Integrals of the form. Notes by G.J.O. Jameson. I f (x) = x. f(t) cos t dt, S f (x) = x

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1 Integrals of the form f(t)eit dt Notes by G.J.O. Jameson Basic results We consider integrals of the form with real and imaginary parts I f () = f(t)e it dt, C f () = f(t) cos t dt, S f () = f(t) sin t dt. where and f(t) is real, continuous and non-negative on (, ). Our conditions will ensure that I f () eists for all >. If I f () eists (it will in some cases, not others) it is called the complete integral of this kind. First, some elementary observations, presupposing the eistence of I f () for > : (E1) By the fundamental theorem of calculus, I f () = f()ei. (E2) (E3) (E4) I f () = e i f( + u)e iu du. I f ( + 2kπ) = I g (), where g(t) = f(t + 2kπ). For λ >, we have f(t)eiλt dt = 1 I λ g(λ), where g(t) = f(t/λ). Our general assumption is that f(t) is defined and non-negative for all t > (and possibly also for t = ), and that f(t) is completely monotonic, that is: (CM) lim t f(t) = and ( 1) n f (n) (t) for all n and t >. This implies automatically that ( 1) n f (n) (t) is decreasing and tends to as t. So ( 1) n f (n) (t) also satisfies (CM). The basic eample is f(t) = 1/t p for any p > : by a slight abuse of notation, we use the notation I p (), C p (), S p () for this case. Cases of particular interest are p = 1, defining the sine and cosine integrals and p = 1, defining the Fresnel integrals. There 2 are well-known values for the complete integrals S 1 () = π/2 and I p () = Γ(q)e 1 2 qπi, where < p < 1 and q = 1 p. Methods for these evaluations, and other results specific to these cases, are presented in companion notes [Jam1], [Jam2]. Another eample satisfying (CM) is f(t) = e at, where a >. In this case, we can 1

2 evaluate I f () eplicitly: I f () = e (i a)t dt = e(i a) a i = a + i a e a e i. If f is completely monotonic, then so are f(t + a) and f(t) f(t + a) for a >. Also, if f and g are completely monotonic, then so is fg. Numerous further eamples of completely monotonic functions are given in [AB] and references listed there. There is, in fact, a welldeveloped general theory of such functions (e.g. [Wid, chap. 4]), but we do not need it for our purposes. We denote by (CM k ) condition (CM) restricted to the first k derivatives, with f (k) assumed continuous. To start the investigation of these integrals, we write I f (, y) = y f(t)e it dt and integrate by parts, temporarily assuming only (CM 1 ): I f (, y) = [ ] y if(t)e it + i Since f (t), we have f (t)e it f (t), hence I f (, y) y f (t)e it dt = if()e i if(y)e iy + ii f (, y). (1) y ( f (t)) dt = f() f(y), (2) which converges to f() as y. Hence the integral defining I f () is convergent, with absolute value not greater than f() (distinguish between I f () and I f ()!). Now taking the limit as y, we obtain the basic result on convergence, together with a long list of identities and inequalities, which we summarise as follows: PROPOSITION 1. If f satisfies (CM 1 ), then the integrals defining I f () and I f () are convergent for all >, and the following statements apply: I f () = if()e i + ii f (), (3) C f () = f() sin S f (), (4) S f () = f() cos + C f (), (5) I f () f(), (6) 2

3 I f () 2f(), (7) f()(1 + sin ) C f () f()(1 sin ), (8) f()(cos 1) S f () f()(cos + 1). (9) Note. Using (1) and (2), one can easily derive corresponding statements for I f (, y), C f (, y) and S f (, y), without any assumptions about f(t) outside the interval [, y]. For eample, I f (, y) f()+f(y)+f() f(y) = 2f(). By writing out the real and imaginary parts in (1), one can show that (8) and (9) still apply for all y >. If y = + 2kπ, then e iy = e i, and we see that I f (, y) 2[f() f(y)]. Our standing assumption from now on will be that f satisfies condition (CM), though it will be clear that particular results only require (CM k ) for suitable k. Then (7) applies to f to give I f () 2f (). Typically, (and in particular for f() = p ), f () will be of a smaller order of magnitude than f() for large, so this will be a better estimation of I f () than (6). It then makes sense to restate (3) in the form I f () = if()e i + r f (), where r f () 2f () (and similarly for (4) and (5)). So under the condition f ()/f() as, we have I f () f() as. This suggests that the natural comparison in (7) is really with f() rather than 2f(). It certainly shows that if C is the best constant in the inequality I f () Cf(), then C 1. We will see later that the factor 2 can indeed be removed in (7). At the same time, (6) will typically give a better estimation of I f () than (7) for close to. Repetition of the process that gave (3) leads at once to the following identities. PROPOSITION 2. If f is completely monotonic, then for all >, I f () = if()e i f ()e i I f (), (1) C f () = f() sin f () cos C f (), (11) S f () = f() cos f () sin S f (). (12) Further, I f () = i [f() f ()] e i [ f () f (3) () ] e i + I f (4)(), (13) C f () = [f() f ()] sin [ f () f (3) () ] cos + C f (4)(), (14) S f () = [f() f ()] cos [ f () f (3) () ] sin + S f (4)(). (15) 3

