The real exponential integrals. Notes by G.J.O. Jameson. e t dt, E (x) = x. 1 e t t

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1 The real eponenial inegrals Noes by G.J.O. Jameson The funcions E() and E () Define, for suiable, E() e d, E () e d. () E(), as well as various muaions and equivalen forms, is known as he eponenial inegral. Noaion someimes used in specialis lieraure is E () for our E() and Ein() for E (). In he noaion of he incomplee gamma funcion, E() is Γ(, ). Clearly, E() is defined for >, and E () for (noe ha ( e )/ as ). Since he inegrands are posiive, boh funcions are non-negaive, E() is decreasing and E () is increasing. (E ) () ( e )/. A simple inequaliy for E() is By he fundamenal heorem of calculus, E () e / and E() e d e. (2) Since < e for >, we have < E () for >. A beer inequaliy for > is By inegraing he series E () < E () + e ( ) n we obain a power series epression for E (): E () ( ) n n n d < log +. (3) n n n!, n!n 2 2! !3, (4) from which, in principle, values can be calculaed (hough in pracice he calculaion is only pleasan for fairly small ). For eample, we find E () For a >, he subsiuion a u gives e a d e u a u du E(a), (5)

2 e a d a e u u du E (a), (6) The numbers E() and E () have paricular significance. Inegraion by pars gives E() [ ] e log + e log d Since ( e ) log as +, we have similarly E () [ ] ( e ) log To relae E () and E(), observe ha e log d e log d. (7) e log d. (8) so E () E () e d log e d log E() + E(), E() E () log + c, (9) where c is consan, in fac c E() E (). By (7) and (8), we have c e log d. () Wihou knowing he value of c, we can deduce ha E() log as +, hence E() as +. Anoher deducion from (9) is he following inegral: EXPINT. For a, b >, e a e b d log b log a. () Proof. By (6) and (9), e a e b d E (b) E (a) E(b) E(a) + log b log a E(b) E(a) + log b log a log b log a as. An alernaive proof of () is by epressing he inegrand as b a e y dy and reversing he double inegral. I is also a special case of he Frullani inegral (see [Fer, p ] or [Jam2]). 2

3 However, for a fully saisfacory version of (9), and for he calculaion of E(), of course we need o know he value of c. This will be our main resul, and we reurn o i shorly. Bu firs we record a number of equivalen epressions for E(). The subsiuion + u in () gives E() e e u du. (2) + u This epression can be used o define E(z) for any comple z no on he negaive real line. As a conour inegral, i equaes o he inegral of e ζ /ζ on he horizonal half-line defined by ζ z +u for u. No very helpfully, he noaion Ei(z) is someimes used for E( z). We will no discuss he comple funcion E(z) in hese noes. Subsiuing e u v in (2), we obain E() e Subsiuing u in (), hen u + v, we obain E() e u u dv. (3) log v du e e v dv. (4) + v Inegraing he firs of hese epressions by pars, we have also he following ideniy, generalising (7): E() e u log u du. (5) Noe also ha E() 2 (e u2 /u) du. The number δ ee() has been called he Euler-Gomperz consan. By (2) and (3), δ e u + u du log v dv. Deerminaion of c By (9), we have wo ways o epress c as a limi. Since lim + E (), we have Also, since lim E(), we have We will use (7) o evaluae c. c lim +[E() + log ]. (6) c lim [E () log ]. (7) 3

4 EXPINT2 LEMMA. For n, ( 2 n ) e ( n) n e. (8) Proof. From he series for e and /( ), we have + e /( ), and hence ( 2 )e e for < <. Subsiue /n and ake he nh power o obain ( ) n ( 2 e n e n n). 2 for n. Now ( a) n na for a, so (8) follows. EXPINT3 THEOREM. We have c γ, hence E() E () log γ, (9) e log d γ. (2) Proof. By (7), i is enough o show ha E (n) log n γ as n. Since e lim n ( n )n, i seems a leas plausible ha E (n) is approimaed by K n, where K n so ha K n log n γ as n. n [ ( ) n ] n u [ ( u)n ] du v n v dv d ( + v + + v n ) dv n, To make his precise, we have by (8) ( e ) n e + n n 2 e. I follows ha E (n) K n E (n) + n, where n n n e d < n. Hence, for eample, E() E () γ

5 The inegral in (2) is also called he eponenial inegral. Readers familiar wih he gamma funcion will recognise ha i equaes o Γ (), and indeed he heory of he gamma funcion provides an alernaive mehod for is evaluaion. We menion some inegrals derived from (2). Firs, he subsiuion a u (where a > ) gives e a log d e u (log u log a) a du (γ + log a), (2) a which we can rewrie raher pleasanly as e a (log +γ)d log a. Ne, subsiuing a e u, we obain Thirdly, we have ( log log u) du γ. (22) e log d γ (23) To prove his, noe ha d d (e ) ( )e. So inegraion by pars gives [ ( )e log d ] e log + hence (23). Noe ha his inegral equaes o Γ (2). e d, Anoher inegral epression for γ can be derived direcly from (7): EXPINT4. We have γ ( ) + e d. (24) Proof. By (7), γ lim [E () log( + )] lim ( e ) + ) d. ( ( + ) e d This is he case in Dirichle s inegral epression for ψ() Γ ()/Γ() [AAR, p. 26]. Calculaion of γ. By (9), we have γ E () E() log. This has been used for he calculaion of γ o grea degrees of accuracy. For a suiably chosen, one can calculae E () and log, and esimae E() by is asympoic epansion (discussed below), or simply choose so ha E() is suiably small. For a survey of his opic, see [GS] or [BM, chap. 9]. 5

