## 177. Infinite Integrals.

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

**.**

*infinite integral*So far we have supposed \(\phi(t)\) 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 \(\phi(t)\) is a function of \(t\) continuous when \(t \geq a\), and \[\lim_{x \to \infty} \int_{a}^{x} \phi(t)\, dt = l,\] then we shall say that the infinite integral \[\begin{equation*} \int_{a}^{\infty}\phi(t)\, dt \tag{1} \end{equation*}\] 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 \[\int_{a}^{x}\phi(t)\, dt \to \infty,\] 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 \[\int_{a}^{x}\phi(t)\, dt = \Phi(x),\] then the integral converges, diverges, or oscillates according as \(\Phi(x)\) tends to a limit, tends to \(\infty\) (or to \(-\infty\)), or oscillates, as \(x \to \infty\). If \(\Phi(x)\) tends to a limit, which we may denote by \(\Phi(\infty)\), then the value of the integral is \(\Phi(\infty)\). More generally, if \(\Phi(x)\) is any integral function of \(\phi(x)\), then the value of the integral is \(\Phi(\infty) – \Phi(a)\).

(ii) In the special case in which \(\phi(t)\) is always positive it is clear that \(\Phi(x)\) 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 \(\frac{1}{2}\pi\). 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 \(\int_{a}^{x} \phi(t)\, dt\) 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 \(\phi(x)\) is positive and steadily decreases as \(x\) increases, then the infinite series \(\sum\phi(n)\) and the infinite integral \(\int_{1}^{\infty} \phi(x)\, dx\) 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 \(\int_{a}^{\infty} \phi(x)\, dx\) is convergent, and \(b > a\), then \(\int_{b}^{\infty} \phi(x)\, dx\) is convergent and \[\int_{a}^{\infty} \phi(x)\, dx = \int_{a}^{b} \phi(x)\, dx + \int_{b}^{\infty}\phi(x)\, dx.\]

## 178. The case in which \(\phi(x)\) 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 \[\int_{a}^{x} \phi(t)\, dt < K\] for all values of \(x\) greater than \(a\).*

Similarly, corresponding to C, we have the theorem:

*if \(\int_{a}^{\infty} \phi(x)\, dx\) is convergent, and \(\psi(x) \leq K\phi(x)\) for all values of \(x\) greater than \(a\), then \(\int_{a}^{\infty} \psi(x)\, dx\) is convergent and*

*\[\int_{a}^{\infty} \psi(x)\, dx \leq K\int_{a}^{\infty} \phi(x)\, dx.\] 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 \(\phi(x) = r^{x}\), as we shall do in Ch. IX. The most important special tests are obtained by comparison with the integral \[\int_{a}^{\infty} \frac{dx}{x^{s}}\quad (a > 0),\] whose convergence or divergence we have investigated in § 175, and are as follows:

*if \(\phi(x) < Kx^{-s}\), where \(s > 1\), when \(x \geq a\), then \(\int_{a}^{\infty} \phi(x)\, dx\) is convergent; and if \(\phi(x) > Kx^{-s}\), where \(s \leq 1\), when \(x \geq a\), then the integral is divergent; and in particular, if \(\lim x^{s}\phi(x) = l\), 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 \(\sum \phi(n)\) is convergent then \(\phi(n) \to 0\); but it is *not* always true, even when \(\phi(x)\) is always positive, that if \(\int_{a}^{\infty} \phi(x)\, dx\) is convergent then \(\phi(x) \to 0\).

Consider for example the function \(\phi(x)\) 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 \(\xi\), \[\int_{0}^{\xi} \phi(x)\, dx < \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}},\] so that \(\int_{0}^{\infty} \phi(x)\, dx\) is convergent; but it is not true that \(\phi(x) \to 0\)

## 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 \[\begin{equation*} \int_{a}^{\infty} \phi(x)\, dx \tag{1} \end{equation*}\] is convergent. Further suppose that, for any value of \(\xi\) greater than \(a\), we have, as in § 161, \[\begin{equation*} \int_{a}^{\xi} \phi(x)\, dx = \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt, \tag{2} \end{equation*}\] where \(a = f(b)\), \(\xi = f(\tau)\). Finally suppose that the functional relation \(x = f(t)\) is such that \(x \to \infty\) as \(t \to \infty\). Then, making \(\tau\) and so \(\xi\) tend to \(\infty\) in (2), we see that the integral \[\begin{equation*} \int_{b}^{\infty} \phi\{f(t)\}f'(t)\, dt \tag{3} \end{equation*}\] is convergent and equal to the integral .

On the other hand it may happen that \(\xi \to \infty\) as \(\tau \to -\infty\) or as \(\tau \to c\). In the first case we obtain \[\begin{aligned} {2} \int_{a}^{\infty} \phi(x)\, dx &= &&\lim_{\tau\to-\infty} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt\\ &= -&&\lim_{\tau\to-\infty} \int_{\tau}^{b} \phi\{f(t)\}f'(t)\, dt = -\int_{-\infty}^{b} \phi\{f(t)\}f'(t)\, dt.\end{aligned}\] In the second case we obtain \[\begin{equation*} \int_{a}^{\infty} \phi(x)\, dx = \lim_{\tau\to c} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt. \tag{4} \end{equation*}\] We shall return to this equation in § 181.

