Evaluation of some non-elementary integrals involving sine, cosine, exponential and logarithmic integrals: Part II Текст научной статьи по специальности «Математика»

URAL MATHEMATICAL JOURNAL, 2018, Vol. 4, No. 1, pp. 43-55

DOI: 10.15826/umj.2018.1.004

EVALUATION OF SOME NON-ELEMENTARY INTEGRALS INVOLVING SINE, COSINE, EXPONENTIAL AND LOGARITHMIC INTEGRALS: PART II

Victor Nijimbere

School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada victornijimbere@gmail.com

Abstract: The non-elementary integrals = J[sin(Ax^)/(Axa)]dx, 3 > 1, a > 3 +1 and =

J[cos(Ax^)/(Axa)]dx, 3 > 1, a > 2, + 1, where {3,a} € R, are evaluated in terms of the hypergeometric function 2F3. On the other hand, the exponential integral = /xa)dx, 3 > 1, a > 3 + 1is

expressed in terms of 2F2. The method used to evaluate these integrals consists of expanding the integrand as a Taylor series and integrating the series term by term.

Key words: Non-elementary integrals, Sine integral, Cosine integral, Exponential integral, Hyperbolic sine integral, Hyperbolic cosine integral, Hypergeometric functions.

1. Introduction

Let us first give the definition of the non-elementary integral. This definition is also given in Part I [6], we repeat it here for reference.

Definition 1. An elementary function is a function of one variable constructed using that variable and constants, and by performing a finite number of repeated algebraic operations involving exponentials and logarithms. An indefinite integral which can be expressed in terms of elementary functions is an elementary integral. And if, on the other hand, it cannot be evaluated in terms of elementary functions, then it is non-elementary [4, 9].

The cases consisting of the non-elementary integrals Si,,a = /[sin (Ax3)/(Axa)]dx, p > 1, a < p + 1 and Ci3,a = f [cos (Ax3)/(Axa)]dx, p > 1, a < 2p + 1, where {p,a} € R, were considered and evaluated in terms of the hypergeometric functions 1F2 and 2F3 in Part I [6], and their asymptotic expressions for |x| ^ 1 were derived too in Part I [6]. The exponential integral Ei,,a = f(eAx^/xa)dx where p > 1 and a < p + 1 was expressed in terms of 2F2, and its asymptotic expression for |x| » 1 was derived as well in Part I [6].

Here, we investigate other cases which were not treated neither in Part I [6] nor elsewhere. We evaluate Si,,a = J[sin(Ax3)/(Axa)]dx, p > 1, a > p + 1 and Ci,,a = J[cos(Ax3)/(Axa)]dx, P > 1, a > 2p + 1 and Ei,,a = f(eXx/S /xa)dx, p > 1, a > p + 1. In order to take into account all possibilities, we write these integrals as Si,,3+a = /[sin (Ax3)/(Ax3+a)]dx, p > 1, a > 1, Ci3,23+a = f [cos (Ax3)/(Ax23+a)]dx, p > 1,a > 1, and Ei3,3+a = f (eAx^/x3+a)dx, p > 1, a > 1 where {p,a} € R. On one hand, Si3,3+a and Ci3,23+a are expressed in terms of the hypergeometric function 2F3, while on another hand, Ei3,3+a is expressed in terms of the hypergeometric function 2F2.

These integrals involving a power function x3 in the argument of the numerator are the generalizations of the exponential, sine and cosine integrals in [7] (see sections 8.19 and 8.21 respectively),

which have applications in different fields in science, applied sciences and engineering including physics, nuclear technology, mathematics, probability, statistics, and so on. For instance, the generalized exponential integral E1)1+a is used in fluidodynamics and transport theory, where it is applied to the solution of Milne's integral equations [2], there are also used in modeling radiative transfer processes in the atmosphere and in nuclear reactors [10], etc. Exponential asymptotics involving generalized exponential integrals are used in probability theory, see for example [3]. On the hand, generalized sine and cosine integrals are frequently utilized in Fourier analysis and related domains [8]. Therefore, we are justified to further generalize these functions and their connections to hypergeometric functions.

Before we proceed to the main objectives of this paper consisting of evaluating the above interesting cases of non-elementary integrals (see sections 2, 3 and 4), we first define the generalized hypergeometric function as it is an important tool that we are going to use in the paper.

Definition 2. The generalized hypergeometric function, denoted as pFq, is a special function given by the series [1, 7]

rw 7 7 7 \ (a1)n(a2)n ■ ■ ■ (ap)n xn

(b1 )n (b2)n ■■■ (bq )n n\

where a1,a2, ■ ■ ■ , ap and ; b1,b2, ■ ■ ■ ,bq are arbitrary constants, ($)n = + n)/r($) (Pochham-mer's notation [1, 7]) for any complex •&, with ($)0 = 1, and r is the standard gamma, function [1].

