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

URAL MATHEMATICAL JOURNAL, 2018, Vol. 4, No. 1, pp. 24-42

DOI: 10.15826/umj.2018.1.003

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

Victor Nijimbere

School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada [email protected]

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

/[cos(Ax^)/(Axa)]dx, £ > 1, a < 2£ + 1, where {£,a} € R, are evaluated in terms of the hypergeometric functions 1F2 and 2F3, and their asymptotic expressions for |x| ^ 1 are also derived. The integrals of the form J[sinn (Ax^)/(Axa)]dx and J[cosn (Ax^)/(Axa)]dx, where n is a positive integer, are expressed in terms and Ci^Q,, and then evaluated. and are also evaluated in terms of the hypergeometric function 2F2.

And so, the hypergeometric functions, 1F2 and 2F3, are expressed in terms of 2F2. The exponential integral Ei^Q, = f(e^xlS /xa)dx where £ > 1 and a < £ + 1 and the logarithmic integral Li = fx dt/ ln t, ^ > 1, are also expressed in terms of 2F2, and their asymptotic expressions are investigated. For instance, it is found that for x » 2, Li ~ x/ln x + ln(ln x/ln2) - 2 - ln2 2F2(1, 1; 2, 2;ln2), where the term ln (ln x/ln2) - 2 -ln2 2^2(1, 1; 2, 2;ln2) is added to the known expression in mathematical literature Li ~ x/ln x. The method used in this paper consists of expanding the integrand as a Taylor and integrating the series term by term, and can be used to evaluate the other cases which are not considered here. This work is motivated by the applications of sine, cosine exponential and logarithmic integrals in Science and Engineering, and some applications are given.

Key words: Non-elementary integrals, Sine integral, Cosine integral, Exponential integral, Logarithmic integral, Hyperbolic sine integral, Hyperbolic cosine integral, Hypergeometric functions, Asymptotic evaluation, Fundamental theorem of calculus.

1. Introduction

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 [6, 10].

Liouville 1938's Theorem gives conditions to determine whether a given integral is elementary or non-elementary [6, 10]. For instance, it was shown in [6, 10], using Liouville 1938's Theorem, that the integral Si1;1 = /(sin x/x)dx is non-elementary. With similar arguments as in [6, 10], One can show that Ci1;1 = /(cosx/x)dx is also non-elementary. Using the Euler formulas e±ix = cos x ± i sin x, and noticing that if the integral of a function g(x) is elementary, then both its real and imaginary parts are elementary [6], one can, for instance, prove that the integrals Si£,a = /[sin(Ax£)/(Axa)]dx, ,0 > 1,a > 1, and Ci£,a = /[cos(Ax£)/(Axa)]dx, where ft > 1 and a > 1, are non-elementary by using the fact that their real and imaginary parts are non-elementary. The integrals /[sinn (Ax£)/(Axa)]dx and /[cosn (Ax£)/(Axa)]dx, where n is a positive integer, are also non-elementary since they can be expressed in terms of Si£,a and Ci£,a.

To my knowledge, no one has evaluated these integrals before. To this end, in this paper, formulas for these non-elementary integrals are expressed in terms of the hypergeometric functions 1F2 and 2F3 whose properties, for example, the asymptotic expansions for large argument (|Ax| » 1),

are known [9]. We do so by expanding the integrand in terms of its Taylor series and by integrating the series term by term as in [7]. And therefore, their corresponding definite integrals can be evaluated using the Fundamental Theorem of Calculus (FTC). For example, the sine integral

B

f sin(Axg) ,

Si/3,« = J dx, ft > 1, a <¡3 + 1,

A

is evaluated for any A and B using the FTC.

On the other hand, the integrals Eig;Q, = /(eAxl /xa)dx and f dx/ ln x, are expressed in terms of the hypergeometric function 2F2. This is quite important since one may re-investigate the asymptotic behavior of the exponential (Ei) and logarithmic (Li) integrals [3] using the asymptotic expressions of the hypergeometric function 2F2 which are known [9].

Some other non-elementary integrals which can be written in terms of Eig;Q, or J dx/ ln x are also evaluated. For instance, as a result of substitution, the integral / eAelx dx is written in terms of Ei^i = /(eAxl/x)dx and then evaluated in terms of 2F2, and using integration by parts, the integral / ln(ln x)dx is written in terms of J" dx/ ln x and then evaluated in terms of 2F2 as well.

Using the Euler identity e±ix = cos(x) ± i sin(x) or the hyperbolic identity e±x = cosh(x) ± sinh(x), Sig;Q, and Cig;Q, are evaluated in terms Eig;Q,. And hence, the hypergeometric functions 1F2 and 2F3 are expressed in terms of the hypergeometric 2F2.