6 Asympoic epression for E() The inequaliy (2) is enough for many purposes, bu here is an asympoic epression giving a beer esimaion for large, if waned. I is helpful o describe he process more generally, as follows. Suppose ha f() and all is derivaives end o as, and wrie I f () e f() d. Inegraing by pars, we have [ ] I f () e f() + e f () d e f() + I f (). Applying his o f () and subsiuing, and repeaing he process, we obain for all k I f () e [f() + f () + + f (k ) ()] + I f (k)(). If also ( ) k f (k) () is non-negaive and decreasing, hen ( ) k I f (k)() ( ) k e f (k) (). For E(), we apply his wih f() /. The condiions are saisfied, so we obain ( E() e + 2 ) (k )! + 2 ( )k + ( ) k r 3 k k (), where r k () k! k+. For large, we obain an accurae esimaion of E() by choosing a suiable number of erms. For eample E() ce, where.9 < c <.92. However, wha his generaes is an asympoic epression, no a convergen series, because for any paricular, he successive derivaives ulimaely grow large. Some inegrals involving E() Since E () e /, we have d d [E()] E() e, so an aniderivaive of E() is E() e. Now E() as + and as, so we have E() d [ E() e ] An alernaive roue o (25) is by reversal of he implied double inegral:. (25) E() d e e e d. d d d d 6

7 We describe some furher inegrals. In each case, eiher of hese mehods can be used. We will se ou one mehod; he reader migh care o invesigae he oher one. EXPINT5. For a >, a E() d Γ(a) a. (26) Proof. Inegrae by pars: a E() d a [ a a E() ] Γ(a) a. + a e d a a e d In paricular, for posiive inegers n, n E() d n!/(n + ). EXPINT6. For a >, e a E() d log( + a). (27) a Proof. The inegral is e a e d d The saed value now follows by (). e e a d d e ( e a ) a d. EXPINT7. We have E() 2 d 2 log 2. (28) Proof. Inegrae by pars:.e() 2 d 2 [ ] E() 2 2 log 2, + 2 e E() d E() e d in which we used (27) and lim +[E() 2 ] (which follows from (9)). Noe ha we can epress E() 2 as an inegral: E() 2 2E()E () d 2 E() e d. 7

8 The double-inegral mehod hen gives again E() 2 d 2 e E() d. Applicaion o he cosine inegral Consider he inegrals corresponding o () wih e replaced by cos : C() cos d, C () cos d. (29) Convergence of he inegral defining C(), ogeher wih he fac ha lim C(), is easily esablished by inegraion by pars. We also define (following sandard noaion) Si() We will use a conour inegral o relae C () and E (), enabling us derive a resul analogous o (9) for C(). Eacly as for E(), we have where c C() C (). By (9) and (3), We will prove: sin d. C() C () log + c, (3) C() E() C () E () + c c. (3) EXPINT8 THEOREM. We have C () E () as, hence c γ and C() C () log γ, (32) cos e d. (33) Once we have shown ha C () E () as, i follows from (3) ha c c, so c γ, hence (32). Also, i follows ha C() E() as +, hence (33) (for his, we do no need o know ha c γ). EXPINT9 LEMMA. We have π/2 e R sin θ dθ π 2R. Proof. Since he funcion sin θ is concave on [, π 2θ ], we have sin θ 2 π π/2 e R sin θ dθ π/2 e 2Rθ/π dθ [ π ] π/2 2R e 2Rθ/π π 2R ( e R ). 8 on [, π ]. Hence 2

9 Proof of EXPINT8. Le C R be he circular arc of radius R in he posiive quadran, represened by z Re iθ for θ π 2. Denoe by Γ he closed conour consising of C R ogeher wih he real inerval [, R], and he imaginary ais from ir o. Le f(z) eiz. z Then f has no pole a, since f(z) i + z +. 2 f(z) dz. The conribuion of he real ais is Γ By Cauchy s inegral heorem, R e i d C (R) isi(r). The conribuion of he imaginary ais, aken owards he origin, is R e d E (R). Now consider C R. The conribuion of he erm is π i. The conribuion of he oher erm z 2 is I R : CR e iz z dz π/2 ie ireiθ dθ. Now e ireiθ e R sin θ, so by Lemma EXPINT9, I R π/(2r). Now considering he real par, we deduce C (R) E (R) π 2R, so C (R) E (R) as R. A he same ime, we deduce from he imaginary par ha Si(R) π 2 π 2R, hereby proving he sine inegral sin d π 2. An alernaive proof of EXPINT8 by double inegrals is given in [Jam3]. A direc proof of (32) can be seen in [Jam]. References [AAR] [BM] George E. Andrews, Richard Askey and Ranjan Roy, Special Funcions, Cambridge Univ. Press (999). George Boros and Vicor H. Moll, Irresisible Inegrals, Cambridge Univ. Press (24). [Fer] W. L. Ferrar, Inegral Calculus, Oford Univ. Press (958). 9

10 [GS] Xavier Gourdon and Pascal Sebah, The Euler consan γ, a hp://numbers.compuaion.free.fr/consans/consans.hml. [Jam] G. J. O. Jameson, Sine, cosine and eponenial inegrals, Mah. Gazee 99 (25), [Jam2] [Jam3] G. J. O. Jameson, The Frullani inegrals, a jameson/ G. J. O. Jameson, Using double inegrals o solve single inegrals, Mah. Gazee (26). updaed January 26

Lectures # 5 and 6: The Prime Number Theorem.

Lectures # 5 and 6: The Prime Number Theorem. Lecures # 5 and 6: The Prime Number Theorem Noah Snyder July 8, 22 Riemann s Argumen Riemann used his analyically coninued ζ-funcion o skech an argumen which would give an acual formula for π( and sugges