2. Evaluation of the sine integral [ > 1, a > 1

Theorem 1. Let ft > 1 and a > 1, and let a = mft + e, where m is an integer (m € N) and

—ft < e < ft.

1. If e = 0, then

sin(Xx3) ^ , lAn+m X2n+2m x23n+1

%/?+« = f^* = Y, ("!)

n=-m

\m \ 2m.

r(2n + 2m + 2) 2ftn + 1

(—1)mX2mx / 1 3 1 X2 x23

+ 2^W(m+ l)r (ml 3/2) (2/?+ 1) ^ I,1'1 + + + ^ "^J +

(2.1)

where m = a/ft. 2. If e = 1, then

/sjn(\x3) X2m — 1 X 2n+2m x23n

{ N + ragm(-1)ra+mr(2n + 2m + 2) 2^

(-1)m+1X2m+2x23 / 5 x2 x23\ ^

(2.2)

where m = (a — 1)/ft.

3. Finally, if e € (-p, 0) U (0,1) U (1, p), we have

i sin (Ax3) A2m ™1—e — 1 \2ra+2m ™23«_e+1

I sin^AX ' _ ^ -yyn A___|_ A x

J Ax3+a v ' r(2m + 2) 1 - e n=-m r(2n + 2m + 2) 2pn - e + 1

(_ 1)m+1 A2m+2x23-e+1

+ 22m+3y^T(m + 2)r (m + 5/2) (2/3 - e + 1) X / 1 - e 5 1 - e A2x23\ ^

(2.3)

where m = (a - e)/p.

Proof. We proceed as in [5, 6]. We expand g(x) as Taylor series and integrate the series term by term. We use the gamma duplication formula[1], the gamma property r(a + 1) = ar(a) and Pochhammer's notation (see Definition 2). We also set a = mp + e, and then we obtain

fsir^3) i —l— ^ (Ax3)2ra+1 _ f- \2n ,n_

J \xJ3+a ax-J ij (277 + 1)! ax~ J j (277,+ 1)!* ax

fm-1 \2ra f ™ \2ra

EC IV" 2/3(ra-m)-ej . / V^ / -i yra A 2(3(n-m)-tj

1 j (277, + 1)! aX+ ^ ' (277 + 1)! ^

n=0 v y n=m v y

• -1 \2ra+2m /• ^ \2ra+2m

V {-i)n+m-—---x2fin-€dx + / V(-l)ra+m7—---x2f}n~€dx

J (2n + 2m + 1)! J ^ ' (2n + 2m +1)!

n= — m v j -j ra=0 v '

-1 \2ra+2m /• ^ \2ra+2m

^ v 7 r(2n + 2m + 2) J ^ ' r(2n + 2m + 2)

n=- m v j -j ra=0 v '

\ 2m f I" \2ra+2m

= ( pm A / _ i / ( jY»+m A_ 2Pn-e^

y ' PC9m -1-9^ / y ' TC9n -I-9m -I-9V

, m— 1 , 2n ^ ro

r(2m + 2W xe J ^ v ' r(2n + 2m + 2)'

A2n+2m

\ " c -1)ra+m_-_x2,3n~edx

^ ' r(2n + 2m + 2)

n=1

A 2m r dx f _1 A2n+2m

— C Dm / _ i / C 1 V"+m _ 2/3ra-ej

j r(2m + 2) J x.t + J ¿J ' T(2n + 2m + 2)X

(■ ^ A 2n+2m+2

I ; r(2n + 2m + 4)

n=0

A2m [■ _1 A2n+2m ™23n—e+1

\ '«+m

r(2m + 2) 7 ' ^ v r(2n + 2m + 2) 2/3n -e + 1

n=—m

^ A2n+2m+2 ™23ra+23—e+1 I V^c iyra+m+1 A__±__| p

' T(2n + 2m + 4) 2/3n + 2/i - e + 1

n=0

A2m p dx _1 x2ra+2m ™23«_e+1

_ / i \m A_ I , \ ^ /_i y«+m______

~ r(2m + 2) J ; r(2n + 2m + 2) 2/3n -e + 1

(2.4)

(-ir+1A ~ (l)ra(l+(l-e)/(2ft))ra (—A2.t2/3/4)"