This type of integrals find applications in many fields in Science and Engineering. For instance, in wireless telecommunications, the random attenuation capacity of a channel, known as fading capacity, is calculated as [11]

oo

^fading = £[log2(l + mi2)] = J log2(l + POe-*d£ = ¿e1^ Ehl (oo) - Ehl (i)

0

where the fading coefficient H is a complex Gaussian random variable, and E (|X|2 < P) is the maximum average transmitted power of a complex-valued channel input X. In number theory, the prime number theorem states that [3]

x

/dx

■-—, a > 1, ln x

where n(x) denotes the number of primes small than or equal to x. Moreover, there are applications of sine and cosine integrals in electromagnetic theory, see for example Lebedev [5]. Therefore, it is quite important to adequately evaluate these integrals.

For that reason, the main goal of this paper is to evaluate non-elementary integrals of sine, cosine, exponential and logarithmic integrals type in terms of elementary and special functions with well known properties so that the fundamental theorem of calculus can be used so that we can avoid to use numerical integration.

Part I is indeed devoted to the cases = J[sin(Ax^)/(Axa)]dx, ^ > 1, a < ft + 1,

Cig,« = /[cos (Ax^)/(Axa)]dx, ft > 1, a < 2ft + 1 and Ei^ = /(eAxl/xa)dx where ft > 1, a < ft + 1, where {ft, a} € R. The other cases Sig,a ^[sin(Axg)/(Axa)]dx, ft > 1, a > ft + 1, Cig,a = /[cos (Axg)/(Axa)]dx, ft > 1, a > 2ft + 1 and Eig,a = /(eAxl/xa)dx where ft > 1, a > ft + 1, where {ft, a} € R, which may involve series whose properties are not necessary known will be considered in Part 2 [8].

Before we proceed to the objectives of this paper (see sections 2, 3, 4 and 5), we first define the generalized hypogeometric function as it is an important mathematical that we are going to use throughout the paper.

Definition 2. The generalized hypergeometric function, denoted as p, is a special function given by the series [1, 9]

TP (n n n -h h h -rr\ - ST • • • \Uy)n ■>-

where a1, a2, ■ ■ ■ , ap and ; b1, b2, ■ ■ ■ , are arbitrary constants, ($)n = r($ + n)/r($) (Pochham-mer's notation [1]) for any complex •&, with ($)0 = 1, and r is the standard gamma function [1, 9].

2. Evaluation of the sine integral and related integrals

/13 3 A2x2 \

Proposition 1. The function G(x) = x iF2 ( -; -, -;--— J , w/iere iF2 is a hypergeometric

function [1] and A is an arbitrarily constant, is the antiderivative of the function g(x) = .

Thus, X

f sin (Ax) , n/l 3 3 A2x2\ ^

1 -dx = x1F2 -;-,-;--

Ax 1 2 V 2' 2' 2' 4

Proof. To prove Proposition 1, we expand g(x) as Taylor series and integrate the series term by term. We also use the gamma duplication formula [1]

r(2a) = (27r)-^22°-^r(a)r + ^ > « G C>

the Pochhammer's notation for the gamma function [1],

(a)n = a(a + 1) • • • (a + n — 1) = —^tt-t—a € C,

r(a)

and the property of the gamma function r(a + 1) = oT(o:) (eg., r (n + 3/2) = (n + 1/2) r (n + 1/2) for any real n). We then obtain

n=0 n=0

x A, A2n x2n x ^ r (n + 1/2)

_ _ V^f IV» A__x 1 r* = Z. , \2 2\n 1 ^

; (2n + l)!n + l/2 2 r(2??. + 2)T (n + 3/2) ;

n=0 n=0

^ (1/2)„ (-A2x2/4)n /13 3 A2x2\ r, r„ , r.

n=0

□

In the following lemma, we assume that the function G(x) is unknown and therefore we establish its properties such as the inflection points and its behaviour as x ^

sin x

Lemma 1. Let G(x) be the antiderivative for g(x) = — (A = 1), and G(0) = 0.

1. Then G(x) is linear around x = 0 and the point (0, G(0)) = (0,0) is an inflection point of the curve Y = G(x), x € R.

2. And lim G(x) = —6 while lim G(x) = 6, where 6 is a positive finite constant.

X^ — TO X^ + TO

Proof. 1. The series

sin x ^ (Ax)

g(x) = ^ = £(-1)

2n

x ' (2n + 1)!

<?-(*)=✓(*) = = -AEf-«"^

n=0

gives G'(0) = g(0) = 1. Then, around x = 0, G(x) ~ x since G'(0) = g(0) = 1 and G(0) = 0. Moreover,

........... \2n

> i — ii —

(277 + 3)!

n=0

and so G''(0) = g'(0) = 0. Hence, by the second derivative test, the point (0, G(0)) = (0,0) is an inflection point of the curve Y = G(x).