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

(2) **Integration by parts.** The formula for integration by parts (§ 161) is \[\int_{a}^{\xi} f(x)\phi'(x)\, dx = f(\xi)\phi(\xi) – f(a)\phi(a) – \int_{a}^{\xi} f'(x)\phi(x)\, dx.\]

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

## 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 \(\phi(x)\) is continuous throughout the range of integration \({[a, A]}\) except for a finite number of values of \(x\), say \(x = \xi_{1}\), \(\xi_{2}\), …, and that \(\phi(x) \to \infty\) or \(\phi(x) \to -\infty\) 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 \(\xi\). 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 \(\xi\) 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 \(\int_{a}^{A} \phi(x)\, dx\) as \[\int_{a}^{\xi} \phi(x)\, dx + \int_{\xi}^{A} \phi(x)\, dx,\] assuming each of these integrals to have been satisfactorily defined. We shall suppose, then, that \(\xi = a\); it is evident that the definitions to which we are led will apply, with trifling changes, to the case in which \(\xi = A\).

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

The integral \(\int_{1/A}^{\infty} y^{s-2}\, dy\) is convergent if \(s < 1\) (§ 175) and means \(\lim\limits_{\eta\to\infty} \int_{1/A}^{\eta} y^{s-2}\, dy\). But if we make the substitution \(y = 1/x\), we obtain \[\int_{1/A}^{\eta} y^{s-2}\, dy = \int_{1/\eta}^{A} x^{-s}\, dx.\] Thus \(\lim\limits_{\eta\to\infty} \int_{1/\eta}^{A} x^{-s}\, dx\), or, what is the same thing, \[\lim_{\epsilon\to +0} \int_{\epsilon}^{A} x^{-s}\, dx,\] 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 \(\int_{a}^{A} (x – a)^{-s}\, dx\) by the equation \[\int_{a}^{A} (x – a)^{-s}\, dx = \lim_{\epsilon\to +0} \int_{a+\epsilon}^{A} (x – a)^{-s}\, dx.\]

We are thus led to the following general definition:

*if the integral \[\int_{a+\epsilon}^{A} \phi(x)\, dx\] tends to a limit \(l\) as \(\epsilon \to +0\), we shall say that the integral \[\int_{a}^{A} \phi(x)\, dx\] is convergent and has the value \(l\)*.

Similarly, when \(\phi(x) \to \infty\) as \(x\) tends to the upper limit \(A\), we define \(\int_{a}^{A} \phi(x)\, dx\) as being \[\lim_{\epsilon \to +0} \int_{a}^{A-\epsilon} \phi(x)\, dx:\] 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 \(\phi(x)\).

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 \[\begin{equation*} \int_{a}^{\infty} \phi(x)\, dx = \int_{b}^{c} \phi\{f(t)\}f'(t)\, dt. \tag{1} \end{equation*}\] The integral on the right-hand side is defined as the limit, as \(\tau \to c\), of the corresponding integral over the range \({[b, \tau]}\), *i.e.* as an infinite integral of the second kind. And when \(\phi\{f(t)\}f'(t)\) has an infinity at \(t = c\) the integral is essentially an infinite integral. Suppose for example, that \(\phi(x) = (1 + x)^{-m}\), where \(1 < m <2\), and \(a = 0\), and that \(f(t) = t/(1 – t)\). Then \(b = 0\), \(c = 1\), and becomes \[\begin{equation*} \int_{0}^{\infty} \frac{dx}{(1 + x)^{m}} = \int_{0}^{1} (1 – t)^{m-2}\, dt; \tag{2} \end{equation*}\] and the integral on the right-hand side is an infinite integral of the second kind.

On the other hand it may happen that \(\phi\{f(t)\}f'(t)\) is continuous for \(t = c\). In this case \[\int_{b}^{c} \phi\{f(t)\}f'(t)\, dt\] is a finite integral, and \[\lim_{\tau \to c} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt = \int_{b}^{c} \phi\{f(t)\}f'(t)\, dt,\] 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.

## 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 \[J = \int_{1}^{7} (x^{2} – 6x + 13)\, dx.\] We find by direct integration that \(J = 48\). Now let us apply the substitution \[y = x^{2} – 6x + 13,\] 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 \[J = \int_{8}^{20} y\frac{dx}{dy}\, dy = \pm\tfrac{1}{2}\int_{8}^{20} \frac{y\, dy}{\sqrt{y – 4}}.\] The indefinite integral is \[\tfrac{1}{3}(y – 4)^{3/2} + 4(y – 4)^{1/2},\] 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 \(x^{2} – 6x + 13\) has a minimum for \(x = 3\), when \(y = 4\). As \(x\) increases from \(1\) to \(3\), \(y\) decreases from \(8\) to \(4\), and \(dx/dy\) is negative, so that \[\frac{dx}{dy} = -\frac{1}{2\sqrt{y – 4}}.\] As \(x\) increases from \(3\) to \(7\), \(y\) increases from \(4\) to \(20\), and the other sign must be chosen. Thus \[J = \int_{1}^{7} y\, dx = \int_{8}^{4} \left\{-\frac{y}{2\sqrt{y – 4}}\right\} dy + \int_{4}^{20} \frac{y}{2\sqrt{y – 4}}\, dy,\] a formula which will be found to lead to the correct result.

Similarly, if we transform the integral \(\int_{0}^{\pi} dx = \pi\) by the substitution \(x = \arcsin y\), we must observe that \(dx/dy = 1/\sqrt{1 – y^{2}}\) or \(dx/dy = -1/\sqrt{1 – y^{2}}\) according as \(0 \leq x < \frac{1}{2}\pi\) or \(\frac{1}{2}\pi < x \leq \pi\).

Verify the results of transforming the integrals \[\int_{0}^{1} (4x^{2} – x + \tfrac{1}{16})\, dx,\quad \int_{0}^{\pi} \cos^{2}x\, dx\] by the substitutions \(4x^{2} – x + \frac{1}{16} = y\), \(x = \arcsin y\) respectively.

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