+ 22m+3 0fr(m+2)r (m+5/2) (2ft-e+l) ^ (m+2)„ (m+5/2)„ (2+(l-e)/(2ft))„ n! + 1

x2m r dx 1 \2n+2m ™23n—e+1 _ I i \m X I dx | \ i i \n+m X x

TV9m + 91 / ^^

r(2m + 2) 7 xe n=—m r(2n + 2m + 2) 2ftn — e + 1

( — 1)m+1x2m+2x23—e+1

+ 22m+3 v/vFri'm + 2)T (m + 5/2) (2ft - e + 1) * ^ ( 1 — e 5 1 — e X2x23\ ^

1. For e = 0, we substitute e = 0 in (2.4), and hence, we obtain

[ sin(A..3)dr= f 1 ^ iy»(^)2ra+1ir

/ \rrfirra ' (On i

J Xx3+a J Xx3xa y (2n + 1)!

n=0 v '

— 1 \ 2n+2m r ^ \2n+2m

£ (-i)*t*'r,- j.j.+ /D-1)"^,,,^,^

r(2n + 2m + 2) J ' r(2n + 2m + 2)

n= -m n=o

— 1 \2n+2m x23n+1 ^ \2n+2m x23n+1

X . v -> , „ \ n_|_m A X

E( nn+m_____I V^/]^

V ' ro.n + 9 m + 91 9 fin. + 1 '

r(2n + 2m + 2) 2ftn + 1 ^=0 r(2n + 2m + 2) 2ftn + 1

X2n+2m x23n+1

n=o

-1

yra+m

E (—1)

r(2n + 2m + 2) 2ftn + 1

_(—l)mA2m.r_ ~ (l)ra (1 + 1/(2ft))ra (-AV3/4)'

2m+10F(2/i + l)r(m + l)r (m + 3/2) (m + 1)„ (m + 3/2)„ (2 + 1/(2/3))„ n!

-1 X2n+2m x23n+1 _ ^ ^ ^_-^yra+m______

n= -m

m 2m

r(2n + 2m + 2) 2ftn + 1

(—1)m X2mx / 1 3 1 X2x23

+ 2^Ar(m + l)r(m + 3/2)(2/j + l) i1'1 + 20;m +1>m +2'2 +20;~— ' +C'

which is (2.1), and where m = a/ft. 2. For e = 1, we set e = 1 in (2) and obtain

/ain(\x3) X 2m _1 \ 2n+2m x23n

1 Jdr-(-l)m_-_In Irl + V (-l)n+m_-_-_

Xx,3+a l> T{2m + 2) m+ ¿J ' T(2n + 2m, + 2)2f3n

(—1)m+1 X2m+2x23 / 5 X2x23\ ^

+ 2^Vir(m + 2)r (m + 5/2) ft ^ ^ m + 2'm + 2'2; " —) +

which is (2.2), and where m = (a — 1)/ft. 3. For e € (—ft, 0) U (0,1) U (1,ft), (2) gives

r oin(\x3) X2m x1—e — 1 \2n+2m x23n—e+1 / °1U \AX ! j _ / i \m A_ x__| V^ / i yra+m A__x

/ W/3+« ~ 1 ; Ff9m -1-911 — c 2-^t 1 ' 'VO.n -I- 9m

Xx3+a v 7 r(2m + 2) 1 — e „=—m r(2n + 2m + 2) 2ftn — e + 1

+-

n= - m

( —1)m+1 X2m+2x23—e+1

22m+3 0fr('m + 2)r (m + 5/2) (2ft - e + 1)

1 — e 5 ^ 1 — e X2 x23\ ^

X2F3[l>l + ^-;m + 2>m + ->2 + —J+C,

n= -m

n= -m

which is (2.3), and where m = (a - e)/p. □

Example 1. In this example, we evaluate / [sin(x2)/x3-5] dx. We first observe that A = 1 and p = 2. We also have 3.5 = p + a = 2 + 1.5 = 2 + (1)2 - 0.5 = p + mp + e, and so m = 1 and e = -0.5. Substituting A = 1,p = 2, m = 1 and e = -0.5 in (2.3) gives

f sin(x2) , x1-5 x—2-5 x5-5 / 9 7 17 x4\ ^

We can use the same procedure for the hyperbolic sine integral, the results are stated in the following theorem. Its proof is similar to that of Theorem 1, we will omit it.

Theorem 2. Let p > 1 and a > 1, and let a = mp + e, where m is an integer (m € N) and

-p < e < p.