2. It is straight forward, using Squeeze theorem, to obtain lim g(x) = lim g(x) = 0. And

X^ — TO X^+TO

since both g(x) and G(x) are analytic on R, then G(x) has to be constant as x — by Liouville Theorem (section 3.1.3 in [4]) since if a complex function is entire on C then both its imaginary and real parts are analytic on the real line R including at x — Also,

there exists some numbers 5 > 0 and e such that if |x| > 5 then || sinx|/x — 1/x| < e, and lim (| sinx|/x)/(1/x) = lim (| sinx|/x)/(1/x) = ±1. This makes the function gi(x) =

X^ — TO X^ + TO

—1/x an envelop of g(x) away from x = 0 if sin x < 0 and g2(x) = 1/x an envelop of g(x) away from x = 0 if sin x > 0. Moreover, on one hand, g2 < G'' < g' if x < —5, and g' and g2 do not change signs. While on another hand, g' < G'' < g2 if x > 5, and also g' and g2 do not change signs. Therefore there exists some number 6 > 0 such G(x) oscillates about 6 if x > 5 and G(x) oscillates about —6 if x < —5. And |G(x)| < 6 if |x| < 5. □

Example 1. For instance, if A = 1, then

f sin X n /1 3 3 x2\ ^ . .

J ~HT = X 2 \2' 2' 2' ~~J + (2-D

sin x 1 3 3 x2

By Proposition 1, the antiderivative of g(x) = —- is G(x) = x 1F2 ( -;-,-;— — I, and the

x \ 2 2 2 4 /

graph of G(x) is shown in Figure 1. It is in agreement with Lemma 1. It is seen in Figure 1 that (0, G(0)) = (0,0) is an inflection point and that G attains some constants as x — as predicted by Lemma 1.

In the following lemma, we obtain the values of G(x), the antiderivative of the function g(x) = sin (Ax)/(Ax), as x — using the asymptotic expansion of the hypergeometric function 1F2.

Lemma 2. Consider G(x) in Proposition 1,and preferably assume that A > 0.

1. Then,,

and

G(x)

-9 = -u 2

x

Figure 1. G(x) is the antiderivative of sin(x)/x given in (2.1).

G(-oo) = lim G(x) = lim x iF2

x^—œ x^œ \ 2 2 2 4

G(+oo) = Jim^GOr) = Jim x XF2 Q; ||

n

2À'

n 2Â

2. And by the FTC,

(2.2) (2.3)

sin (Ax) , , , 7T / Jtt

n

(2.4)

P r o o f.

1. To prove (2.2) and (2.3), we use the asymptotic formula for the hypergeometric function 1F2 which is valid for |z| » 1 and —n < arg z < n, where arg z is the argument of z in the complex plane. It can be derived using formulas (16.11.1), (16.11.2) and (16.11.8) in [9] and is given by

'R— 1

r(b1)r(b2)z—£

iF2 (ai; 61,62; —z) = (ai)n

(-z)"

+ O(|z|—R )

, n=0

r(61 — a1 — n)r(62 — a1 — n) n!

+ e"1/3e'W3(-'701"bl"b3+1/2)/2 (e )- + O(lzl-)'

(2.5)

r(a1)

„ n=0

'S— 1

+

WW ' , E +0(lirs)

r(a1)

„ n=0

where a1, 61 and 62 are constants and the coefficient is given by formula (16.11.4) in [9]. We then set z = A2x2/4, a1 = 1/2, 61 = 3/2 and 62 = 3/2, and we obtain

Ax

—2R

e

2

œ

A

—œ

A e

A2x2 2 I ^ 2n

n=0

-2n

2

Ax

-2S>

/— i\x (S-l / \ \ —2n

Vn e ■■

+0

A2x2 2 ^ 2n V 2 /

n=0

Ax

-2SN

Then, for |x| » 1,

ln=0 '

-2n

Ax

't)

Ax ~2~

n

2A| x|

while

iAx ' S-1

A2x2 ~2~ I ^ 2"

V^r e*-

££Hf) -E-i'-i +o

n=0

\ \ -2n S-1 / , ■> -2n

. Ax \ / .Ax

n=0 iAx i „-iAx

Vtt e + e (Xxy

7t-

2

cos (Ax)

(Ax)2

Ax

-2S^

We then obtain,

^,133 A2x2\ 7T x v/ttcos (Ax) . .

2' 2' 2' 4 J ~ 2A]xj A AX ' N

Hence,

n( \ V P i 1 3 3 AV G(-oo) = hm x\F'2 ( -;-,-;--—

x——^o \ 2 2 2 4

lim

7t x y/7TCOS(Ax) 2A |x| ~ ~A Ax

n 2A

and

G(+oo)=^limcoX1F2Q;||

A2x2

lim , x—■y2A |x|

7t x v/7tcos(Ax)\ 7t

A Ax

2A

2. By the FTC,

0 y

f S1D ¿x = iim f sm(Xx) dx + lim f sin (Ax) = G(+oo) - G(-oc) J Ax y—-^ y Ax y—y Ax

y 0

/13 3 A2y2\ /13 3 A2 y2

= y^oo^F2 2;2'2;-^ K^co^2 2;2'2;-^

n A'

We now verify whether this is correct or not using Fubini's Theorem [2]. We first observe that

f sin (Ax) , f sin (Ax) ,

' v -dx = 2 / —r—-dx

Ax

Ax

since the integrand is an even function. We have in terms of double integrals and using Fubini's Theorem that

f dx = f f e_i'Tsin (Xx)dsdx = f f e~sxs'm(Xx)dxds.

