PROOF OF FIRST STANDARD FORM OF NONELEMENTARY FUNCTIONS


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1 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) PROOF OF FIRST STANDARD FORM OF NONELEMENTARY FUNCTIONS 1 Dharmndra Kumar Yadav, Dipak Kumar Sn ABSTRACT 1 Associat Profssor in Applid Mathmatics HMR Institut of Tchnology & Managmnt, Hamidpur, Dlhi36(India) Associat Profssor in Mathmatics R. S. Mor Collg, Govindpur, Dhanbad88109, Jharkhand (India) In th prsnt papr w hav provd on of th si standard forms of indfinit nonintgrabl functions (classically known as nonlmntary functions) and thir ampls givn by Yadav & Sn by applying strong Liouvill s thorm, its spcial cas, som wllknown nonlmntary functions and two proprtis du to Marchisotto & Zakri. Ky Words: Nonlmntary functions, strong Liouvill s thorm tc. 010 AMS Subjct Classification: 6A09, 6B I INTRODUCTION A natural qury ariss in calculus that what typ of functions cannot b intgratd? or which indfinit intgrals ar not lmntary? Th first ampl which lads us byond th rgion of lmntary functions is th lliptic intgrals du to John Wallis (1655). Such intgrals cannot b valuatd in trms of th lmntary functions was provd by Josph Liouvill in 1833 and th main rsults on functions with nonlmntary intgrals bgan with Strong Liouvill Thorm (1835) and Strong Liouvill Thorm (spcial cas, 1835). Marchisotto & Zakri (1994) studid nonlmntary functions and mntiond two important ampls 4 and 5 [5, pp ], which ar tratd as proprtis in proving th functions lmntary and nonlmntary. By applying th abov proprtis w gt th following wllknown nonlmntary functions: d, a d,(a 0), d, d, d, sin d, cos d Proofs of Nonlmntary Functions: Yadav & Sn [6] hav givn si standard forms of indfinit nonintgrabl functions out of which form1 is as follows: 1
2 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) An indfinit intgral of th form f ( ) d f '( ), whr f() is a polynomial function of dgr, or a trigonomtric (not invrs trigonomtric) function, or a hyprbolic (not invrs hyprbolic) function is always nonintgrabl i.. nonlmntary. Proof: W will prov it taking diffrnt possibl cass as follows: Cas I: Whn f() is an algbraic function (polynomial) of dgr : W hav Lt g() f ( ) f '( ) p() R() q() d g() 1 d, [Taking g() ]. From strong Liouvill thorm (spcial cas), d is lmntary if and only if thr ists a rational function R() such that g() R '() R(), whr g.c.d.(p(), q())=1. Thn w hav 1 R '() 1 R '() R() R() q()p'() p()q'() [] p()q() [q()] (1.1) p()q '() p'() q() [] p() q() Which implis q() f () as q() cannot divid p() and q (). In this cas ithr q()=k, a constant or a polynomial of dgr lss than or qual to th dgr of f (). For q()=k, from (1.1) w hav kp'() [] p()k k (1.) Comparing dgrs of in (1.) rsults out in a contradiction. Hnc q() cannot b a constant. For q() a polynomial of dgr lss than or qual to th dgr of f (), w hav sinc q() f (), lt us assum that f ()=q().h(). Thn from (1.1) q()h()q()p'() q()h()p()q'() [q()h()] p()q() [q()] h()p()q '() h()p'() 1 q()[h()] p(). (1.3) q() Which implis q() h(), sinc q() cannot divid p() and q (). Lt h()=q().ξ(). Thn from (1.3), w hav