1. If e = 0, then

f sinh (Ax3) v1 A2n+2m x23n+1

-dx = ^

J Ax3+a ^ r(2n + 2m + 2) 2pn + 1

A2mx 1 3 1 A2x23

+ 2^ Ar(m + l)r(m + 3/2)(2/j + l) ^ ^ 1 + + + ~J +

where m = a/p. 2. If e = 1, then

f sinh (Ax3) , A2m , . . ^^ A2n+2m x23n

/ —;—it,-dx = —-- In s + > —--——

J Ax3+a r(2m + 2) 1 1 =_ r(2n + 2m + 2) 2pn

A2m+2x23 / 5 A2x23\ ^

+ 22^ Ar(m + 2)r(m + 5/2)/j ^ m + 2, m + - , 2; — j + C,

where m = (a - 1)/p. 3. Finally, if e € (-p, 0) U (0,1) U (1,p), we have

sinh (Ax3) , A2m x1—e V1 A2n+2m x23n—e+1 ----dx =---h >--

\xJ3+a " r(2m + 2) 1 - e T(2n + 2m + 2) 2/3??, - e + 1

A2m+2 x23—e+1 j__L_x

22m+3 v/vFr(m + 2)T (m + 5/2) (2/3 - e + 1) / 1 - e 5 1 - e A2x23\ ^

where m = (a - e)/p.

3. Evaluation of the cosine integral Ci3,23+a, [ > 1, a > 1

Theorem 3. Let ft > 1 and a > 1, and let a = 2ftm + e, where m is an integer (m € N) and

—2ft < e < 2ft.

1. If e = 0, then

COS (Xx'3) _ 1 .T1"23"0 , -nn+m+l A2ra+2m+1 x2'3n+1

CilV2l3+a = / \J+a dx = T~ + ("1)

Xx23+a X 1 — 2ft — a ^ v 7 r(2n + 2m + 3) 2ftn + 1

n= — m

(—1)mX2mx /13 1 X2x23

+ 2»^ Ar(m + 3/2)r(m + 2)(2/i + l) 2i?3 I,1' 1+2ft; ™ + 2' m+2' 2+2ft;--~) +C'

(3.5)

where m = a/(2ft). 2. If e = 1, then

fcosiXx'3) 1 .t1-23-» (—l)mA2m+1 y , iyra+m+i A2ra+2m+1 ,t23»

J Xx2'3+a XI-2fi- a r(2m + 3) 11 ; r(2n + 2m + 3) 2 fin

(—1)m+1x2m+3x23 ( 5 X2x23\ ^

+ 2^ Ar(m + 5/2)r(m + 3)/j m + 2'''m + 3' 2; ~T"J +

(3.6)

where m = (a — 1)/(2ft). 3. Finally, if e € (—2ft, 0) U (0,1) U (1, 2ft), we have

cos (Xx3) 1 x1—23—a (—1)mx2m+1 x

1—e

1 x ( 1) X x dx = — --—--h

Xx23+a X 1 — 2ft — a r(2m + 3) 1 — e

— 1 X 2n+2m+1 x23n—e+1 ^ jjre+m+1 A__±_

+

r(2n + 2m + 3) 2ftn — e + 1

(_1)m+1 x2m+3x23—e+1

_____x

22m+4 v/vFr (m + 5/2) T(m + 3)(2/3 - e + 1)

(3.7)

1 — e 5 1 — e X2x23

x2F3^l + —;m + -,m + 3,2 + —J+C, where m = (a — e)/(2ft).

Proof. We proceed as in Theorem 1. We have

f cos (Xx1'3) _ f 1 ^ lT^2ndr

J Xx2/3+a J Xx2/3+a } (2n)!

f 1 1 f ^ \2n

Xx23+a X J ' (2n)!

n=1

1 1 /• ^ X 2n+2 1 -dr + - / Vf-l)ra+1—__2/3)i-o J

J Xx23+a Xj ^ 7 (2n + 2)!

n=o

1 /•m—1 X 2n+1 /• ^ X 2n+1

1 X X

x d,x+ V(-l)ra+17^--x2l3n-2l3m-"dx+ V(-ir+17^--x2f3n-2f3m-edx

Xx2>3+a J ; (277+2)! J ' (2n+2)!