(2.6)

00

00

2

2

2

4

Now using the fact that the inside integral in (2.6) is the Laplace transform of sin (Ax) [1] yields

77-..........

s2 + A2

0 0 0

+TO +TO +TO +TO

f sin (Ax) ^ = ^ i sin (Ax) ^ = 2 i i e-s.Tgin (Xx-jdgdx = 2-Z1 =

Hence,

+to

f si:

J Ax "" ~ J Ax A7 J —v-y-— ^

— TO 0 0 0

as before. □

Setting A = 1 as in Example 1, Lemma 2 gives lim G(x) = —0 = —n/2 while lim G(x) =

X^ — TO X^+TO

0 = n/2. And these are the exact values of G(x) as x ^ in Figure 1.

Theorem 1. If ß > 1 and a < ß + 1, then the function

_ xß~a+1 i J_ 1 J_ 33 A2x2/3\

~ ß-a + 1 12 V 2ft + 2ft + 2; "2ft + 2ft + 2' 2; T~J '

where i F2 is a hypergeometric function [1] and A is an arbitrarily constant, is the antiderivative of

the function g(x) = S1D^^ ^. Thus,

sin (Axß) xß—"+1 / a 1 1 a 1 3 3 A2x2ß

Siß'a ~ J ^ ft - a + 1 li?2 V 2ft + 2ft + 2' ~2ft + 2ft + 2' 2'--~) +

(2.7)

And for |x| » 1,

xß-a+i ( a_ J_ 1. J_ 3 3. A2x2/3\

ft-a + 11 2 V 2ft + 2ft+ 2'~2ft + 2ft+ 2'2' 4~) {2/X)1+1/ß~a/ß T {—a/(2ft) + 1/(2/3) + 3/2) 0F x/3"°+1 ft - a + 1 0Fcos (Ax/3)

ß — a + 1 r(1 + a/(2ß) — 1/(2ß)) 2 |x|ß—"+1 ß A2 xß+a—1

Proof.

sin (Axß) i 1 lAn (Axß)2n+1

SW = /9(x)dx = f = f D"1 )ra7^tT)!

TO \2n r TO \2n f

n=0 ^ ' n=0 ^ '

TO \ 2n x2ßn+ß—a+1

EC IV" A__l__I /-<

V 7 (2??, + l)!2ft??, + ft-a: + l

n=0

TO

n

„ß—a+1 TO \ 2n x2ßn

1_V^C 1V"__-__I- r

2/3 ; (2n + l)!n-a/(2ft)+ 1/(2/3)+ 1/2 ^

x£-a+1

V r (n — a/(2/j) + 1/(2/j) + 1/2) 2 2/3

x

20 n=0 r(2n + 2)r (n - a/(20) + 1/(2,0) + 3/2)

£-a+1 rc (_ a/(20) + 1 /(20) + 1 /2) f- A2

v (~a/( 2/?) + 1/(2/?) + l/2)w (—A2x2£/4)"

, - a + 1 n=0 (3/2)n (-a/(2,) + 1/(20) + 3/2)n n'

/3-a + l 1 2 V 20+ 20 + 2'~20+ 20 + 2'2'

To prove (2.8), we use the asymptotic formula for the hypergeometric function 1F2 in equation (2.5), and proceed as in Lemma 2. □

Beside, we can show as above that if 0 > 1 and a < 0 + 1, then

f sinh (Ax£), x£-a+1 / a 1 1 a 1 3 3 A2x2£ \ ^ . .

Corollary 1. Let 0 = a. If a > 1, then

0 1 /

t S'" = G(0> - G'(-co) = (T) ' J™±I> (2.10)

Axa ' \A) r (3/2 - 1/(2a)) 2

+oo ,

[ «¡¡iMU = G(+00> - C(0) = (f) ' r/i/W + i) ^ (2.n)

J Xxa K ' y ' \\) r (3/2 — 1/(2ck)) 2 v y 0

and

f sin (Ax") /2\1/o r(l/(2a) + l)

J ^-dX = G{+00)-G{-00) = {x) r (3/2- 1/(2«)) ^ (2"12)

Proof. If 0 = a, Theorem 1 gives

^ n . 1 11,3 A2x2a \

f (2\ 1/a 0F r (l/(2a) + 1) x _ yft cos (Xxa)\ = _ (2\ 1/a r(l/(2a) + l) yfr 2 r(3/2-l/(2a))|x| aA2 x2«"1 J ~ VV r (3/2 - l/(2a)) 2

^ n ^/11 ,3 A2x2a\

((2\1/o 0F r(l/(2a) + l) x 0Fcos(Ax»)\ = (2\1/a r(l/(2a) + l) yfr .t4+co^aJ 2 r(3/2-l/(2a))|x| aA2 x2«"1 y U/ r (3/2 - l/(2a)) 2 "

Hence, by the FTC, 0

Ax" v ' v ' \ \A) r (3/2 — 1/(2a)) 2

•2\1/o r(l/(2a) + l) y/ir

A) r (3/2 — 1/(2a)) 2 '

Tsi..= G(+co)_ = mr(i/(2a) + i;I ^_0

(2.13)

0

Ax" W r (3/2 — 1/(2a)) 2

2\1/o r(l/(2a) + l) 0F

A) r (3/2 — l/{2a)) 2 ' (2'14)

And combining (2.13) and (2.14) gives (2.12). □

Theorem 2. If ft > 1 and a < ft + 1, then the FTC gives

b r

f sin(Aar)

J Axa

A

dx = G(B) — G(A), (2.15)

for any A and any B, and where G is given in Theorem 1.