3 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) 3 q() ()p'() p()q'() () [q()] [ ()] p() 1 (1.4) Comparing th dgrs of in both sids in (1.4) rsults out in a contradiction. Thrfor such R() dos not ist, i.., th givn function is nonlmntary. Cas II: Whn f() b a trigonomtric (not invrs trigonomtric) function: Lt us considr thm on by on. 1.1 For sin function, w hav sin () d d '()cos () whr φ() b any polynomial of dgr 1. On putting sinφ()=, w hav, sin () d d '()cos () [ '()] (1 ) (1.1.1) Subcas I: Whn φ() is linar, lt φ()=+b. Thn from (1.1.1) w hav d d [ '()] (1 ) (1 ) 1 d d (1 ) (1 ), whr p d dp,[putting1 p] (1 ) p and d 1 p dp,[putting1 p] (1 ) p. Both ar nonlmntary from ampl4 du to Marchisotto t al [5, pp.300]. Thrfor th givn function is also nonlmntary. Subcas II: Whn φ() is a polynomial of dgr. Lt us considr φ()= +b+c. Thn w hav from (1.1.1), on putting sinφ()= d [ '()] (1 ) 1 d b 4c, whr k 1 4 [sin k](1 ) 4 1 d 4 1 [sin k] (1 ) (1 ) 1 F,, 1,sin d 1 3 F, y, y, y d 3
4 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) dy1 dy dy3 1 1 y 1,, d d y 1 d 1 y Applying strong Liouvill thorm, part (b), w find that it is lmntary if and only if thr ists an idntity of th form n d 1 U 0 c i log U i i1 4[sin k](1 ) d n du0 U' i c 1 i 4[sin k](1 ) d i1 Ui whr ach U j is a function of, y 1, y, and y 3. Considring diffrnt forms of U j lik 1 1 log[(sin k) ], log(sin k) w find that no such U j ist. Hnc th givn function is nonlmntary. Similarly w can prov it nonlmntary for highr dgr polynomials φ()., 1. For cosin function, w hav cos () d d '()sin () whr φ() b any polynomial of dgr 1. On putting cosφ()=, w hav, cos () d d d '()sin () [ '()] (1 ) [ '()] (1 ) (1..1) Subcas I: Whn φ() is linar, lt φ()=+b. Thn from (1..1) w hav d d [ '()] (1 ) (1 ) which is nonlmntary provd in sction 1.1 subcasi. Subcas II: Whn φ() is a polynomial of dgr. Thn from (1..1) w hav d 1 d 1 [ '()] (1 ) 4 (cos k)(1 ) 1 F[,, 1,cos ]d A similar argumnt will hold as in sction 1.1 to prov it nonlmntary., 4
5 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) 1.3. For tangnt function, w hav on putting tanφ()=. tan () d d '()sc () d [ '()] (1 ). (1.3.1) Subcas I: Whn φ() is linar, lt φ()=+b. Thn from (1.3.1) w hav d d [ '()] (1 ) (1 ) 1 d d (1 i) (1 i) Now i ip d dp, putting (1 i) p (1 i) i p By strong Liouvill thorm (spcial cas), it is lmntary if and only if thr ists a rational function R() such that it satisfis th idntity 1 R '(p) ir(p) p 1 R(p) 0 and R '(p) p But R(p) cannot b ro, so such R(p) dos not ist. Hnc it is nonlmntary. Also d (1 i) p ip i i dp, putting (1 i) p Again by strong Liouvill thorm (spcial cas), it is lmntary if and only if thr ists a rational function R() which satisfis th idntity 1 R '(p) ir(p) p 1 R(p) 0 and R '(p) p But R(p) cannot b ro, so such R(p) dos not ist. Hnc it is nonlmntary. Thrfor th givn function is nonlmntary in this cas. Subcas II: Whn φ()= +b+c. Thn from (1.3.1) w hav d 1 d b 4c,k 1 [ '()] (1 ) 4 [tan k](1 ) 4 1 F[,,(1 ),tan ]d 1 3 F[, y, y, y ]d dy1 dy dy3 1 1 y 1,, d d d 1 y 5