4 Proof. Applying (3) to f (t) and substituting back into (3), we obtain (1). Now apply (1) to f (t) and substitute, obtaining: I f () = if()e i f ()e i if ()e i + f (3) ()e i + I f (4)(). which equates to (13). The other statements are derived by taking real and imaginary parts. The reader is urged not to be daunted by this proliferation of formulae! It should be clear that they simply record the consequences of repeating a very simple integration by parts, with versions for each of the three integrals. Bounds for the remainder terms convert these identities into inequalities. As before, we have alternative bounds. For eample, I f (4)() is bounded both by f (3) () and by 2f (4) (). As an illustration, we restate (13) eplicitly for the case f(t) = 1/t: ( 1 I 1 () = 2 ) ( 1 ie i ) e i + 24I (), in which 24I 5 () is bounded both by 6/ 4 and 48/ 5. It is quite easy to deduce that I 1 () < 1/ for all > 4 (see [Jam1]). For S 1 (), the method is developed further in [JLM] to derive the stronger inequality S 1 () π 2 tan 1 (note that this is eact at ). Of course, the process can be continued: successive derivatives of f() appear in the epressions multiplying e i and ie i. The outcome is an asymptotic epansion for I f (). However, this does not simply deliver ever-closer approimations, because typically for a fied, the derivatives f (n) () will ultimately grow large in magnitude. The functions S f () and C f () (The reader could omit, or defer, this subsection). We now present a number of identities and inequalities for S f () and C f (), particularly relating to the points nπ and (n 1)π 2 (which we denote by u n to shorten the formulae). Typically, inequalities relating to S f (nπ) reverse according to whether n is even or odd: this is handled most efficiently by stating them in terms of ( 1) n S f (nπ). Since S f () = f() sin and sin t on intervals [2nπ, (2n+1)π], S f() is decreasing on such intervals, and similarly it is increasing on intervals [(2n 1)π, 2nπ]. Hence it has maima at the points 2nπ and minima at (2n + 1)π. Similarly, C f () has maima at u 2n and minima at u 2n+1. 4

5 In particular, S f () is decreasing on (, π], so it either tends to infinity or to a finite limit (which is then S f ()) as +. Similarly for C f (). Assuming no more than that f(t) is decreasing, we can prove: PROPOSITION 3. Suppose that f(t) is non-negative and decreasing, and that S f () eists for all >. Write S f (nπ) = S n. Then for n, ( 1) n (S n S n+2 ) (16) ( 1) n S n. (17) For all 2nπ, we have S 2n+1 S f () S 2n. The least value of S f () is S 1. If S is finite, it is the greatest value. Proof. By substituting t = u + π on [(n + 1)π, (n + 2)π] and recombining, we see that S n S n+2 = (n+2)π nπ f(t) sin t dt = (n+1)π nπ [f(t) f(t + π)] sin t dt. Since f(t) f(t + π) and ( 1) n sin t on [nπ, (n + 1)π], this implies (16). Hence ( 1) n (S n S n+2k ) : taking the limit as k, we have ( 1) n S n. Now take between (2n + 2k)π and (2n + 2k + 2)π. By the monotonicity properties, we have S f () S 2n+2k S 2n and S f () S 2n+2k+1 S 2n+1. The statements on greatest and least values follow. Similar statements apply to C f (), with nπ replaced by u n, ecept that if C f () is finite, it can be either greater or less than C f (u 2 ), the other candidate for the greatest value. (See [Jam2]). We remark that (17) also follows from (9), applied to nπ for even and odd n. Our earlier identities simplify pleasantly at the points nπ and u n, because sin nπ = cos u n =. Also, cos nπ = ( 1) n and sin u n = ( 1) n+1. In particular, (12) and (11) become: S f (nπ) = ( 1) n f(nπ) S f (nπ), (18) C f (u n ) = ( 1) n f(u n ) C f (u n ). (19) PROPOSITION 4. If f is completely monotonic, then for n 1, ( 1) n S f (nπ) f(nπ), (2) ( 1) n C f (u n ) f(u n ). (21) 5

