then Take the logarithms on both sides, and divide by n; This demonstration of a theorem due to Euler* is given by Mr Leslie Ellis in the Cam. Math. Journal, Vol. 11. p. 282. log 2. we find 80 = π log 2 = ƒa dy (cot-1y)2, dx x log x (1 − x2)3 by parts; then fda a log a = (1 − a2) − log {1 + (1 − a®)3} + {1− (1 −æ°)}} log æ. (1 − x2)} - Acta Petrop. Vol. 1. p. 2. On taking this between the limits 0 and 1, we find since {1-(1-x)} log a vanishes at both limits, and therefore taking it between the limits 0 and 1, Euler, Nov. Com. Petrop. Vol. xix. p. 30. To find the value of da log x (16) 1 Since = 1 x + x2 1 + x and fdx x" log x = +1 log x n 1 (n + 1)2 ; which, when taken between the limits 0 and 1, is reduced to when a <1. If a >1, the only difference is that the preceding result is to be multiplied by a-2". Poisson, Journal de l'Ecole Polytechnique, Cah. xvII. p. 614. dxx sin x (18) To find the value of On expanding the denominator, we have a series consisting of the even powers of cos x. Take one of these as (cos x)2r: by integration by parts. In taking the limits between 0 and fo da (cos x)2+1 vanishes as 2r+1 is odd; therefore There are here three cases to be considered-according as a is less than b, equal to b or some multiple of b, and greater than b, not being a multiple of it. We have therefore to integrate a series of functions of 1 Now if we decompose (1 + x2) (m2 − x2) into quadratic factors and integrate from 0 to , observing that we have ∞, Hence if we have a function of a which can be decomposed into a sum of partial fractions of the form m2 1 so that In the case under consideration, therefore, we have Then as If a be greater than b and not a multiple of it, let a = 2rb + c where r is an integer, and c < a. sin (2rb+c) x−sin {2 (r−1) b+c} x=2 sin bæ cos {(2r−1)b+c} x, +2{cos(a−b) x+cos(a−3b)x+...+cos(b+c)x } . |