tion of the following statement: " “「1-. is always finite if a be any quantity less than unity, and by taking a sufficiently near to unity, we can make the value of the integral differ as little as we please from 2." 46. Next take ୮" a dx 1 Xx the value of this integral is — log (1 − a) which increases indefinitely as a approaches to unity. Hence in this case we may write 1 dx 0 = ∞ pro vided that we regard this as an abbreviation of the following increases indefinitely as a approaches to unity, and by taking a sufficiently near to unity we can make the integral greater than any assigned quantity.” 47. Next consider If without remarking that the function to be integrated becomes infinite when x=1, we propose to find the value of the integral between the limits 0 and 2, we obtain — 1 — 1, that is -2. But this is obviously false, for in this case every term of the series indicated by Σ(x) Ax is positive, and therefore the limit cannot be negative. In fact are both infinite. dx (1 − x)2 This example shews that the ordinary rules for integrating between assigned limits cannot be used when the function to be integrated becomes infinite between those limits. rb 48. In the fundamental investigation in Art. 2, of the value of (x) dæ, the limits a and b are supposed to be a finite as well as the function (x). But we shall often find it convenient to suppose one or both of the limits infinite, as we will now indicate by examples. =tana; the larger a becomes, the nearer tan a approaches to, and by taking a sufficiently large, we can make tan ́ɑ enough we can make log (1 + a) greater than any assigned quantity. Hence for abbreviation we may write 49. Suppose the function () to become infinite once between the limits a and b, namely, when x= c. then apply the ordinary rules of integration to but we may apply those rules to We cannot "$(x) de; a for any assigned value of μ however small. The limit of the last expression when u is diminished indefinitely is called by Cauchy the principal value of the integral | (x) dx. •b a for die α log [_de_ -- [ _dar - - log b = 0; dx сних с hence the principal value is log c-a log Students are sometimes doubtful respecting the value which is to be assigned to sin1 (1) and sin(-1) in such a result as the above. Suppose we assume x = a sin; thus the integral becomes São or 0. Now x increases from -a to a, hence the limits assigned to 0 must be such as correspond to this range of values of x. When xa then may have any π value contained in the formula (4n-1), where n is any integer. Suppose we take the value (4n-1), where n is some definite integer, then corresponding to the value x=a we must take 0 = (4n π 1); this will be obvious on 2 examination, because x is to change from a to +a, so that it continually increases and only once passes through the value therefore therefore π dx + √" log sin x' dæ 0 1 — — π log 2. 2 = [* log sin x da, by equation (4) of Art. 41 ; Again, [ log sine de = (π — 9)2 log sin ê de, by equaf 02 tion (3) of Årt. 41; therefore Required flog (1+x) 1+x2 0 0 dx. Put x=tany, and the integral becomes [*log (1+tan y) dy; but by equation (3) of Art. 41 log (1 + tan y) dy = ["log {1 + tan (y See Cambridge Mathematical Journal, Vol. III. p. 168. 52. The remainder after n+1 terms of the expansion of (a + h) in powers of h, may be expressed by a definite integral. For let $ F (z) = 0 (x − 2) + z¤′ (x − 2) + 1⁄2 4′′ (x − 2)......+ " (x-2). [n 4” (x−z). Differentiate with respect to z, then Integrate both members of this equation between the limits O and h; thus h2 n $ (a + h) = 4 (a) + hp' (a) + 120′′ (a)..........+ 1" (a) 1μn+1 (a + h − z) dz. |