Page:EB1911 - Volume 10.djvu/776

Rh hence the sine series is For general values of x, the series represents the ordinates of the row of broken lines in fig. 3. 3.

The cosine series, which represents the same function for the interval 0 to l, may be found to be This series represents for general values of x the ordinate of the set of broken lines in fig. 4.

4.

Dirichlet’s Integral.—The method indicated by Fourier, but first carried out rigorously by Dirichlet, of proving that, with certain restrictions as to the nature of the function ƒ(x), that function is in general represented by the series (3), consists in finding the sum of n+1 terms of that series, and then investigating the limiting value of the sum, when n is increased indefinitely. It thus appears that the series is convergent, and that the value towards which its sum converges is {ƒ(x + 0) + ƒ(x − 0)}, which is in general equal to ƒ(x). It will be convenient throughout to take − to as the given interval; any interval −l to l may be reduced to this by changing x into lx/, and thus there is no loss of generality.

We find by an elementary process that Hence, with the new notation, the sum of the first n+1 terms of (3) is If we suppose ƒ(x) to be continued beyond the interval − to, in such a way that ƒ(x) = ƒ(x + 2), we may replace the limits in this integral by x + , x − respectively; if we then put x′ − x = 2z, and let ƒ(x′) = F(z), the expression becomes $$\frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} F(z)\frac{\text{sin } mz}{\text{sin } z} dz$$, where $$m = 2n + 1$$; this expression may be written in the form

We require therefore to find the limiting value, when m is indefinitely increased, of $$\int_{0}^{\frac{\pi}{2}} F(z)\frac{\text{sin } mz}{\text{sin } z} dz$$; the form of the second integral being essentially the same. This integral, or rather the slightly more general one $$\int_{0}^{h} F(z)\frac{\text{sin } mz}{\text{sin } z} dz$$, when 0 < h ≦, is known as Dirichlet’s integral. If we write $$X(z) = F(z) \frac{z}{\text{sin } z}$$, the integral becomes $$\int_{0}^{h} X(z)\frac{\text{sin } mz}{\text{sin } z} dz$$, which is the form in which the integral is frequently considered.

The Second Mean-Value Theorem.—The limiting value of Dirichlet’s integral may be conveniently investigated by means of a theorem in the integral calculus known as the second mean-value theorem. Let a, b be two fixed finite numbers such that a < b, and suppose ƒ(x), (x) are two functions which have finite and determinate values everywhere in the interval except for a finite number of points; suppose further that the functions ƒ(x), (x) are integrable throughout the interval, and that as x increases from a to b the function ƒ(x) is monotone, i.e. either never diminishes or never increases; the theorem is that $\int_{a}^{b}f(x)\phi(x)dx=f(a+0)\int_{a}^{\xi}\phi (x)dx+f(b-0) \int_{\xi}^{b}\phi (x)dx$

when is some point between a and b, and ƒ(a), ƒ(b) may be written for ƒ(a + 0), ƒ(b − 0) unless a or b is a point of discontinuity of the function ƒ(x).

To prove this theorem, we observe that, since the product of two integrable functions is an integrable function, ∫ b a ƒ(x) (x) dx exists, and may be regarded as the limit of the sum of a series

ƒ(x0) (x0) (x1 − x0) + ƒ(x1) (x1) (x2 − x1) +. . . + ƒ(xn−1) (xn−1) (xn − xn−1)

where x0 = a, xn = b and x1, x2. . . xn−1 are n − 1 intermediate points. We can express (xr) (xr+1 − xr) in the form Yr+1 − Yr, by putting

Yr = K=r K=1  (x K−1 ) (x K − x K−1 ), Y0 = 0.

Writing Xr for ƒ(xr), the series becomes

X0 (Y1 − Y0) + X1 (Y2 − Y1) + ... + Xn−1 (Yn − Yn−1)

or

Y1 (X0 − X1) + Y2 (X1 − X2) + ... + Yn(Xn−1 − Xn) + YnXn.