Proof. Equation (2.15) holds by Theorem 1, Corollary 1 and Lemma 2. Since the FTC works for A = —to and B = 0 in (2.10), A = 0 and B = +to in (2.11) and A = —to and B = +to in (2.12) by Corollary 1 for any ft = a > 1 and by Lemma 2 for ft = a = 1, then it has to work for other values of A, B € R and for ft > 1 and a < ft + 1 since the case with ft = a > 1 is derived from the case with ft > 1 and a < ft + 1. □

Theorem 3. Let ft > 1, then the function

fAxr/2) / 5 A2x2r

G{x) = In |.r| - V J 2F3 1,1; 2, 2, —

6ft 2 3 V 2' 4

where 2F3 is a hypergeometric function [1] and A is an arbitrarily constant, is the antiderivative of

, . sin (Axr) the function g(x) = /j+1 . Ihus,

TO

P r o o f.

sin (Axr) r 1 ^ (Axr )2n+1

(2n + l)\dX

/r sin (Axr) f 1 ^

= / ay/3+i dx = / w3tt z^"1) Ax Ax n=0

TO \2n r r TO \2n r

- ^ - /f + B-D-fsr+iji/

n=0 v ' n=1 v '

TO \ 2n+2 x2rn+2r

n=0

\2™2r TO \2n x2rn

= in M _ y (-1)"—^----+ C

11 2¡3 ' (2n + 3)!n + l

n=0

_lDh, (r (n +1))2 (-\*x*T

11 2 0 ¿¿r(2n + 4)r(n + 2) n\

_ ill (A-^/2)2- (i)w(i)w (-AV/V4)" H 60 ¿¿(2)n(2)n(5/2)n n!

In M - ^^ 2F3 (l, 1; 2,2, | +C = G(x) + C.

3. Evaluation of the cosine integral and related integrals

Theorem 4. If 0 > 1 and a < 20 + 1, then the function

, 1 x1-a Ax2£-a+1 / a 1 a 1 3 A2x2£

G(x) = T i--™-rr 2^3 1, -77-T + -7-T + 1; -77-T + -7-T + 2, 2; -

□

A1 - a 20 - a + 1 2 20 20 ' 20 20 '2' ' 4

where 2F3 is a hypergeometric function [1] and A is an arbitrarily constant, is the antiderivative of the function g(x) = ^' Thus,

f cos (Ax£) , 1 x1-a 1 Ax2£-a+1 a 1 a 1 „ 3 A2x2£\ ^

and for |x| » 1,

Ax2£-a+1 a 1 a 1 „ 3 A2x2£

2F3 1, -— + — + 1; -— + — + 2, 2; -

(3.16)

20 - a + 1 2 3 V 20 20 ' 20 20 ' 2' ' 4 AAr /__a + J_ + A (2\ 2 + + 2 cos(A^)

(3.17)

20 V 20 20 y VV A0 A20 x£+a-1

P r o o f. If 0 > 1 and a < 20 + 1,

/"1 1 /" ^ A 2™

Axa A J ' (2n)'

n=1

1 x1-a 1 ^ A2n+2 /•

"Al-a ^ (2n + 2)\JX

1 x1-a ^ a 2n x2£ra+2£-a+1

A N V 1 Y"__-__I-r*

Z^1 > (On I 9M Ofti I OR _ nz J_1 +

A 1 - a ' (2n + 2)'20n + 20 - a + 1

n=0

V r (n — a/(20) + 1/(20) + 1) 2 w

.TlTVn. - n/om -1-1 /(OR-) + 9V ;

1 x 1—a Ax 2^—a+1

' ' X"

A 1 - a 20 r(2n + 3)r (n - a/(20) + 1/(20) + 2)

la.i-a i ~ (l)ra (-«/(20) + 1/(20) + 1)„ (—A2.T2/3/4)"

Al-a 2 20 - a + 1 ^ (3/2)„ (2)„ (-a/(20) + 1/(20) + 2)„ n!

1 x1-a 1 Ax2£-a+1 /. a 1 a 1 3 A2x2£ \

= Al^" 220-a + l2jP3 ^'"20 + 20 + 1;"20 + 20 +2'2'2; ~J +

To prove (3.17), we use the asymptotic expression of 2F3 (a1,a2; 61, 62,63; —z) for |z| » 1, where a1, a2, b1, 62 and 63 are constants, and —n < arg z < n . It can be obtained using formulas 16.11.1, 16.11.2 and 16.11.8 in [9] and is given by

2F3 (01,02; 61,62,63; —z) =

_(ai)nT(a1-a2-n)_(-z)~n R\

r(a2) * | T(6i — a,i — ??)r(62 — a,i — ??)r(63 — a,i — n) nl lH 'J

/V _(fl2№2-Ol-n)_(-*)-"

r(ai) * |^r(6i-a2-7i)r(62-a2-7i)r(63-a2-7i) n! lH jJ

, r(6j)r(62)r(63) e2zl/2e~i7r/2 (Ze~i7T')(al+af2-bl-bf2-b3 + 1/'2)/'2 ^ w |

+ nai)T(a.2) A lto2ra+1 J

+ r(fll)r(a2) 0F \h2n+1 J'

(3.18)

where the coefficient is given by formula 16.11.4 in [9].

