Advanced Calculus fi..

450 Advanced Calculus, Fifth EditionThe latter condition can be satisfied by the nth term test. If xl > bN, then-b, 5 xl 5b,+l for some n 2 N. Henceandthe last step is justified, since f (x) does not change sign between b, and b,+l. HenceIt follows thatliliXIf(x)dx = I.COROLLARY Let f (x) be continuous for a 5 x < oo, let f (x) decrease as xincreases, and let lim,,, f (x) = 0. Then the integralsconverge.Jm (x)sinx dx, Jw f (x) cos x dxProof. We consider the sine integral, the cosine integral being similar. Under theassumptions made, the alternating series of Theorem 5 1 is, except perhaps for thefirst term, of the form(n+l)rr , , " :.,,z.q* ?*f-$.Fan, an=ln f (x)sinx dx. -- -n=kSince f (x) decreases as x increases, la, 1 is decreasing; since f (x) has limit 0 asx -+ oo, a, converges to 0. The alternating series test (Theorem 18) then guaranteesconvergence.EXAMPLES The integralssin xlw dx, dx, l m y d xall exist, by virtue of the preceding corollary.