6 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) By strong Liouvill thorm part (b), it is lmntary if and only if thr ists an idntity of th form containing U j, a function of, y 1, y, and y 3 du U i' ci 4[tan k](1 ) d U n i 1 i1 Considring diffrnt forms of U j lik log(tan 1 +k), log[ (tan 1 +k)], tc. w find that no such U j ist, i.., no such idntity ist. Hnc th givn function is nonlmntary. Similarly w can prov it nonlmntary for highr dgr polynomials φ(). i 1.4. For cotangnt function, w hav on putting cotφ()= cot () d d '()cosc () d [ '()] (1 ).. (1.4.1) SubcasI: For φ()=+b, w hav from (1.4.1) d d [ '()] (1 ) (1 ) Which is nonlmntary, provd in sction 1.3, subcasi. SubcasII: For φ()= +b+c, w hav from (1.4.1) d 1 d b 4c,k 1 [ '()] (1 ) 4 (cot k)(1 ) 4 1 F[,,(1 ),cot ]d A similar argumnt will hold as in sction 1.3 to prov it nonlmntary For coscant function, w hav on putting coscφ()= cosc () d d '()cosc ()cot () d [ '()] ( 1) (1.5.1) Subcas I: Whn φ() is linar, lt φ()=+b. Thn from (1.5.1) w hav d d [ '()] ( 1) ( 1) d d ( 1) 6
7 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) Whr th first intgral d ( 1) is nonlmntary as provd in sction 1.1, subcasi and th scond intgral d d is also nonlmntary from ampl5 du to Marchisotto t al [5, pp.301]. Subcas II: Whn φ()= +b+c, thn from (1.5.1) w hav d [ '()] ( 1) d [ b] ( 1) d b 4c,k 1 4[cos c k] ( 1) 4 1 F[,, 1,cosc ]d 1 3 F[, y, y, y ]d dy1 dy dy3 1 1 y 1,, d d y 1 d 1 y By strong Liouvill thorm part (b), this is lmntary if and only if thr ists an idntity of th form containg U j, a function of, y 1, y, and y 3 as follows du U i' ci 4[cos c k] ( 1) d U n i 1 i1 Considring diffrnt forms of U j lik log[cosc 1 +k], log[ (cosc 1 +k)], tc., w find that no such U j ist, which satisfy th abov idntity. Hnc th givn function is nonlmntary. Similarly w can prov it for highr dgr polynomial φ() For scant function, w hav on putting scφ()= i sc () d d '()sc () tan () SubcasI: For φ()=+b, w hav from (1.6.1) d [ '()] ( 1). (1.6.1) d d [ '()] ( 1) ( 1) 7
8 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) Which is nonlmntary, provd in sction 1.5, subcasi. SubcasII: For φ()= +b+c, w hav from (1.6.1) d 1 d b 4c,k 1 [ '()] ( 1) 4 [sc k] ( 1) 4 1 F[,, 1,sc ]d It can b provd nonlmntary by th similar procdur as has bn don in sction 1.5. Cas III: Whn f() b a hyprbolic (not invrs hyprbolic) function. Lt us considr thm on by on For sin hyprbolic function, w hav on putting sinhφ()= sinh () d d '()cosh () d [ '()] (1 ) (1.7.1) Subcas I: Whn φ() is linar, lt φ()=+b. Thn from (1.7.1) w hav d d [ '()] (1 ) (1 ) which is nonlmntary, provd in sction 1.3, subcasi. Subcas II: Whn φ()= +b+c. Thn from (1.7.1), w hav d d [ '()] (1 ) [ b] (1 ) d b 4c, whr k 1 4(sinh k)(1 ) 4 1 F[,, 1,sinh ]d 1 3 F[, y, y, y ]d dy1 dy dy3 1 1 y 1,, d d y 1 d 1 y Applying strong Liouvill thorm part (b), it is lmntary if and only if thr ists an idntity of th form du U' n 0 i c 1 i 4(sinh k)(1 ) d i1 Ui Considring diffrnt possibl forms of U j lik log[sinh 1 +k], log[ (sinh 1 +k)], tc. w find that no such U j ist. Hnc th givn function is nonlmntary. Similarly w can prov it nonlmntary for highr dgr polynomials.. 8
