $\int_{a}^{A} (x - a)^{- s} d x = lim_{ϵ \to + 0} \int_{a + ϵ}^{A} (x - a)^{- s} d x .$

We are thus led to the following general definition:

if the integral $\int_{a + ϵ}^{A} ϕ (x) d x$ tends to a limit $l$ as $ϵ \to + 0$ , we shall say that the integral $\int_{a}^{A} ϕ (x) d x$ is convergent and has the value $l$ .

Similarly, when $ϕ (x) \to \infty$ as $x$ tends to the upper limit $A$ , we define $\int_{a}^{A} ϕ (x) d x$ as being $lim_{ϵ \to + 0} \int_{a}^{A - ϵ} ϕ (x) d x :$ and then, as we explained above, we can extend our definitions to cover the case in which the interval $[a, A]$ contains any finite number of infinities of $ϕ (x)$ .

An integral in which the subject of integration tends to $\infty$ or to $- \infty$ as $x$ tends to some value or values included in the range of integration will be called an infinite integral of the second kind: the first kind of infinite integrals being the class discussed in § 177 et seq. Nearly all the remarks (i)–(vii) made at the end of § 177 apply to infinite integrals of the second kind as well as to those of the first.

181.

We may now write the equation (4) of § 179 in the form $\begin{matrix} (1) & \int_{a}^{\infty} ϕ (x) d x = \int_{b}^{c} ϕ {f (t)} f^{'} (t) d t . \end{matrix}$ The integral on the right-hand side is defined as the limit, as $τ \to c$ , of the corresponding integral over the range $[b, τ]$ , i.e. as an infinite integral of the second kind. And when $ϕ {f (t)} f^{'} (t)$ has an infinity at $t = c$ the integral is essentially an infinite integral. Suppose for example, that $ϕ (x) = (1 + x)^{- m}$ , where $1 < m < 2$ , and $a = 0$ , and that $f (t) = t / (1 - t)$ . Then $b = 0$ , $c = 1$ , and becomes $\begin{matrix} (2) & \int_{0}^{\infty} \frac{d x}{(1 + x)^{m}} = \int_{0}^{1} (1 - t)^{m - 2} d t; \end{matrix}$ and the integral on the right-hand side is an infinite integral of the second kind.

On the other hand it may happen that $ϕ {f (t)} f^{'} (t)$ is continuous for $t = c$ . In this case $\int_{b}^{c} ϕ {f (t)} f^{'} (t) d t$ is a finite integral, and $lim_{τ \to c} \int_{b}^{τ} ϕ {f (t)} f^{'} (t) d t = \int_{b}^{c} ϕ {f (t)} f^{'} (t) d t,$ in virtue of the corollary to Theorem (10) of § 160. In this case the substitution $x = f (t)$ transforms an infinite into a finite integral. This case arises if $m \geq 2$ in the example considered a moment ago.

Example LXXVI

1. If

ϕ (x)

is continuous except for

x = a

, while

ϕ (x) \to \infty

x \to a

, then the necessary and sufficient condition that

\int_{a}^{A} ϕ (x) d x

should be convergent is that we can find a constant

K

such that

\int_{a + ϵ}^{A} ϕ (x) d x < K

for all values of

ϵ

, however small (cf. § 178).

It is clear that we can choose a number $A^{'}$ between $a$ and $A$ , such that $ϕ (x)$ is positive throughout $[a, A^{'}]$ . If $ϕ (x)$ is positive throughout the whole interval $[a, A]$ then we can of course identify $A^{'}$ and $A$ . Now $\int_{a - ϵ}^{A} ϕ (x) d x = \int_{a - ϵ}^{A^{'}} ϕ (x) d x + \int_{A^{'}}^{A} ϕ (x) d x .$ The first integral on the right-hand side of the above equation increases as $ϵ$ decreases, and therefore tends to a limit or to $\infty$ ; and the truth of the result stated becomes evident.