A2x2r a 1 a 1 3

We now set z = 4 , a 1 = 1, a2 = + ^ + 1, 61 = -— + — + 2, 62 = - and 63 = 2 m (3.18) to obtain

/ a 1 a 1 3 A2x2r \

+ + 2)+- + ^r (-SL + J. + 2) f A.) ~a/l3+l/l3+2 (3.19) A2 V ¡3 ¡3 J 2 V 2ft + 2ft + 7 V Ax/3/ V 7

+A f+ I + cos(Aa;/3)

A3 ft ft x3r

Hence, multiplying (3.19) with Ax2r-"+1/(2ft — a + 1) gives (3.17). □

On the other hand, we can show as above that if ft > 1 and a < 2ft + 1, then

f cosh (Axr) , 1 x1-" 1 Ax2r-"+1 / a 1 a 1 3 A2x2r\ ^

J = aT^+2W^TT{i>-2p + 2e + 1''-2e + 2e + 2' 2'2; — J

Theorem 5. Let ft > 1, then the function

1 A, , , A /Axr\2 „ / „ 5 A2x2r

= -oxo - ö ln N + Mr 2^3 1,1; 2, 3; -

2Aßx2ß 2 1 1 6ß V 4 y > 2> > 4

where 2F3 is a hypergeometric function [1] and A is an arbitrarily constant, is the antiderivative of

, . cos (Axß) the function g(x) = 2/3+1 • ¿hus,

f cos (Axß), 1 A. . . A / Axß\2 / 1 5 0 A2x2ß\ ^

(3.20)

We also have,

" cos(Axß) 1 Ax2ß / 3 oo A2x2ßs

a/3a = J \x dX=\lnlx]~^2F3{1A]2'V2]--(3"21)

Proof.

Xx'3f 1 v^v -, V» (Xx'3)2n

(2 n)l

f f c0s (Axß) f 1 v-A

Ci/3,2/3+1 = / iK* = / 2/3+1 dx = / T^+IzJ"1)

J J J n=Q

1 1 f ^ \2n

Ax2ß+! A J 7 (2n)

i i ^ \2ra+2 /•

_____i Vi-l)ra —_ / r2ßn~ldr

2\ßx2ß } (277 + 2)! J X

n=Q

1 A f dx , A

i2ra

---7

2Aßx2ß 2 J x 7 (2n + 2)!

n= 1

1 \ ^ \ 2n /■

-—- - In |X| + A3 y(-l)" A „ / .T2/3n+2/3-ld

2Xßx2li 2 1 1 y (2n + 4)! J

1 A ^ A 2" x2ßn+2ß

" " 2 ln N + 2(_1)n(2n + 4)!20n + 20 + C

1 A3x2/3 ^ (r (??< +1))2 (—A2x2/3)ra

9 9R TO.n + Mr in + 9^ n! +

2Aßx2ß 2 1 1 2ß "=Q r(2n + 5)r (n + 2) n

1 + il-C).

m |X| + i ¿^(o\(z/<>\ „I

2Aßx2ß 2 11 6 A W "=Q (2)n (5/2)" (3)" n! 1 A. . . A /Axß\2 / 1 0 5 _ A2x2ß

A, , , A /Axß\2 / , 5 „ A2x2ß\ ^

2Aßx2ß 2 ' ' 6ß V 4 / 2 V ' ' 2' ' 4 The proof of (3.21) is similar, we do not show it here. □

4. Evaluation of some integrals involving Sia,ß and Cia,ß

/cos" (Axß)

-—dx, where n is a positive integer and ß > 1, a < 2ß + 1, can be

written in terms of (3.16) and then evaluated.

cos \ ax )

Example 2. In this example, the integral f ——dx is evaluated by linearizing cos4 (Ax/3).

f cos4 (Ax") , 1 f cos (4Ax") , 1 f cos (2Ax") , 3 f ,

/ -H--d® = - / -;--dx + - / -;--dx + - dx =

J Axa 8 7 Axa 2 J Axa 8 7

1 x1-a 1 Ax2^-a+1 ^ / a 1 a 1 3 - 2«

2^3 1, -77^ + —j + 1; --r- + — + 2, -, 2; —4A x '

(Axß)

8A1 - a 4 2ß - a + 1 2ß 2ß ' 2ß 2ß '2'

1 x1-a 1 Ax2ß-a+1 / a 1 a 1 3 ,2 2ß\ 3x ^

sin™ (Ax")

If /3 > 1 and a < /3 + 1, the integral J ——dx, where n is a positive integer, can be written

either in terms of (2.7) if n odd, and then evaluated.