9 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) 1.8. For cosin hyprbolic function, w hav on putting coshφ()= cosh () d d '()sinh () SubcasI: For φ()=+b, w hav from (1.8.1) d [ '()] ( 1). (1.8.1) Which is nonlmntary provd in sction 1.1, subcasi. SubcasII: For φ()= +b+c, w hav from (1.8.1) d d [ '()] ( 1) ( 1) d 1 d b 4c,k 1 [ '()] ( 1) 4 (cosh k)( 1) 4 1 F[,, 1, cosh ]d It can now b provd nonlmntary by strong Liouvill thorm part (b). Similarly w can prov it for highr dgr polynomial φ() For tangnt hyprbolic function, w hav on putting tanhφ()= tanh () d d '()sch () SubcasI: For φ()=+b, w hav from (1.9.1) d [ '()] (1 ) (1.9.1) d d [ '()] (1 ) (1 ) Which is nonlmntary, provd in sction 1.1, subcasi. SubcasII: For φ()= +b+c, w hav from (1.9.1) d 1 d b 4c,k 1 [ '()] (1 ) 4 (tanh k)(1 ) 4 F[,,(1 ),tanh 1 It can now b provd nonlmntary by strong Liouvill thorm part (b). Similarly w can prov it for highr dgr polynomial φ(). ]d 9
10 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) For cotangnt hyprbolic function, w hav on putting cothφ()= SubcasI: For φ()=+b, w hav from (1.10.1) cot h() d d '()cosch () d [ '()] ( 1).. (1.10.1) d [ '()] ( 1) d ( 1) Which is nonlmntary provd in sction 1.1, subcasi. SubcasII: For φ()= +b+c, w hav from (1.10.1) d 1 d b 4c,k 1 [ '()] ( 1) 4 (coth k)( 1) 4 1 F[,,( 1),coth ]d It can now b provd nonlmntary by strong Liouvill thorm part (b). Similarly w can prov it for highr dgr polynomial φ() For coscant hyprbolic function, w hav on putting coschφ()= cosch() d d '()cosch ()coth () SubcasI: For φ()=+b, w hav from (1.11.1) d [ '()] ( 1) (1.11.1) d d [ '()] ( 1) ( 1) d d 1 Both ar nonlmntary provd in sction 1.5, subcasi and sction 1.7, subcasi rspctivly. SubcasII: For φ()= +b+c, w hav from (1.11.1) d 1 d b 4c,k 1 [ '()] ( 1) 4 (cos ch k) ( 1) 4 1 F[,, 1,cosch ]d It can now b provd nonlmntary by strong Liouvill thorm part (b). Similarly w can prov it for highr dgr polynomial φ() For scant hyprbolic function, w hav on putting schφ()= 10
11 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) sch () d d '()sch () tanh () SubcasI: For φ()=+b, w hav from (1.1.1) d [ '()] (1 ) (1.1.1) d d [ '()] (1 ) (1 ) d d 1 Both ar nonlmntary provd in sction 1.5, subcasi and sction 1.1, subcasi rspctivly. SubcasII: For φ()= +b+c, w hav from (1.1.1) d 1 d 1 [ '()] (1 ) 4 (sch k) (1 ) 1 F[,, 1,sch ]d It can now b provd nonlmntary by strong Liouvill thorm part (b). Similarly w can prov it for highr dgr polynomial φ(). Lt us considr som ampls on this standard form of nonlmntary functions: a b Eampl 1: Show that th intgral d, a 0 is nonlmntary. Proof: W hav a b a b d d d Now for scond intgral, putting a = w hav b log a a a d a a d d d a which is nonlmntary from ampl5 du to Marchisotto t al [5, pp.301]. Eampl : Show that th intgral Proof: W hav sin d cos sin sin cos d cos cos d is nonlmntary. d (1 ) On putting sin=. Which is nonlmntary, provd in sction 1.1, subcasi. cos Eampl 3: Show that th intgral d is nonlmntary. sin 11