If the condition is not satisfied then $\int_{a - ϵ}^{A} ϕ (x) d x \to \infty$ . We shall then say that the integral $\int_{a}^{A} ϕ (x) d x$ diverges to $\infty$ . It is clear that, if $ϕ (x) \to \infty$ as $x \to a + 0$ , then convergence and divergence to $\infty$ are the only alternatives for the integral. We may discuss similarly the case in which $ϕ (x) \to - \infty$ .

2. Prove that $\int_{a}^{A} (x - a)^{- s} d x = \frac{(A - a)^{1 - s}}{1 - s}$ if $s < 1$ , while the integral is divergent if $s \geq 1$ .

3. If $ϕ (x) \to \infty$ as $x \to a + 0$ and $ϕ (x) < K (x - a)^{- s}$ , where $s < 1$ , then $\int_{a}^{A} ϕ (x) d x$ is convergent; and if $ϕ (x) > K (x - a)^{- s}$ , where $s \geq 1$ , then the integral is divergent. [This is merely a particular case of a general comparison theorem analogous to that stated in § 178.]

4. Are the integrals $\begin{matrix} \int_{a}^{A} \frac{d x}{\sqrt{(x - a) (A - x)}}, \int_{a}^{A} \frac{d x}{(A - x) \sqrt[3]{x - a}}, \int_{a}^{A} \frac{d x}{(A - x) \sqrt[3]{A - x}}, \\ \int_{a}^{A} \frac{d x}{\sqrt{x^{2} - a^{2}}}, \int_{a}^{A} \frac{d x}{\sqrt[3]{A^{3} - x^{3}}}, \int_{a}^{A} \frac{d x}{x^{2} - a^{2}}, \int_{a}^{A} \frac{d x}{A^{3} - x^{3}} \end{matrix}$ convergent or divergent?

5. The integrals $\int_{- 1}^{1} \frac{d x}{\sqrt[3]{x}}, \int_{a - 1}^{a + 1} \frac{d x}{\sqrt[3]{x - a}}$ are convergent, and the value of each is zero.

6. The integral $\int_{0}^{π} \frac{d x}{\sqrt{\sin x}}$ is convergent. [The subject of integration tends to $\infty$ as $x$ tends to either limit.]

7. The integral $\int_{0}^{π} \frac{d x}{(\sin x)^{s}}$ is convergent if and only if $s < 1$ .

8. The integral $\int_{0}^{\frac{1}{2} π} \frac{x^{s}}{(\sin x)^{t}} d x$ is convergent if $t < s + 1$ .

9. Show that $\int_{0}^{h} \frac{\sin x}{x^{p}} d x,$ where $h > 0$ , is convergent if $p < 2$ . Show also that, if $0 < p < 2$ , the integrals $\int_{0}^{π} \frac{\sin x}{x^{p}} d x, \int_{π}^{2 π} \frac{\sin x}{x^{p}} d x, \int_{2 π}^{3 π} \frac{\sin x}{x^{p}} d x, \dots$ alternate in sign and steadily decrease in absolute value. [Transform the integral whose limits are $k π$ and $(k + 1) π$ by the substitution $x = k π + y$ .]

10. Show that $\int_{0}^{h} \frac{\sin x}{x^{p}} d x,$ where $0 < p < 2$ , attains its greatest value when $h = π$ .

11. The integral $\int_{0}^{\frac{1}{2} π} (\cos x)^{l} (\sin x)^{m} d x$ is convergent if and only if $l > - 1$ , $m > - 1$ .

12. Such an integral as $\int_{0}^{\infty} \frac{x^{s - 1} d x}{1 + x},$ where $s < 1$ , does not fall directly under any of our previous definitions. For the range of integration is infinite and the subject of integration tends to $\infty$ as $x \to + 0$ . It is natural to define this integral as being equal to the sum $\int_{0}^{1} \frac{x^{s - 1} d x}{1 + x} + \int_{1}^{\infty} \frac{x^{s - 1} d x}{1 + x},$ provided that these two integrals are both convergent.