Example 3. In this example, the integral f ^-j^-^-dx is evaluated by linearizing sin3 (Ax'3).

r sin3 {XxJ3) 1 r sin (3Xxl3) 3 r sin (Ax?)

J Xxa 'X ~ 4 J Xxa 'X + 4 J Xxa 'X 1 xr-"+1 „ / a 1 1 a 1 3 3 9A2x2r \

4f3-a + llF'2\ 2/3 + 2/3 + 2' 2/3 + 2/3 + 2' 2' 4 J 3 xr-"+1 / a 1 1 a 13 3 A2x2r \ ^

i---, R,---1---1--•---1---1--_•--J- C

4/3-a + l V 2/3 2/3 2' 2/3 2/3 2' 2' 4 y '

Example 4- Let us now evaluate the integrals J" sin(A/xM)dx and f cos(A/xM)dx.

1. The integral J" sin(A/xM)dx is evaluated using the substitution u = 1/x and Theorem 1 if I > 1. Then, we have

f ( A N f sin (AuM) , Au^-i / 1 1 1 3 3 A2u2^\

/ sin — \ax = — / ---au =--\I2---h -;---1— —, — 1---—

J VW J u2 JL4 — 1 V 2M 2 2'2' 4 J

A(1/x)"-1 / 1113 3 A2 \ ^

I — 1 V 2i 2' 2i 2' 2' 4x2^ /

(4.22)

The integral sin(A/xM)dx is evaluated using the substitution u = 1/x and Theorem 3 if I = 1. Then, we have

f /AN f sin(Au) , , . . (Au/2)2 / 5 A2u2

/sin (-)dx = - / -^dt/, = - In |u| + v ; ; 2F3 1,1; 2, 2, -; -

x u2 6 2 3 2 4

2. Making the substitution u = 1/x and applying Theorem 4 gives

f ( AN, [ cos (Au") , 1 Au2"-1 / 1 1 3 A2u2"

(4.23)

Making the substitution u = 1/x and applying Theorem 5, then for i = 1, we have

i (A V f cos(Au) , 1 A. . . A (Au\2 ^ / 5 0 A2x2

/ cos - tte = - / -V^dt/, = —- + - In |m| - - — 2F3 1,1; 2, - , 3; -

^y J u2 2Au2 2 1 1 6 V 4 y ; 2' ' 4

x2 A, . . A / A \2 5 0 A2 \ ^

= —--In b--— 2E3 ( 1,1; 2, -, 3;--^ + C.

2A 2 11 6 \Ax) \ ' ' ' 21 1 Ax-2 J

5. Evaluation of exponential (Ei) and logarithmic (Li) integrals

Theorem 6. If ß > 1, then for any constant A,

gA^ß Axß

-dx = In Id H—

xß

--dx = In |.r| + — 2F2( 1,1; 2, 2; Xxß) + C,

xß

and

Xx'3 2^2(1,1; 2, 2; Xx13) ~ -2 + j-^, \x\ > 1. (5.24)

P r o o f.

6—dx = ( - Y^ ^—Q—dx = [ — + [ Y^ -dx = In Id + V ^ I x^^dx

x x n' x n' n'

n=0 n=1 n=1

^ A™ x"™ ^ A™+1

n=1 n=0

='»N + f E + - ='» N + ^(M;, * A,', + c.

To derive the asymptotic expression of Ax" 2F2(1,1; 2,2; Ax"), |x| » 1, we use the asymptotic expression of the hypergeometric function 2F2 (a1,a2; b1,b2; z) for |z| » 1, where z € C, and a1,a2,61 and 62 are constants. It can be obtained using formulas 16.11.1, 16.11.2 and 16.11.7 in [9] and is given by

2F2 (01,02; 61,62; z) =

_r(bi)T(b2) ^ f^1 (fli)nr(ai-fl2-«) (ze^)~n . n(H-R)\ r(a2) j ^r^-ai-njr^-ai-n)« n\ + jJ

r(61)r(62) Oo f^1 Wfe-fli-n) (ze^)-» 1 (5.25)

+ r(ai) Ue j ^r(6i-a2-n)r(62-a2-n)n n! + jJ

+ r(fti)r(b2;1 +a,3_bl_b3 f y ^ G(| !_S) | r(ai)r(a2)

where the coefficient is given by formula 16.11.4. And the upper or lower signs are chosen according as z lies in the upper (above the real axis) or lower half-plane (below the real axis). Setting z = Ax", a1 = 1, a2 = 1,61 =2 and 62 = 2 in (5.25) yields

2F2(1,1;2,2;A*")~^ + ^, N » 1.

Hence,

Xx1'3 2F2(l,l-,2,2-,Xx) ~ -2 + j-^, \x\ > 1.