1 r m—1 A2n+1 r ^ A2n+1

dx+ (-1

Ml I * X t II -T Ml II X t II

A" J J (277+2)!' " J ^ (277+2)!

n=0 n=m v '

f 1 — 1 a 2n+2m+1

___rlr i / V^ ('_lV«+m+1___ 2/3ra-ej

Xx2,3+a aX + J l> (2n + 2777 + 2)! *

n — — m v '

n=—m

^ A 2n+2m+1 i / iyra+m+1 _ 2/3ra-ej

+ I ij (277 + 2/77 + 2)! ^

n=0

_1

1 f —1 a 2n+2m+1

+ / Y^ (i\ra+m+l_-_ 2/3ra-ej

J Ax23+a J ^ v y r(2n + 2m + 3)'

\2n+2m+1

Y^ ( — 1) «+«1+1 A_2/3n-e^

^ ' T(277 + 2777 + 3)'

n=0 v '

1 A2m+1 r dx r _L A2n+2m+1 ___^xfj^+l^_ I ___ I I f_iy»+m+l_^_ 2/3n-ej

Xx2l3+* ax + ^ r(2m + 3) J xe + J J^J l> r(2n + 2m + 3)

\2n+2m+1

Y^ ( 1 V"+m+l _ 2/3ra-e j

^ ' r(277 + I'Til + 3) ax

1 A2m+1 f dx f _L A2n+2m+1

ax + [ l) TO.rn+M l> re?,! ax

Ax23+a 7 r(2m + 3W xe J ^ v 7 r(2n + 2m + 3)

\2n+2m+3 t A_

T(277 + 2m + 5)

\2ra+2m+3

yra+m A_ 2/3ra+2/3-e j

2-J* ij m«. + 9m + M * aX

n=0

1 x1-23—a \2m+1 f dx —1 \2n+2m+1 ™23ra—e+1

1 x A / dx \ > , ^ N n+m+1 a x

A 1 - 2p - a v ' r(2m + 3^ xe y ' r(2n + 2m + 3) 2pn - e + 1

n=—m

^ \2n+2m+3 ™23ra+23—e+1

I Y"V ]\n+m A_____I n

' r(277 + 2m + 5) 2/377 + 2/3 - e + 1

n=0

1 x1 —23—a \2m+1 r dx —1 \2n+2m+1 ™23ra—e+1

1 x ./-\m + 1 A / dx \ > / ^ xn+m+1 A x

, / 1 w+1 A2m+1 f dx

A 1 - 2p - a v ' r(2m + 3) J xe ^t—m r(2n + 2m + 3) 2pn - e + 1

(_1)m+iA2m+3x2/3-e+i ~ (l)ra(l+(l-e)/(2/3))ra (-AV3/4)"

22m+4 (m+5/2) r(m+3)(2/3—e+1) ^ (m+5/2)„ (m+3)„ (2+(l-e)/(2/3))„ 77! 1

1 x1—23—a \2m+1 f dx — 1 \2n+2m+1 ™23ra— e+1

1 x / i\m A / dx 1 \ ^ / i\ra+m+1 A x

A 1 - 2p - a v ' r(2m + 3^ xe n=—m r(2n + 2m + 3) 2pn - e + 1

(-1)m+1A2m+3x23—e+1 / 1-e 5 1-e A2x23\ ^ + 22m+4 Ar (m + 5/2) r(m + 3)(2/3 - e + 1) ^ m+2' m+3'--^J

(3.8)

1. For e = 0, we substitute e = 0 in (3.8), and hence, we obtain

—1

Xx2/3+a - J xx2/3+a ^ J Y, r(2n + 2m + 3)'

\2n+2m+1 1 x1—23—c

COS (A.T3) _ f dx f iy»+m+l A2ra+2m+1 20nj

\T2/3+aaX- W2/3+« / Z^ ^ ^ m«. + 9m + .TT

n=—m

y,______I I ^ T1*

V^/ lY»+m+l A_r23'Vr - -——

j r(277 + 2m + 3) A 1 - 2/3 - a

n=0

— 1 X 2n+2m+1 x23n+1 ^ \2n+2m+1 x23n+1

, / i yra+m+1 A__±__| V"V lV»+™+l____

^ r(2n + 2m + 3) 2fin + 1 ^ r(2n + 2m + 3) 2fin + 1

n=—m n=o

1 x1—23—a — 1 X 2n+2m+1 x23n+1

_ ____I ( i yra+m+l A__±_

~ A 1 - 2/3 - a ^ 1 ' r(2n + 2m + 3) 2fin + 1

n= — m

(—1)mX2mx ^ (1)n (1 + 1/(2ft))n (—X2x23/4)'

9m+2 , AFfOfi + 1 ir (m + 3 /91 rim + 91

2m+20F(2ft + l)r (m + 3/2) r(m + 2) ^ (m + 3/2)„ (m + 2)n (2 + l/(2/3))„ n\

1 x1—23—a — 1 x2n+2m+1 x23n+1

1 x x " ' - * ra+m+1 A x

+ E ( —1)

