[PDF] Evaluation of the sine and cosine integrals - Lancaster

View - Download Evaluation of the sine and cosine integrals - Lancaster

Previous PDF

Next PDF

Evaluation of the sine and cosine integrals

Notes by G.J.O. Jameson

The complete sine integral: first method

We shall consider the integrals, in their various appropriate forms, of sintt andcostt . We start with the "complete sine integral":

THEOREM 1.We have?∞

0sintt

dt=π2 .(1) Note first that there is no problem of convergence at 0, because sintt →1 ast→0. A very quick and neat proof of (1) (to be seen, for example, in [Lo]) lies to hand if we assume the following well-known series identity: forx?=kπ,

1sinx=∞?

n=-∞(-1)nx+nπ(2) One proof of (2) [Wa, p. 17-18] is by considering the Fourier series for cosaxon [-π,π]. To derive (1), note first that, since sint/tis an even function, -∞sintt dt= 2?

0sintt

dt.

Denote this byI. The substitutiont=x+nπgives

(n+1)π nπsintt dt= (-1)n?

0sinxx+nπdx.

Assuming that termwise integration of the series is valid, we add these identities for all integersnto obtain at once I=? 0 sinx1sinxdx=π. The termwise integration (for any readers who care) is easily justified by uniform convergence, as follows. By combining the terms fornand-nand multiplying by sinx, we can rewrite the series (2) as sinxx + 2xsinx∞? n=1(-1)nx

2-n2π2= 1.

For 0< x < πandn≥2,????2xsinxx

2-n2π2?

1 Since n=21/(n2-1) is convergent, it follows, by Weierstrass"s "M-test", that the series converges uniformly on the open interval (0,π): this is all we need. We note some immediate variants and consequences of (1). First, for anya >0, the substitutionat=ugives

0sinatt

dt=?

0sinuu

du=π2

In particular,

0sintcostt

dt=12

0sin2tt

dt=π4 .(3)

We will use this several times later.

Next, we can derive the following integral:

0sin 2tt

2dt=π2

.(4) To do this, take 0< δ < Rand integrate by parts on [δ,R]: R

δsin

2tt 2dt=? -sin2tt R R

δ2sintcostt

dt. Now sin2RR →0 asR→ ∞andsin2δδ →0 asδ→0+. Taking limits and applying (3), we obtain (4). This argument is reversible, so (4) equally implies (1). This is a viable alternative, because one can prove (4) in a similar way to (1), using the series 1/sin2x=?∞ n=-∞[1/(x-nπ)2]; this method is followed in [Wa, p. 186-187]. One can develop this process further to evaluate the integrals of sin nx/xmfor various mandn: see [Tr].

The incomplete sine integral

The "incomplete" sine integral is the function

Si(x) =?

x

0sintt

dt. First, some simple facts about it. By the fundamental theorem of calculus, the derivative Si ?(x) is sinx/x. Hence Si(x) is increasing on intervals [2nπ,(2n+ 1)π] and decreasing on intervals [(2n-1)π,2nπ], so it has maxima at the points (2n+ 1)πand minima at the points 2nπ. Now write, temporarily,An=?(n+2)π nπsintt dt. By substitutingt+π=uon [nπ,(n+ 1)π], we see that A n=? (n+1)π nπ? 1t -1t+π? sint dt, 2 in which 1t -1t+π>0. Ifnis even, then sint≥0 on [nπ,(n+ 1)π], soAn≥0, hence

Si[(n+ 2)π]≥Si(nπ). It follows that Si(2nπ)≥Si(0) = 0 for alln, so in fact Si(x)≥0 for

Si(x) is Si(π). Of course, (1) says that Si(x)→π2 asx→ ∞.

By integrating the series

sintt n=0(-1)nt2n(2n+ 1)!, we obtain the explicit series expression

Si(x) =∞?

n=0(-1)nx2n+1(2n+ 1)!(2n+ 1)=x-x33!3 +x55!5 from which, in principle, Si(x) can be calculated, though in practice the calculation is only pleasant for fairly smallx. One finds, for example, Si(π)≈1.85194 (recall that this is the greatest value) and Si(2π)≈1.41816. The complementary sine and cosine integrals, and analogues of (1) forcost We cannot simply replace sintby costin (1), or in the definition of Si(x), because the resulting integral would be divergent at 0. To formulate results that make sense for both sintand cost, we consider instead the complementary integrals

S(x) =?

xsintt dt, C(x) =? xcostt dt. (Here I am departing from the established notation, which is si(x) and ci(x) where we have -S(x) and-C(x)).

By (1), we haveS(0) =π2

andS(x) =π2 -Si(x). By the remarks above,S(x) has maxima at 2nπand minima at (2n-1)π, with greatest valueπ2 and least valueS(π). Also,

S(π)≈ -0.28114 andS(2π)≈0.15264.