12 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) Proof: W hav cos cos ( sin ) d sin ( sin ) d On putting cos=. Which is nonlmntary provd in sction 1.1, subcasi. d (1 ) Eampl 4: Show that th intgral sc tan d is nonlmntary. Proof: W hav, on putting tan= tan 1 d 1 d d d sc (1 ) 4 (i)(1 i) 4 (i)(1 i) (A) W hav on putting 1i = p in th first intgral of (A) ip ip ip d i dp dp dp i (i)(1 i) (1 p) p p.. (B) whr th scond and third intgrals ar nonlmntary from ampl5 du to Marchisotto t al [5, pp.301]. Now putting 1p=X in th first intgral of (B) w hav ip ix dp i dx (1 p) X which is also nonlmntary from ampl5 du to Marchisotto t al [5, pp.301]. Thrfor th first intgral of (A) is nonlmntary. Similarly w can prov that th scond intgral of (A) is also nonlmntary. Thrfor th givn function is nonlmntary. sinh Eampl 5: Show that th intgral d is nonlmntary. cosh Proof: W hav on putting sinh= sinh d cosh (1 ) Which is nonlmntary provd in sction 1.3, subcasi. cot Eampl 6: Show that th intgral d is nonlmntary. cosc Proof: W hav on putting =cot cot d Which is nonlmntary provd in sction 1.3, subcasi. d cosc (1 ) d 1
13 Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) sc Eampl 7: Show that th intgral d sc.tan is nonlmntary. Proof: W hav on putting sc= sc d sc.tan ( 1) d d d ( 1) Which ar nonlmntary, provd in sction 1.5, subcasi. cosc Eampl 8: Show that th intgral d is nonlmntary. cos c.cot Proof: W hav on putting cosc=, cosc d cos c.cot ( 1) Which is nonlmntary, provd in sction 1.5, subcasi. d d d ( 1) sin Eampl 9: Show that th intgral d sin is nonlmntary. Proof: W hav on putting sin =, sin 1 d d sin 4 (1 ) 1 d d 4 (1 ) Whr Now sinc Whr and d is nonlmntary from ampl5 du to Marchisotto t al [5, pp.301]. 1 d 1 d d (1 ) (1 ) (1 ) I p d dp (1 ), on putting 1=p, which is nonlmntary p d 1 p dp on putting 1+=p, which is also nonlmntary (1 ) p from ampl5 du to Marchisotto t al [5, pp.301]. Hnc th givn function is nonlmntary. sin Eampl 10: Show that th intgral.cos d is nonlmntary. 13
14 [ Intrnational Journal Of Advanc Rsarch In Scinc And Enginring IJARSE, Vol. No., Issu No., Fbruary, 013 ISSN (E) Proof: W hav on putting sin =, sin d d 1.cos 4(1 )sin 1 F,, 1,sin d 1 3 F, y, y, y d dy1 dy dy3 1 1 y 1,, d d y 1 d 1 y Applying strong Liouvill thorm, part(b), it is lmntary if and only if thr ists an idntity of th form, containing U i a function of, y 1, y, and y 3 as n duo U' i Ci 1 i1 i d U (1 )sin Taking diffrnt possibl forms of U j w find that no such U j ist. Hnc th givn function is nonlmntary. Rfrncs [1]. Hardy G. H., Th Intgration of Functions of a Singl Variabl, nd Ed., Cambridg Univrsity Prss, London, Rprint 198, 1916 []. Ritt J. F., Intgration in Finit Trms: Liouvill s Thory of Elmntary Mthods, Columbia Univrsity Prss, Nw York, 1948 [3]. Risch R. H., Th Problm of Intgration in Finit Trms, Transactions of th Amrican Mathmatical Socity, 139, , 1969 [4]. Rosnlicht M., Intgration in Finit Trms, Th Amrican Mathmatical Monthly, 79:9, , 197 [5]. Marchisotto E. A. & Zakri G. A., An Invitation to Intgration in Finit Trms, Th Collg Mathmatics Journal, Mathmatical Association of Amrica, 5:4, , 1994 [6]. Yadav D. K. & Sn D. K., Rvisd papr on Indfinit Nonintgrabl Functions, Acta Cincia Indica, 34:3, ,
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