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(4) M. Catalan* has shewn how to evaluate a definite multiple integral which depends on those which have just been considered. It is
in which x 2 = 1 – x;? – X.? ... ?,-1, and the limiting
* Liouville's Journal, Vol. vi. p. 81. The reader is referred to a paper by Mr Boole in the Cambridge Mathematical Journal, Vol. 11. p. 277, entitled “Remarks on a Theorem of M. Catalan,' where the truth of the Theorem is called in question.
917, + 927, + ... + q1, = 0 ) The number of constants in (2) is n (n − 1), the number of conditions in (3) is ln(n + 1), so that there are In(n − 3) arbitrary coefficients. Adding the squares of (2) we have by the conditions (3)
ui? + una + ... + un = x+ x 2 + ... + x 2 = 1. From the same equations we find a,&, + 0,89 + ... + 27% n = u, (a,” + a,” + ... + a,?)2 = Au1, suppose.
Also on changing the differentials by the method given in Chap. u. Sect. 2 of the Diff. Calc. we have
If now we integrate with respect to all the variables except un, the limits being given by the condition
w" + uz? + ... + u’n-1 5 1 - u', we have by (3, c) du, du, ... duni
**(n-1) (1 – – .... , -1)^2 - T44 (m - 1) (1-4) .
Hence, substituting this value and integrating with respect to u, from 0 to 1, we have
or if x, = cos u, x, = sin u cos v, X., = sin u sin v, and if we take the limits from U = 0 to U = 1, and from v = 0 to 19 = 27, it takes the form (d) *12* du dv sin u f(a, cos u + a, sin u cos v + az sin u sin v)
= 27 fo“ do sin of (A cos 6). The formula under this shape was first given by Poisson in the Mémoires de l'Institut, Tom. 11. p. 126.
(5) In the equation
x = 0, $ = 0; X = 1, = 0; and the transformed equation is (a) (*dx _ x02 r(a)f(r – a)
Jo d* (1 + x):= = () The only restriction on the generality of this result is that a must be less than r. If r= 1, we have
1+% sin ar' This last integral may be considered as being made up of two parts, one from x = 0 to % = 1, the other from z = 1 to x = 0. This second part may be reduced to the same limits as the first by assuming x = -, when it becomes
Hence, adding the two parts,
1+% sin ar' In the formula (6) put x = y2 and b = 2a, when we find
z sin bus:
(6) The integral f*dw- may be considered as made up of two parts, one from x = 0 to x = 1, the other from æ = 1 to x = -0. If we put - for æ, the latter part becomes
- fo'da *
Expanding the denominator and integrating from x = 0 to x = 1, we obtain the series 1 1 1 1
sin x = = (1-7) (1+) (-) (1 + )......
which is the difference of two infinite quantities, is easily found. We have
Integrating with respect to n and determining the constant so that the integral shall vanish when n = 1, there results