Meanwhile,C(x) is defined forx >0, but not atx= 0. It has maxima at (2n-12 and minima at (2n+12 )π, with overall least value atπ2 The next result gives pleasantly simple approximations toS(x) andC(x) for largex (it also incorporates the proof that the integrals defining them converge in the first place).

PROPOSITION 1.We have

S(x) =cosxx

+q1(x), C(x) =-sinxx +r1(x),(5) 3 where|q1(x)|and|r1(x)|are not greater than2/x2. HencexS(x)-cosxandxC(x) + sinx tend to 0 asx→ ∞.

Proof. Integrating by parts twice, we obtain

S(x) =?

-costt x-? xcostt 2dt cosxx -?sintt 2? x+q2(x), cosxx +sinxx

2+q2(x),

where q

2(x) =?

x2sintt 3dt. x2t

3dt=1x

2. The stated expression forS(x) follows. The proof forC(x) is

similar: we leave the details to the reader.? Of course, this also shows thatS(x) andC(x) tend to 0 asx→ ∞. The process can be repeated to deliver increasingly accurate asymptotic expressions, and inequalities, for S(x) andC(x). For example, the following inequality forS(x) was established in [JLM]: -tan-1x. Can we find a formula that enables us to calculateC(x), and that opens the way to some kind of analogue of (1)? The key is to introduce the function C ?(x) =? x

01-costt

dt (This function is sometimes denoted by Cin(x)). Since 1-cost=O(t2) ast→0, there is no problem of convergence at 0, and, as with Si(x), we have a power series expression C ?(x) =∞? n=1(-1)n-1x2n(2n)!(2n)=x22!2 -x44!4

We now relateC?(x) andC(x). We have

C ?(x)-C?(1) =? x

11-costt

dt= logx-? x

1costt

dt= logx-C(1) +C(x), so

C(x) =C?(x)-logx+c,(6)

wherecis constant, in factc=C(1)-C?(1). 4 Even without knowingc, we can draw some conclusions from (6). One, which we will use later, isxC(x)→0 asx→0+(since limx→0+(xlogx) = 0). Another is the following integral, which can be regarded as one kind of analogue of (1). It is a special case of the "Frullani integral": see [Fer, p. 133-135] or [Tr], where it is used in the evaluation of the integral of sin nx/xm(I am grateful to Nick Lord for these references).

PROPOSITION 2.Fora, b >0,

0cosat-cosbtt

dt= logb-loga.(7)

Proof. The substitutionat=ugives

x

01-cosatt

dt=? ax

01-cosuu

du=C?(ax). Hence x

0cosat-cosbtt

dt=C?(bx)-C?(ax) =C(bx)-C(ax) + logbx-logax =C(bx)-C(ax) + logb-loga →logb-logaasx→ ∞.? However, for a fully satisfactory version of (6), and for the calculation ofC(x), of course we need to know the value ofc. The answer turns out to be thatc=-γ, whereγis Euler"s constant. Let us state this fact as a theorem:

THEOREM 2.We have

C(x) =C?(x)-logx-γ.(8)

Oddly, this result is not mentioned in the comprehensive article [Lag] on Euler"s con- stant. It can be seen stated without proof in compilations of formulae, such as Wikipedia or [Erd, p. 145]. However, it is not easy to find accessible references with a proof. At the same time, the method will also give a second proof of Theorem 1. Later, we describe an alternative route to both theorems using contour integration. The starting point for the proof will be the following expression forcderived from (6): since lim x→∞C(x) = 0, we have -c= limx→∞[C?(x)-logx].(9) 5 Similarly, sinceC?(x)→0 asx→0+, it will follow from (8), once proved, that C(x)+logx→ -γasx→0+. This describes the nature ofC(x) near 0, so can be regarded as the true analogue of (1). Also, (8) enables us to calculateC(x). We find, for example,C(π2 )≈ -0.47200 (recall that this is the least value) andC(π)≈ -0.07367. Second proof of Theorem 1 and a proof of Theorem 2 We will use the following elementary version of Riemann-Lebesgue Lemma, which is easily proved by integration by parts:iffis continuous on[a,b]and has a continuousquotesdbs_dbs13.pdfusesText_19

[PDF] integrale sin(t)/t

[PDF] procédés théatraux

[PDF] tendinopathie genou traitement

[PDF] tendinite demi membraneux

[PDF] comment soigner une fabella

[PDF] fabella douloureuse

[PDF] tendinite poplité traitement

[PDF] mecanique de fluide resume

[PDF] mécanique des fluides bernoulli exercices corrigés

[PDF] fiche résumé mécanique des fluides

[PDF] mécanique des fluides cours pdf

[PDF] question ? choix multiple culture générale

[PDF] question ? choix multiple definition

[PDF] choix multiple orthographe

[PDF] questions avec reponses multiples synonyme