X 1 — 2ft — a ' r(2n + 2m + 3) 2ftn + 1

n= — m

(—1)mX2mx /13 1 X2x23\

''........2'",+2'2 + M;—r

+!SWM»TWTli 2Fs V1'1 + Wr'm + 7r''+ 2':2 + Wr'~~) + c"

which is (3.5), and where m = a/(2ft). 2. For e = 1, we set e = 1 in (3.8) and obtain

f cos(A.r3) 1 .t1-23-" (—l)mA2m+1 . . ^ , nra+m+1 X2n+2m+l x2f3n J Xx2/3+a A 1 - 2 p-a T(2m + 3) 11 ¿J ' T(2n + 2 m + 3) 2 pn

(—1)m+1X2m+3 x23 / 5 X2 x23\ ^

+ 2^ Ar(m + 5/2)r(m + 3)/j m + 2'''m + 3' 2; ~T" ) +

which is (3.6), and where m = (a — 1)/(2ft).

3. For e € (—2ft, 0) U (0,1) U (1, 2ft), (3.8) gives

[ cos (Xx3) , 1 x1—23—a (—1)mX2m+1 x1—e -dx = ---—--h

Xx23+a X 1 — 2ft — a r(2m + 3) 1 — e —1

+

— 1 x2n+2m+1 x23n—e+1

r(2n + 2m + 3) 2f3n -e + 1

(_1)m+1 X2m+3 x23—e+1

_____x

22m+4 v/viT (m + 5/2) r(m + 3)(2/3 - e + 1)

/ 1 — e 5 1 — e X2x23\ ^

which is (3.7), and where m = (a — e)/(2ft). □

Example 2. In this example, we evaluate / [cos(x)/x5] dx. We first observe that X = 1 and ft = 1. We also have 5 = 2ft + a = 2 + 3 = 2 + 2(1)(1) + 1 = ft + 2ftm + e, and so m = 1 and e = 1.

Substituting X = 1, ft = 1, m = 1 and e = 1 in (3.6) gives

f cos(x) , x—4 x—2 ln Ixl x2 ^ / _ 7 , „ x2\

We can use the same procedure for the hyperbolic cosine integral, the results are stated in the

next theorem. Its proof is similar to Theorem 3's proof, we will omit it.

Theorem 4. Let ft > 1 and a > 1, and let a = 2ftm + e, where m is an integer (m € N) and

—2ft < e < 2ft.

1. If e = 0, then

cosh (Ax3) , 1 x1—23—a v1 A2n+2m+1 x23n+1

v Jdx = -----h >

A 1 2p a

J Ax23+a A 1 - 2p - a ^ r(2n + 2m + 3) 2pn + 1

A2mx 1 3 1 A2x23

^2m+2y/7rr (m + 3/2) r(m + 2)(2/5 + 1) ^ 1 + 2^; m + 2' m + 2'2 + 2^; ^J + where m = a/(2p). 2. If e = 1, then

f cosh (Ax3) , 1 x1—23—a A2m+1 , . . ^ A2n+2m+1 x23n -dx = -----h --ln|x| + x

J Ax23+a A 1 - 2p - a r(2m + 3) 1 1 n=—m r(2n + 2m + 3) 2pn

A2m+3x23 ^ ( 5 A2x23\ ^

+ 22^0Fr(m + 5/2)r(m + 3)/3 (M;m + ^+ 3,2; —J + C,

where m = (a - 1)/(2p).

3. Finally, if e € (-2p, 0) U (0,1) U (1, 2p), we have

f COS (Ax3) _1 a;1"23"" (-l)mA2m+1 x1"6 y, A2n+2m+l x2/3n-e+l J Ax23+" a;_Al-2/3-a+ r(2m + 3) 1 - e + T(2n + 2m + 3) 2/?n - e + 1

A2m+3x23—e+1

J__x

22m+4y/7rr (m + 5/2) r(m + 3)(2/5 - e + 1) / 1 - e 5 1 - e A2x23\ ^

where m = (a - e)/(2p).

4. Evaluation of the exponential integral Ei3 3+a, ft > 1, a > 1

Theorem 5. Let p > 1 and a > 1, and let a = pm + e, where m is an integer (m € N) and

-p < e < p.

1. If e = 0, then

/gAx^ 1 x1—3—a 1 \n+m x3n+1

\^dX=\l-f3-a+J^mT(n + m + 2) fin + l

(4.9)

Amx /1 1 \

+ —I^TflTn 2^2 l,l + -r;m + 2,2 + -;Ax3 +C, r(m + 2)(p +1) V p p J

where m = a/(p). 2. If e = 1, then

1 x1—3—a \m 1 \™+m x3™

~ J \x?+*dX ~ A 1 - /3 - a + T(m + 2) ln N + T(n + m + 2) f3n ^

Am+1x3

m

r(m + 3)p

where m = (a - 1)/(p).