Now, by supposition, all the numbers Y1, Y2 ... Yn are finite, and all the numbers X r−1 − Xr are of the same sign, hence by a known algebraical theorem the series is equal to M (X0 − Xn) + YnXn, where M is a number intermediate between the greatest and the least of the numbers Y1, Y2, ... Yn. This remains true however many partial intervals are taken, and therefore, when their number is increased indefinitely, and their breadths are diminished indefinitely according to any law, we have $$\int_{a}^{b} f(x) \phi (x)dx=\{f(a)-f(b)\} \bar \text{M} + f(b) \int_{a}^{b} \phi (x)dx$$

when M is intermediate between the greatest and least values which $$\int_{a}^{x} \phi (x)dx$$ can have, when x is in the given integral. Now this integral is a continuous function of its upper limit x, and therefore there is a value of x in the interval, for which it takes any particular value between the greatest and least values that it has. There is therefore a value between a and b, such that $$\bar \text{M} = \int_{a}^{\xi}\phi (x)dx$$, hence

$$\int_{a}^{b}f(x)\phi(x)dx=\{f(a) - f(b)\} \int_{a}^{\xi} \phi(x)dx+f(b) \int_{a}^{b}\phi(x)dx = f(a)\int_{a}^{\xi} \phi(x)dx+f(b)\int_{\xi}^{b}\phi(x)dx.$$ If the interval contains any finite numbers of points of discontinuity of ƒ(x) or (x), the method of proof still holds good, provided these points are avoided in making the subdivisions; in particular if either of the ends be a point of discontinuity of ƒ(x), we write ƒ(a + 0) or ƒ(b − 0), for ƒ(a) or ƒ(b), it being assumed that these limits exist.

Functions, with Limited Variation.—The condition that ƒ(x), in the mean-value theorem, either never increases or never diminishes as x increases from a to b, places a restriction upon the applications of the theorem. We can, however, show that a function ƒ(x) which is finite and continuous between a and b, except for a finite number of ordinary discontinuities, and which only changes from increasing to diminishing or vice versa, a finite number of times, as x increases from a to b, may be expressed as the difference of two functions ƒ1(x), ƒ2(x), neither of which ever diminishes as x passes from a to b, and that these functions are finite and continuous, except that one or both of them are discontinuous at the points where the given function is discontinuous. Let, be two consecutive points at which ƒ(x) is discontinuous, consider any point x1, such that  ≦ x1 ≦ , and suppose that at the points M1, M2 ... Mr between and x1, ƒ(x) is a maximum, and at m1, m2 ... mr, it is a minimum; we will suppose, for example, that the ascending order of values is, M1, m1, M2, m2 ... Mr, mr, x1; it will make no essential difference in the argument if m1 comes before M1, or if Mr immediately precedes x1, Mr−1 being then the last minimum.

Let

(x1) = [ƒ(M1) − ƒ( + 0)] + [ƒ(M2) − ƒ(m1)] + ... + [ƒ(Mr) − ƒ(m r−1 )] + [ƒ(x1) − ƒ(mr)];

now let (x1) increase until it reaches the value (Mr+1) at which ƒ(x) is again a maximum, then let

(x1) = [ƒ(M1) − ƒ( + 0)] + [ƒ(M2) − ƒ(m1)] +. . . + [ƒ(Mr) − ƒ(mr−1)] + [ƒ(Mr+1) − ƒ(mr)];

and suppose as x increases beyond the value Mr+1, (x1) remains constant until the next minimum mr+1 is reached, when it again becomes variable; we see that (x1) is essentially positive and never diminishes as x increases.

Let

(x1) = [ƒ(M1) − f(m1)] + [ƒ(M2) − ƒ(m1)] + ... + [ƒ(Mr) − ƒ(mr)],

then let x1 increase until it is beyond the next maximum Mr+1, and then let

(x1) = [ƒ(M1) − ƒ(m1)] + [ƒ(M2) − ƒ(m1)] +. . . + [ƒ(Mr) − ƒ(mr)] + [ƒ(Mr+1) − ƒ(x1)]

thus (x1) never diminishes, and is alternately constant and variable. We see that (x1) − (x1) is continuous as x1 increases from to , and that (x1) − (x1) = ƒ(x1) − ƒ( + 0), and when x1 reaches, we have − (x1) = ƒ( − 0) − ƒ( + 0). Hence it is seen that between and , ƒ(x) = [(x) + ƒ( + 0)] − (x), where (x) + ƒ( + 0), (x) are continuous and never diminish as x increases; the same reasoning