The first integral is a convergent infinite integral of the second kind if $0 < s < 1$ . The second is a convergent infinite integral of the first kind if $s < 1$ . It should be noted that when $s > 1$ the first integral is an ordinary finite integral; but then the second is divergent. Thus the integral from $0$ to $\infty$ is convergent if and only if $0 < s < 1$ .

13. Prove that $\int_{0}^{\infty} \frac{x^{s - 1}}{1 + x^{t}} d x$ is convergent if and only if $0 < s < t$ .

14. The integral $\int_{0}^{\infty} \frac{x^{s - 1} - x^{t - 1}}{1 - x} d x$ is convergent if and only if $0 < s < 1$ , $0 < t < 1$ . [It should be noticed that the subject of integration is undefined when $x = 1$ ; but $(x^{s - 1} - x^{t - 1}) / (1 - x) \to t - s$ as $x \to 1$ from either side; so that the subject of integration becomes a continuous function of $x$ if we assign to it the value $t - s$ when $x = 1$ .

It often happens that the subject of integration has a discontinuity which is due simply to a failure in its definition at a particular point in the range of integration, and can be removed by attaching a particular value to it at that point. In this case it is usual to suppose the definition of the subject of integration completed in this way. Thus the integrals $\int_{0}^{\frac{1}{2} π} \frac{\sin m x}{x} d x, \int_{0}^{\frac{1}{2} π} \frac{\sin m x}{\sin x} d x$ are ordinary finite integrals, if the subjects of integration are regarded as having the value $m$ when $x = 0$ .]

15. Substitution and integration by parts. The formulae for transformation by substitution and integration by parts may of course be extended to infinite integrals of the second as well as of the first kind. The reader should formulate the general theorems for himself, on the lines of § 179.

16. Prove by integration by parts that if $s > 0$ , $t > 1$ , then $\int_{0}^{1} x^{s - 1} (1 - x)^{t - 1} d x = \frac{t - 1}{s} \int_{0}^{1} x^{s} (1 - x)^{t - 2} d x .$

17. If $s > 0$ then $\int_{0}^{1} \frac{x^{s - 1} d x}{1 + x} = \int_{1}^{\infty} \frac{t^{- s} d t}{1 + t} .$

[Put

x = 1 / t

18. If $0 < s < 1$ then $\int_{0}^{1} \frac{x^{s - 1} + x^{- s}}{1 + x} d x = \int_{0}^{\infty} \frac{t^{- s} d t}{1 + t} = \int_{0}^{\infty} \frac{t^{s - 1} d t}{1 + t} .$

19. If $a + b > 0$ then $\int_{b}^{\infty} \frac{d x}{(x + a) \sqrt{x - b}} = \frac{π}{\sqrt{a + b}} .$

20. Show, by means of the substitution $x = t / (1 - t)$ , that if $l$ and $m$ are both positive then $\int_{0}^{\infty} \frac{x^{l - 1}}{(1 + x)^{l + m}} d x = \int_{0}^{1} t^{l - 1} (1 - t)^{m - 1} d t .$

21. Show, by means of the substitution $x = p t / (p + 1 - t)$ , that if $l$ , $m$ , and $p$ are all positive then $\int_{0}^{1} x^{l - 1} (1 - x)^{m - 1} \frac{d x}{(x + p)^{l + m}} = \frac{1}{(1 + p)^{l} p^{m}} \int_{0}^{1} t^{l - 1} (1 - t)^{m - 1} d t .$

22. Prove that $\int_{a}^{b} \frac{d x}{\sqrt{(x - a) (b - x)}} = π and \int_{a}^{b} \frac{x d x}{\sqrt{(x - a) (b - x)}} = \frac{1}{2} π (a + b),$ (i) by means of the substitution $x = a + (b - a) t^{2}$ , (ii) by means of the substitution $(b - x) / (x - a) = t$ , and (iii) by means of the substitution $x = a \cos^{2} t + b \sin^{2} t$ .

