We can also write [** dx p (a) sin qx instead of [** daf (a) cos qz, -1 355. The solution of this second problem indicates clearly what the relation is between the definite integrals which we have just employed, and the results of the analysis which we have applied to solids of a definite form. When, in the convergent series which this analysis furnishes, we give to the quantities which denote the dimensions infinite values; each of the terms becomes infinitely small, and the sum of the series is nothing but an integral. We might pass directly in the same manner and without any physical considerations from the different trigonometrical series which we have employed in Chapter III. to definite integrals; it will be sufficient to give some examples of these transformations in which the results are remarkable. 20 we shall write instead of u the quantity; x is a new variable, 1 and n is an infinite number equal to is a quantity formed by dq the successive addition of infinitely small parts equal to dq. We shall represent the variable number i by q dq If in the general sin (2i+1) we put for i and n their values, the term n becomes da sin 2qe. Hence the sum of the series is da dq 2q sin 2qx, the integral being taken from q=0 to q = ∞ ; we have therefore the equationπ = V4 sin 27x which is always true whatever dq q be the positive value of x. Let 2qxr, r being a new varia• dr ble, we have = and = dq_dr r sin r; this value of the defi nite integral sin r has been known for some time. If on fdr supposing r negative we took the same integral from r=0 to r = − ∞, we should evidently have a result of contrary sign – π. 357. The remark which we have just made on the value of whose value we have already found (Article 348) to be equal to 1 or 0 according as x is or is not included between 1 and -1. the first term is equal to or according as x+1 is a dq positive or negative quantity; the second da sin g(-1) is equal toπor-π, according as x-I is a positive or negative quantity. Hence the whole integral is nothing if x+1 and -1 have the same sign; for, in this case, the two terms cancel each other. But if these quantities are of different sign, that is to say if we have at the same time +10 and x-1 <0, the two terms add together and the value of the integral is. Hence the definite integral'da sin q cos que is a function of æ 0 equal to 1 if the variable x has any value included between 1 and -1; and the same function is nothing for every other value of x not included between the limits 1 and - 1. 358. We might deduce also from the transformation of series into integrals the properties of the two expressions* the first (Art. 350) is equivalent to e when x is positive, and to e when a is negative. The second is equivalent to e* if x is positive, and toe if x is negative, so that the two integrals have the same value, when x is positive, and have values of contrary sign when x is negative. One is represented by the line eeee (fig. 19), the other by the line eeee (fig. 20). which we have arrived at (Art. 226), gives immediately the integral 3 2 [ dq sin qπ sin qx; which expression is equivalent to sin æ, if x π 0 1-q is included between 0 and 7, and its value is 0 whenever x exceeds π. 1 At the limiting values of x the value of this integral is ; Riemann, § 15. 2 Cf. Riemann, § 16. ax 3 The substitutions required in the equation are a for x, dq for, q for i α π We then have sin x equal to a series equivalent to the above integral for values of x between 0 and π, the original equation being true for values of x between 0 and a. [A. F.] 359. The same transformation applies to the general equation ‡ π $ (u) = sin u (du & (u) sin u + sin 2u fdu þ (u) sin 2u + &c. Making u == 22, denote 4 (u) or ☀ (~) by ƒ(x), and introduce into the analysis a quantity q which receives infinitely small increq and i to ; substituting ments equal to dq, n will be equal to these values in the general term 1 dq dq qxdx we find dq sin qæ (dæ ƒ (x) sin qe. The integral with respect to u is taken from u=0 to uπ, hence the integration with respect to x must be taken from x=0 to x = n, or from x nothing to x infinite. We thus obtain a general result expressed by the equation § π ƒ (x) = ["dq sin qx ["dxæ ƒ (a) sin qx (e), for which reason, denoting by Q a function of q such that we have qu an equation in which ƒ(u) is a given function, ƒ (u) = fdqQ sin we shall have Q = 2fduf (u) sin qu, the integral being taken from u nothing to u infinite. We have already solved a similar problem (Art. 346) and proved the general equation } π F′(x) = ["dq cos qx · [*d.x F(x) dx F(x) cos qx................ (€), 0 0 which is analogous to the preceding. 360. To give an application of these theorems, let us suppose f(x)=x, the second member of equation (e) by this substitution faq sin qæ fdæ sin qæ a. The integral qx 1 is equivalent to du sin u u, the integral being taken from ɑ nothing to u infinite. it remains to form the integral fdq sin qe, or μa" fdu sin u u−(+1); denoting the last integral by v, taken from u nothing to u infinite, we have as the result of two successive integrations the term xv. We must then have, according to the condition expressed by the equation (e), μν. +πα = μνα or μν = επ; thus the product of the two transcendants and from these two equations we might also conclude the following', 1 ["dqe-=√, which has been employed for some time. 361. By means of the equations (e) and (e) we may solve the following problem, which belongs also to partial differential analysis. What function Q of the variable q must be placed under 1 The way is simply to use the expressions e+cos √-12+ √-1 sin/-1z, transforming a and b by writing y3 for u and recollecting that √√1 Cf. § 407. [R. I. E.] A = |