Am+1x3 / \

3. Finally, if e € (—ft, 0) U (0,1) U (1, ft), we have

/gAx^ 1 x1—3—a Xm x1—e ~ 1 Xn+m x3n—e+1

-dx =---1----L --

Xx3+a X 1 — ft — a r(m + 2) 1 — e n=—m r(n + m + 2) ftn — e + 1

Xm+1™3—e+1 / 1 _ e 1 _ e \

+r(n, + 3)(;3-t + i) (i.i + —;™ + 8,2 + —;A,") + c,

where m = (a — e)/(ft). Proof. We proceed as before. Then, we have

eXxfi , f 1 A (Aa;3)" _ f 1 I /"vA

n

-da; - / x V v ; da; - / _—da; + - / V —x^'^dx

\xfi+*ax - J \xfi+* n! - J \xfi+*ax \ J n\

n=0 n=1

1 v i ^ \ ^ Ara+1 fin—a j

+ A J ^ (n + l)\X

n=0

1 /"m 1 \n ^ \n

dx+ Y -——rX^-^-'dx + / V -——.x^-^-'dx

Xx?+a J Z^ (n + l)! J ^ (n+l)\

n=0 n= m

1 ,,m—1 >n ^ \n

Aa;3+"+J (n + 1)! +J^(n + l)l

1 r — 1 \n+m r ^ \n+m

dx + / V ——--x^-'dx + / V ——--x^-'dx

(n + m + 1)! / (n + m + 1)!

n= -m n=0

1 /■ ~ 1 Xn+m ^ ^^ Xn+m

Xx3+a J ^ (n + m + 1)! J ^ (n + m + 1)!

n= -m n=0

1 — 1 \n+m ^ \n+m

rl-r I / \ " _rpfin-CJ , / \ " A_rpfin-CJ

n i H/«X< / / _ , , «X< / / _ , - «A/ H/«X<

Xx3+a J ^ r(n + m + 2) J ^V(n + m + 2)

n=-m n=0

-1

- f 1 dx ^ f— f V Ara+m

~ J Aa;3+a X + r(m + 2) J xe + J r(n + m + 2)X X

+ i^r(n + m + 2) J Aa;3+" + 2) J x*

/— 1 \n+m /• ^ \n+m+1

^ r(n + m + 2)iC +7^r(n + m + 3)

n=-m n=0

1 x1—3—a Xm /■ dx Xn+m x3n—e+1

X dx

+ fV^TTo) / ^e +

X 1 — ft — a r(m + 2)7 xe n=—m r(n + m + 2) ftn — e + 1

^ Xn+m+1 x3n+3—e+1 , \ " A__^__i

r(n + m + 3) /3n + P - e + 1 1

1 a;1"3"" Am f dx Ara+m ^n-e+i

+ rim + 91 / +

X 1 — ft — a r(m + 2^ xe ^^^ r(n + m + 2) ftn — e + 1

n= -m

n

, Am+1a;3~£+1 ^ (l)n(l + (l-6)//3)w (Aa;3)' T(m + 3)(/3 — e + 1) ^^ (m + 3)ra (2 + (1 — e)//3)n n! + 1

1 x1—3—a Xm /" dx Xn+m x3n—e+1

X dx

+ rim -I- 91 / +

X 1 — ft — a r(m + 2^ xe n=—m r(n + m + 2) ftn — e + 1

(4.11)

(4.12)

дт+1хв-е+1 / ! - £ ! - , \

+ Г(т + 3)(Р-£ + 1) 2i?2 I1' 1 + —; + з, 2 + —; Хх1 j + С\.

1. For е = 0, we substitute е = 0 in (4.12), and hence, we obtain

f 0Axe r d™ r — 1 \ra+m г \ra+m

/ + / V _-_r/3radr + / V_-_r/3radr

~ J \xP+« Г(?г + m + 2)Ж ^ ' ./ Г(тг + m + 2)X ™

Ахв+а J Ахв+а J ^ Г(п + m + 2) J ^ Г(п + m + 2)'

n= —m 4 ' n=0

1 x1—в—a — 1 дга+m дга+m

Ea x ^—л

Г in + m + /ill + 1 +

A 1 - в - a ^ Г(п + m + 2) вп + 1 ^ Г(п + m + 2) вп + 1

n=—m v ' n=0

1 x 1—в—a — 1 Ara+m хвп+1

_ Л1 -P-a + E Г(?г + m + 2) fin + 1 A"\r ^ (1)„(1 + l//3)„ (Аж/З)" _ 1 x1-'3-»