6 form Proof. By (17), applied to f, we have ( 1) n S f (nπ). With (18) rewritten in the ( 1) n S f (nπ) = f(nπ) ( 1) n S f (nπ), it follows that ( 1) n S f (nπ) f(nπ). The proof of (21) is similar. Hence, for eample, f[(2n + 1)π S f () f(2nπ) for in [2nπ, (2n + 2)π]. We can derive companion bounds for S f (u n ) and C f (nπ): PROPOSITION 5. If f is completely monotonic, then ( 1) n+1 S f (u n ) f (u n ), (22) ( 1) n C f (nπ) f (nπ). (23) Proof. By (4), we have C f (nπ) = S f (nπ). By (2), applied to f, we have ( 1) n+1 S f (nπ) f (nπ), hence (23). The proof of (22) is similar. Of course, we can take these results to a further stage. For = nπ, (15) becomes S f (nπ) = ( 1) n [f(nπ) f (nπ)] + S f (4)(nπ), and one can deduce f(nπ) f (nπ) ( 1) n S f (nπ) f(nπ) f (nπ) + f (4) (nπ). The auiliary functions; the inequality I f () f() Formulae like (3) and (1) simplify pleasantly when epressed in terms of e i I f () and its components, because real and imaginary parts are separated. Write e i I f () = K f (). Note that by (E2), K f () = f(u + )e iu du. (24) We write K f () = V f () + iu f (), so that U f () = S f () cos C f () sin, (25) V f () = C f () cos + S f () sin. (26) U f and V f are called the auiliary functions. (We write U f, V f this way round in a gesture to the customary notation; cf. [DLMF, chapter 6] for the case f(t) = 1/t.) Note that U f (nπ) = ( 1) n S f (nπ) and V f (nπ) = ( 1) n C f (nπ), also U f (u n ) = ( 1) n C f (u n ) and V f (u n ) = ( 1) n+1 S f (u n ). We will see that our results for S f and C f 6

7 at the points nπ and u n can be recaptured as special cases of similar statements for U f () and V f (), now applying for all. Stated for K f () and its components, identity (3) becomes K f () = if() + ik f (), (27) U f () = f() + V f (), (28) V f () = U f (). (29) Similarly, (1) becomes K f () = if() f () K f (), (3) U f () = f() U f (), (31) V f () = f () V f (). (32) The cases = nπ and = u n in (31) reproduce (18) and (19). Meanwhile, differentiation gives K f() = ie i I f () e i e i f() = ik f () f(), (33) hence U f() = V f (), (34) V f() = U f () f(). (35) Note that by (27) and (33), we have K f () = K f (): this also follows formally from (24) by differentiation under the integral sign. Also, from (34) and (35), we see that U f and V f satisfy the differential equations U f +U f = f and V f +V f = f (though we will not use this fact). In fact, they are essentially the solutions of these equations delivered by the technique of variation of parameters. The basic inequalities for U f () and V f () are summarised in the net result. THEOREM 1. If f is completely monotonic, then Further, U f (), V f () and f() U f () are decreasing. U f () f(), (36) V f () f (). (37) Proof. By (6), we have V f () I f () f(), so by (28), U f (). Applied to f, this gives U f (), so by (29), we have V f (). Now applying these statements 7

8 to f, we see that U f () and V f () are non-negative, so by (31) and (32), U f () f() and V f () f (). By (34), we now have U f () = V f() and f () U f () = f () + V f (). By (35), we have V f () = U f() f(). Note. As mentioned previously, weaker conditions are actually sufficient. We require (CM 1 ) for the inequality U f (), hence (CM 2 ) for U f () and V f (), and consequently (CM 4 ) for V f () and V f () f(). Note that (2) and (21) are the cases = nπ and = u n in (36), while (22) and (23) are similarly cases of (37). We have gained other information, for eample ( 1) n S f (nπ) and ( 1) n C f (nπ) decrease with n. We can proceed to further terms, as before. Identity (13) becomes K f () = i [f() f ()] [ f () f (3) () ] + K f (4)(), U f () = f() f () + U f (4)(), V f () = f () f (3) () + V f (4)(), with attendant inequalities derived from U f (4)() f (4) () and V f (4)() f (5) (). We can now show that the factor 2 in (7) can be removed, as stated earlier. THEOREM 2. If f is completely monotonic, then I f () is decreasing and for all >. Proof. Let M() = I f () 2 = C f () 2 + S f () 2. Then I f () f() (38) M () = 2C f ()f() cos 2S f ()f() sin = 2f()[C f () cos + S f () sin ] = 2f()V f (). (This follows equally from (34) and (35).) By (37), we have V f () f (). Hence M (), so M() is decreasing. Also, M () 2f()f (), so f() 2 M() is decreasing. Since it tends to as, it follows that it is non-negative, so M() f() 2, hence I f () f(), for >. Hence C f () and S f () are also bounded by f(). The upper bounds in (36) and (37) are now seen to be special cases of (38), but of course (37) was used in its proof. 8