23. If $s > - 1$ then $\int_{0}^{\frac{1}{2} π} (\sin θ)^{s} d θ = \int_{0}^{1} \frac{x^{s} d x}{\sqrt{1 - x^{2}}} = \frac{1}{2} \int_{0}^{1} \frac{x^{\frac{1}{2} (s - 1)} d x}{\sqrt{1 - x}} = \frac{1}{2} \int_{0}^{1} (1 - x)^{\frac{1}{2} (s - 1)} \frac{d x}{\sqrt{x}} .$

24. Establish the formulae $\begin{aligned} \int_{0}^{1} \frac{f (x) d x}{\sqrt{1 - x^{2}}} = \int_{0}^{\frac{1}{2} π} f (\sin θ) d θ, \\ \int_{a}^{b} \frac{f (x) d x}{\sqrt{(x - a) (b - x)}} = 2 \int_{0}^{\frac{1}{2} π} f (a \cos^{2} θ + b \sin^{2} θ) d θ, \\ \int_{- a}^{a} f {\sqrt{\frac{a - x}{a + x}}} d x = 4 a \int_{0}^{\frac{1}{2} π} f (\tan θ) \cos θ \sin θ d θ . \end{aligned}$

25. Prove that $\int_{0}^{1} \frac{d x}{(1 + x) (2 + x) \sqrt{x (1 - x)}} = π (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}) .$

[Put

x = \sin^{2} θ

and use Ex. LXIII. 8.]

182.

Some care has occasionally to be exercised in applying the rule for transformation by substitution. The following example affords a good illustration of this.

Let $J = \int_{1}^{7} (x^{2} - 6 x + 13) d x .$ We find by direct integration that $J = 48$ . Now let us apply the substitution $y = x^{2} - 6 x + 13,$ which gives $x = 3 \pm \sqrt{y - 4}$ . Since $y = 8$ when $x = 1$ and $y = 20$ when $x = 7$ , we appear to be led to the result $J = \int_{8}^{20} y \frac{d x}{d y} d y = \pm \frac{1}{2} \int_{8}^{20} \frac{y d y}{\sqrt{y - 4}} .$ The indefinite integral is $\frac{1}{3} (y - 4)^{3 / 2} + 4 (y - 4)^{1 / 2},$ and so we obtain the value $\pm \frac{80}{3}$ , which is certainly wrong whichever sign we choose.

The explanation is to be found in a closer consideration of the relation between $x$ and $y$ . The function $x^{2} - 6 x + 13$ has a minimum for $x = 3$ , when $y = 4$ . As $x$ increases from $1$ to $3$ , $y$ decreases from $8$ to $4$ , and $d x / d y$ is negative, so that $\frac{d x}{d y} = - \frac{1}{2 \sqrt{y - 4}} .$ As $x$ increases from $3$ to $7$ , $y$ increases from $4$ to $20$ , and the other sign must be chosen. Thus $J = \int_{1}^{7} y d x = \int_{8}^{4} {- \frac{y}{2 \sqrt{y - 4}}} d y + \int_{4}^{20} \frac{y}{2 \sqrt{y - 4}} d y,$ a formula which will be found to lead to the correct result.

Similarly, if we transform the integral $\int_{0}^{π} d x = π$ by the substitution $x = \arcsin y$ , we must observe that $d x / d y = 1 / \sqrt{1 - y^{2}}$ or $d x / d y = - 1 / \sqrt{1 - y^{2}}$ according as $0 \leq x < \frac{1}{2} π$ or $\frac{1}{2} π < x \leq π$ .

Verify the results of transforming the integrals $\int_{0}^{1} (4 x^{2} - x + \frac{1}{16}) d x, \int_{0}^{π} \cos^{2} x d x$ by the substitutions $4 x^{2} - x + \frac{1}{16} = y$ , $x = \arcsin y$ respectively.

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176. Cauchy’s Condensation Test

Main Page

183. Series of positive and negative terms

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177. Infinite Integrals.

178. The case in which ϕ(x) is positive.

179. Application to infinite integrals of the rules for substitution and integration by parts.

180. Other types of infinite integrals.

181.

182.

178. The case in which $ϕ (x)$ is positive.