r(™ -A- OMR -Ll \

Г(ш, + 2)(в + 1)^(m + 2)n (2 + 1/в)п n! A 1 - в - a

n=0

—1

An+m xen+1 Amx ^ / 1 1 e\ „

+ E TV„ + n, + 91 /fa + 1 + TVn, + 9)f Я + П ^ i1'1 + ™ + 2' 2 + Лж/ +

n=—m r(n + m + 2) pn + 1 r(m + 2)(p + 1) 22^' p' ' p

which is (4.9), and where m = a/p. 2. For e = 1, we set e = 1 in (4.12) and obtain

gAxe 1 x1—в—a Am 1 A™+m хв™ --31—dx = — ——---1----- In L-cl + > —--1-—

Ахв+а A 1 - в - a Hm + 2) 11 ^ Г(п + m + 2) вп

i \ / n=—m v '

Am+V3 Г(т + 3)/3

which is (4.10), and where m = (a - 1)/в-3. For е € (-в, 0) U (0,1) U (1,в), (4.12) gives

Ax

ь-^й + tfi

Axe 1 x1—в—a Am £ 1 A™+m хв™—£+1

dx = — —----h —-- ---h

J Ахв+а A 1 - в - a Г(ш, + 2)1 - е n=—m Г(п + m + 2) вп - е + 1

^m+1xв—e+1 / 1 _ е 1 _ е \

Г(m + 3)(в - е + 1) V в в /

which is (4.11), and where m = (a - е)/в. □

Example 3. In this example, we evaluate /(e—x2/x4)dx. We first observe that A = -1 and p = 2. We also have 4 = p + a = 2 + 2 = 2 + 2(1) + 0 = p + pm + e, and so m = 1 and e = 0. Substituting A = 1, p = 1, m = 1 and e = 0 in (4.9) gives

/e x x 3 1 x

= — 2F2(1,2;3,3;-x2)+C.

Corollary 1. Let a > 1 and let a = m + e, where m is an integer (m € N) and -1 < e < 1.

1. If e = 0 or 1, then

f eAx , 1 Xm , , , Xn+m

i+a = / —n—dx = —---1- —-- m m + >

1'1+a / Xx1+a Xaxa r(m +2) 1 1

Xx1+a Xaxa r(m + 2) 11 ^ r(n + m + 2) n

v ' n=—m v '

Xm+1x

+ r, 2^2 (1,1; m + 3, 2; Xx) + C,

r(m + 3)ft

where m = a - 1. 2. And if e € (—1, 0) U (0,1), we have

eAx , 1 Xm x1—e Xn+m xn—e+1

-(Ix =---1----u >--

Xx1+a Xaxa r(m + 2) 1 — e ^ r(n + m + 2) n — e + 1

\ / <n — — m v '

2—e

Xm+1x

+ 3)(2 — e) 2^ d, 2 - e; m + 3,3 - A,) + C,

where m = a e.

xn

P r o o f.

1. If e = 0 or 1 implies a = m + e is an integer (a € N) since (m € N). Morever, a = m + e implies ft = 1 in Theorem 5. Therefore, we obtain (1) by setting ft = 1 in (4.10).

2. For e € (—1, 0) U (0,1), we set ft = 1 in (4.11) and obtain (2). □

Example 4. In this example, we evaluate f (e—x/x3'7) dx. We first observe that X = —1. We also have 3.7 = 1 + a = 1 + 2.7 = 1 + 2 + 0.7 = 1 + m + e, and so m = 2 and e = 0.7. Substituting X = —1, m = 2 and e = 0.7 in (2) gives

/e x x 2.7 x0.3 x 1-7 x 0.7 x1-3

e x x x x x ____ , _ _ _ _ ,

= — L8 - TT + TT - 3L2 f1' * 2'3; -T) + C'

5. Conclusion

Formulas for the non-elementary integrals Si3,a = /[sin (Xx3)/(Xxa)]dx, ft > 1, a > ft + 1, and Ci3,a = /[cos (Xx3)/(Xxa)]dx, ft > 1, a > 2ft + 1, were explicitly derived in terms of the hyperge-ometric function 2F3 (see Theorems 1 and 2). Once derived, formulas for the hyperbolic sine and hyperbolic cosine integrals were deduced from those of the sine and cosine integrals (see Theorems 2 and 4). On the other hand, the exponential integral Ei3,a = /(eAx^/xa)dx, ft > 1, a > ft + 1 was expressed in terms of the hypergeometric function 2F2 (see Theorem 5 and Corollary 1).

Beside, illustrative examples were given. Therefore, their corresponding definite integrals can now be evaluated using the FTC rather than using numerical integration.

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