9 We mention two corollaries. COROLLARY 2.1. If y = + 2kπ, then I f (, y) f() f(y). Proof. By (E3), I f (y) = I g (), where g(t) = f(t + 2kπ). So I f (, y) = I f () I f (y) = I h (), where h(t) = f(t) f(t + 2kπ). The statement follows by applying Theorem 2 to h. COROLLARY 2.2. If λ >, then f(t)e iλt dt f() λ. Proof. This is an immediate consequence of Theorem 2 and (E4). Theorem 2 has appeared in [Jam3]. Although these integrals are a highly classical topic, not much attention seems to have been given to inequalities of this type. I am not aware of any earlier mention of (38) in the literature, although for the case f(t) = 1/t p, it is an easy consequence of a known epression for I p (): we return to this below. As noted earlier, the condition really required for (37), and hence (38), is (CM 4 ), Without any differentiability conditions, a trivial eample shows that factor 2 in (7) can be attained: take f(t) to be 1 for t π and for t > π: then I f () = 2i. We now give an eample satisfying (CM 2 ) in which I f (), and indeed S f (), is greater than f(), and (37) fails. Eample. Fi b > 2π, and let f(t) = { (b t) 2 for < t b, for t > b. Let a = b 2π. Then f (t) = 2 for a t < b, so I f (a) = b a 2eit dt =, so (1) applies to give I f (a) = [if(a) f (a)]e ia = (4π 2 i + 4π)e ia, hence I f (a) > 4π 2 = f(a). Also, S f (a) = 4π 2 cos a + 4π sin a, which is greater than 4π 2 at least when < a < π 6. Further, one checks easily that V f() = 2(b ) 2 sin(b ), which is greater than f () = 2(b ) when a < < a + π. (Strictly, this eample does not satisfy (CM 2 ) because f has a discontinuity, but this can be rectified by a small perturbation, or by a natural relaation of the definition of (CM 2 ) to restrict to left derivatives at a point b such that f(t) = for all t b.) 9

10 An alternative method for Theorems 1 and 2 We sketch a superficially very quick alternative proof of these theorems using a theorem of Bernstein [Wid, p. 16], which states that all completely monotonic functions can be epressed as f(t) = e ut dµ(u) for some non-negative measure µ. As an illustration, the epression for 1/t p derived directly from the definition of the gamma function, is 1 t p = 1 Γ(p) u p 1 e ut du. Proof of Theorems 1 and 2. Assuming validity of the reversal of integration, we have I f () = = = e it e ut dµ(u) dt e (u i) u i e (u i)t dt dµ(u) dµ(u). Since u i 1, we deduce I f () e u dµ(u) = f(). Also, we can derive the following epressions for the auiliary functions: U f () = e u u dµ(u), V f () = ue u u dµ(u), from which it is clear that U f (), V f () and f() U f () are decreasing. For the case f(t) = 1/t p, this is indeed an appealing method, though the reversal of integration requires justification, at least by first considering the integral on a bounded interval [, y] for t. The implied epression for I p () is surely well known. However, for general completely monotonic functions, this method depends on heavy machinery in the form of Bernstein s theorem. Also, it does not apply to functions satisfying only (CM 4 ). 1

11 References [AB] [DLMF] [Jam1] [Jam2] [Jam3] [JLM] Horst Alzer and Christian Berg, Some classes of completely monotonic functions II, Ramanujan J. 11 (26), NIST Digital Library of Mathematical Functions at dlmf.nist.gov G. J. O. Jameson, The sine and cosine integrals, at jameson G. J. O. Jameson, The integral at jameson e it t p dt; Fresnel-type integrals, G. J. O. Jameson, An inequality for integrals of the form f(t)eit dt, J. Math. Ineq. 1 (216), Graham Jameson, Nick Lord and James McKee, An inequality for Si(), Math. Gazette 99 (215), [Wid] D. V. Widder, The Laplace Transform, Princeton Univ. Press (1946). updated 21 January