Page images

which integrals have already been evaluated in (97) and (98), Art. 98.

Again multiplying (140) by V-1, and adding (139); and replacing the circular functions by their exponential equivalents, we have

[ocr errors]
[ocr errors]


Let a = 0; in which case a = 5: then we have

lam-lebxv=1 dx = emr(m); (142) This definite integral is of a form which will hereafter be considered at length.

27-1 For another application of (135) let f'(x) = 1 no, where 2 > m > 1; -also let' xn = 0, yn= b; wo = y = 0; so that f'(z) does not become infinite for any value of its subject-variable within the range of integration, and f'(o) = f'(0) = 0; consequently from (135), 10 am-1dx

. (ax)m-1 : (a+b =1)

-d(ax) Jo 1 + ax + bxv-1 JO 1+ ax

= sin ma, by (72), Art. 94.

sin ma

[ocr errors]
[ocr errors]
[ocr errors]
[blocks in formation]

105.] The preceding process can always be safely employed, so long as f'(z) does not become infinite or discontinuous within the limits of integration. If however f'(2) does become infinite for particular values of x and y within the range of integration, say when x = , and y=n; then the circumstance requires close examination ; and to simplify the inquiry I will take that particular form of z given in Art. 103 ; viz.,

z = x+yV1; in which case f'(&+n V-1) = 0.

Let the definite integral be divided into two parts with the following limits; for the former part let the range of the x-integration extend from X, to f-i; and for the latter part from f+i to Xw, i being an infinitesimal according to the theory of Art. 89; the range of the y-integration being in both parts from y, to yo. Thus the part of the integral corresponding to the infinite value of the element-function will be excluded ; but ultimately if i = 0, the whole value will be included. Under these circumstances the integral given in (127) consists of the two following integrals: {'(x+ynV – 1)—f'(x+yov – 1)} dx

= v=1/" {(E=i+yN=1)F"(x+y=1)} dy ; 186+y»v=13–8(6+%ov –1)} dx

=v=1/"{$(x+y N=1)f"(E+i+yV –1)} dy ; Now let these two integrals be added; then if i = 0, the sum of the left-hand members is

.* {f'(x+ywv=1)f'(x+9oV])} d«, and is determinate, because the range of integration does not include the values of the variables for which the element-function is infinite. Also the sum of the right-hand members is V-1/"{$"(xn+y1=1)—f'(x+y— 1)} dy

- V-1/""{8(6+i+y1 –1)fi+y1 =1)}dy, of which the former part is determinate; the latter part is not

[ocr errors]


[ocr errors]

so; it would be equal to 0, if i = 0, were it not that within the range of the y-integration y takes the value n, when x has that of $; and consequently the variables have those values for which the element-function is infinite. This latter part may therefore be finite, and must be determined in each case. Thus the definite integral found in the ordinary way is to be diminished by this quantity ; let us denote it by a; so that A = V-1/" {f ($+i+yV–1)-f(x-i+yv — 1)}dy. (147)

This quantity is called the correction for infinity or for discontinuity; and its value may be determined as follows.

(1) When n, the value of y for which the element-function becomes infinite, is between Yn and yo, so that the preceding definite integral includes values of the element-function on both sides of that value; then as for all other values of y when i = 0, the whole function under the sign of integration in (147) is zero, the limits may be extended from Yn and yo to too and - 00 without any change of value of the integral, so that A = V

($+i+yV-1)-f'(8-i+yV-1)}dy. (148) (2) If n = yo, the inferior limit, only those values of the element-function which lie on the positive side of y=n= y, are included in the definite integral; so that all those lying beyond that limit, and up to infinity, may be included without change of value of the integral : in this case A= v=1 / {f(E+i+y1=1)—f'(4-i+y V-1)} dy. (149)

(3) If n = Yn, the superior limit, only those values of the element-function which lie on the negative side of y=n= yn are included in the definite integral: in this case the lower limit of integration may be extended to -00 without any change of value of the integral; and we have a = V-1/" {f(E+i+y V-1)-f(8-i+y1 – 1)}dy. (150)

I may observe in passing that the correctness of this change of limits in the value of a given in (147), without any change of value in the integral, may also be demonstrated by a transformation of variable in (147), by replacing y by n+iu, where i is an infinitesimal, and u is the variable. I have preferred however the general reasoning given above.

These values of a may be further simplified. Since

f'(x+yV-1) = 0, when x = & and y = n, f'(x) = oo when z = $+nv -1 = $, say: Now the factor 2–5 may enter into f'(z) in any power: I shall assume that it enters in only the first power, leaving the student to refer to the original memoirs of Cauchy for the more complicated case; and accordingly I shall suppose (2-08'(z) to be finite when z = 5 Let (2-05'(z) = (z).

(151) Then, omitting the limits of a, which will be hereafter supplied according to (148) or (149) or (150),

y (+i+yN-1) (f-i+yV-1)
Mli(y-n) V-1 -+ (y-n) V

"{dy. (152)

II As this integral vanishes, when i =, for all values of y except those near to ne *(&ti+n V-1) = *($-i+n V-1) = y($+n V-1).

= E (say); (153) and we may place this quantity outside the sign of integration : so that

1 A = EN

li + (y-n) V-1 - i+(y-n) V

[ocr errors]
[ocr errors][merged small]

Hence if the value of a is that given in (148), the limits of integration are too and —00; and A = 27E-1;

(155) and if A has the value given in (149) or in (150), so that the limits are so and n, or in and -00, then A = TEV -1;

(156) thus if the value of y, which makes the element-function equal to infinity, is a limit of the y-integration, the correction for infinity or for discontinuity is only one, half of its value when this is not the case.

In the preceding inquiry we have investigated the circumstances and the necessary correction when the element-function



becomes infinite at given values of its subject-variables. This may occur however many times within the range of integration; and in this case the correction must be introduced for each set of variables. Thus, suppose f'(z)= oo, when x = $1, y=ni; x = {z, y = ng;... X = Ero y = Nn; and the corrections corresponding to these values and determined as above to be A1, A2, ... An: then the whole correction for infinity or discontinuity is the sum of these several corrections; and if E1, Eg, ... E, correspond to them, the whole correction

a = 297–1{E,+Eg+...+ En}
= 27-1z.E;

(157) remembering always to diminish any one of these partial corrections by one half, if the variable corresponding to it is a limit of the y-integration.

Hence, finally, we have *{(x+y.V~])=f(x+yov ])} dx = V-1/""{f'(xn+yV-1)-f'(x+yV-1)} dy

-2711 .E. (158) 106.] The preceding equation is too general for our present purpose ; and I propose to take certain particular cases of it, from which definite integrals may be determined.

Let Xn = too, Xo = -00; Yn = +00, y = 0; so that the range of the x-integration includes all positive and negative values of x, and that of the y-integration includes all positive but no negative values of y: and thus in seeking the values of x and y, viz., & and , for which f'(x+yV-1) = 0, $ may be negative as well as positive, but n must always be positive.

Let us moreover suppose that f'(x +yv - 1) = 0, when x = +0, whatever is the value of y; and = 0, when y = +0, whatever is the value of x. Then under all these circumstances, we have from (158)

2) dx = 27 V-12.E.

(159) The following are examples of this formula. po $(x)

dx. Here, when x is replaced by x +yV-1, J- 1 + x2

[ocr errors]
[ocr errors]
« PreviousContinue »