Hence if the following integral includes all positive values of the variables within limits of integration assigned by the inequality x +y +2 + ...... <1, (69) then 1= ///... F(x, y, z,...)...dzdydr poopoo sint - F(x, y, ...) cos(x + y +...)t...dy dxdt. (70) To Jo Jo Jo *** This may also be reduced to a form more convenient for the evaluation of the several separate integrations by means of the Gamma-function. Since cos kt is the real part of e-kv-, (70) may be expressed in the form, I = the real part 2 po poros sint of F(x,y....Je-(x+v+...)IV-1 ... dy dx dt. (71) 2 porco de I This last equation, as also (70), states the theorem discovered by Dirichlet for the evaluation of multiple definite integrals. One of the most useful applications of it is to a problem in attractions which occurs in a future part of this treatise. The following examples, which refer to subjects already discussed, are sufficient to exhibit its application. 284.] Ex. 1. Let us take a case of three variables, and assume F(x, y, z) = XP-179-12-12-(x+v+2); and let the limits be assigned by the inequality x+y+z < 1; then |-*(1-x x P-1 y9-12"-10-(2+v+*) dz dy dx = the real part I O 2r(p) sin t dt | XP-1e-ax dx = the real part of og Jo t(a+tv – 1) therefore the real part of To sin t dt tP+q+r-le-al dt; (73) Jo t(a+tV-1)P+9+ 2r(p+9+r) Jo substituting which in (72), mi ri-xi-*- y (1 7-*"70–199-129-12-(x+u+) dz dy dx _(p)r (9) r (-) ({p+e+r-1e-a dt ; (74) r(p+9+r) Jo and thus the triple integral in the left-hand member is expressed in terms of the single integral in the right-hand member. This result is evidently a particular form of (53). If a = 0, 777-***22-149–12r-1dz dy dx = _ r(p)r(9) r(r) (75) r(P+9+9+1) which is the same result as (23), if a, in (23), = 1. Ex. 2. Hereby also we may find the volume of the ellipsoid; and this example is a good illustration of the mode in which these functions are to be treated. In this case the inequality which assigns the limits is . .. v="6. _ 4 7 abc And thus the whole volume = 3 285.] An extension may be given to the process of reduction of a multiple integral by the employment of a discontinuous factor of a more general form than that hitherto introduced. For such a factor is suggested by Fourier's theorem, which is stated in (143), Art. 204; that particular theorem however not being sufficient for the purpose, because limits of a more general form are required. Now if 1 and u are positive quantities, a being greater than M, and if f(t) is finite and continuous for all values of t over the range .-M, then 2 po f(t) cos ku cos tudt du = 1, or = 0, 7 f (k) Jo In according as k falls within or beyond the range 1 - M.] Let = "cos ku du ' 's() cos tu dt -sor"(e) dt ), sin egen Now by the theorem given in Ex. 4, Art. 100, 1* sin ma cos ni də = , = 1; = 0, Jox according as n is less than, is equal to, or is greater than m. Hence applying this theorem to the terms of (82), we have three several cases ; (1) Let k < M < 1, and consequently less than every value of t involved in the last term of (82); so that 1 = f(1) =f(x) - | *f"(e) de (2) Let u<t<n; now the values of the first two terms of (82) are evident; but as t in the last integral ranges from u to 1, and as k falls between and 1, it will be convenient to divide this integral into two parts, as follows; Jo me according as k falls within or beyond the range .-M. In the same manner it may be shewn that f(t) sin ku sin tu dt du = f(K), or = 0, according as k falls within or beyond the range .—M. Hence we have generally 21Sf(t) cos ku cos tu dt du = f(t) sin ku sin t u dt du = f(k), or = 0, $ (84) according as k falls within or beyond the range .-H. In the preceding investigation I have not calculated the values of the definite integral when k=d, and when k = u, because such values are evidently finite, and in the use which will be made of the function each will give only one value of the infinitesimal element; and the omission of this element will not vitiate the definite integral. If, in (84), 1 = 0 and u = 0, the theorem takes the form of Fourier's theorem as given in (145), Art. 204. Also in the first part of (84), if f(t) = 1,1 = 1, u = 0, we have 2 po sin u cos ku, dou du = l, or = 0, according as k is less than or is greater than 1. This is the result already given in (68) of the present chapter, and is the factor of discontinuity which has been heretofore employed. 286.] I propose however now to take the more general factor of discontinuity which is given in the first part of (84); whereby we have ? 17°f(t) cos ku cos tu dt du = f(k), or = 0,Į (85) according as k falls within or beyond the range .-H. J The following are examples in which this factor is employed. I have taken integrals of only two or three variables, so that the formulæ may be shorter; and as the process is the same in all cases, these will exemplify it quite as well as those in which more variables are employed. Let an integral involve three variables, and be of the form where a, b, c are positive quantities, or positive in at least the real parts of them, and where the limits of integration are given by the inequality, u <x+y+z<A. (87) For the sake of abbreviation, let x+y+z = k; and by means of (85) let us replace F (k) in (86) by which = F(k) for all values of k within the range i l; that is, within the range given by the inequality (87); and = 0, for all values of k outside of that range. Now this discontinuous equivalent of F (k) is the real part of = | F(t)e-kuv-l cosutdt du; so that substituting in (86), and extending the ranges of the X-, Y-, 2- integrations to the limits oo and 0, we have I = the real part |