This ends the proof. □

Example 5. The random attenuation capacity of a channel or fading capacity [11] can now be evaluated in terms of the natural logarithm ln and the hypergeometric function 2F2 as

Cbding = E[\og2{l + P\H\2)] = ]Lei/P

£1,1(00) ( i

Lei/p

ln 2

Example 6. One can now evaluate f eAe^x dx in terms of 2F2 using the substitution u = ex, and obtain

' /3 —->-'..... ' ■ - - ■ p

Theorem 7. The logarithmic integral is given by

e Xdx = / --du = lnu + 2F2(1,1; 2, 2; At//3) +C = x + 2F2(1,1; 2, 2; Ae/3'T) + C.

u ft ft

X

dt / ln x \

= J + ln'T 2^2(1,1; 2, 2; ln.T) - 111^2^2(1,1; 2, 2;ln fj, > 1.

m

And for x »

/" r7 + -2-ln^ 2F2(1,1; 2,2; In ¿t). (5.26)

J ln t ln x \ln

Proof. Making the substitution u = ln x and using (4.22) gives

X ln X

^u = [lnu + u2F2(l>l;2>2;u)]K

M ln m

(ln x \

—J + InX 2F2(1,1; 2, 2; ln x) - ln fjL 2F2(1,1; 2,2; ln ¿t)-

Now setting z = lnx, a1 = 1, a2 = 1, 61 = 2 and 62 = 2 in (5.24) or in (4.23) yields

2x

2F2(1,1; 2, 2; ln x) ~ --+ n '' , x > 1.

ln x (ln x)2

This gives

x

lnx 2-F2(l, 1; 2, 2; ln x)--2 + -—, x > 1.

ln x

Hence for x »

X

dt x ln x

u = + www

M

□

We importantly note that Theorem 7 adds the term ln (lnx/ln¡) — 2 — ln12F2(1,1; 2, 2; ln ¡) to the known asymptotic expression of the logarithmic integral in mathematical literature, Li ~ x/ln x [1, 9]. And this term is negligible if x ~ 0(106) or higher.

We can now slightly improve the prime number Theorem [3] as following,

Corollary 2. Let n(x) denotes the number of primes small than or equal to x and i > 1. Then x »

x ln x

^»-^-■»(¡^J-S-M^M^im.

The proof follows directly from equation (5.26) in Theorem 7.

Example 7. One can now evaluate Jln(ln x)dx using integration by parts.

1

/l„ (in *№: = *!„( tax)-/. = x ln (ln x) - ln (ln x) - ln x 2F2(1,1; 2, 2; ln x) + C.

, -dx ln x

Theorem 8. For 0 > 1 and a < 0 + 1, we have

f eAx\ 1 x1-a x"-a+1 / a 1 a 1 -\ „

and for |x| » 1,

Ax"-a+1 ^ / a 1 a 1

2F2 1, -- + - + 1; 2, -- + - + 2; \x

0 - a + 1 2 2 0 0 0 0

Ar (a I \ (_ 1'\ —//3+V/3+i ^ 1 exx?

(5.27)

0V0 0 /\A/ 0 A0x"+a

-1

We also have,

K/3,/3+i = I £mdx = -^+\n(\x\) + ^-2F2(l,l]2,2]\x^+C. (5.28)

Proof.

eAx^ /• 1 (Ax")n 1 /• 1 /• ^ (Ax")n

1/3,a

Ei = / f 1 V^^ir-1 = ^ i^{Xx'3y dr

r±a Xxa " J Xxa ??,! " Ay .t" J n\

n=0 n=1

1 x1-a 1 A An /" «n-a , 1 x1-a A An x"n+"-a+1

, a v — / rl3n~adr = ___ 4- v a x_4. r

\ Z_/ n! / X i ni/fe i « i 1 ^ °

- ---

A 1 -

a1

A 1 - a A^n'y A 1 - a ^ (n + 1)' 0n + 0 - a + 1

n=1 n=0

1 *1_a 4. v rv 0 + 0 + 1/ r ar/3 y , r

+ —~Z^- —7 ^^ \[Xx ) +G

' —n 1 ' " - + -

p p a 1

^ l1'™ I -

x'

'or(,!+2)rHv2)

, ___\ V V / n Vv J , n

A1 — a + 1 \ n!

A 1 - a 0 - a + 1 0 0 0 0

Now setting a,i = 1, a2 = —^ + 4 + 1, b\ = 2, b2 = —^ + 4+2 and 2 = Xx13 in (5.25) gives,

0 0 0 0

a 1 a 1 "

0 0 ) Ax" V 0 0 / VAxV A2x2"

Ax^-"+1

Hence, multiplying (5.29) with —:- gives (5.27). The proof of (5.28) is similar to that

ft—a+1

of (3.20). □

Theorem 9. For any constants a, ft and A,

/ a 1 1 a 1 3 3 A2x2^ 1^2 "773 + 7T3 + "7T3 + 7T3 + 7T> «5 "

or

2ft 2ft 2' 2ft 2ft rr 4

'/a 1 a 1 a 1 a 1

(l, ■-Ti + T) + l; 2.-T) + Tj + 2; i\x' J + 2ii (l, -- + - + 1; 2, -- + - + 2; ^

(5.30)

F, ( — — — — - 3. A2-y2/3

1 2 V 2ft+ + 2'~2ft+ + 2'2' 4

(5.31)

1

2

1

2

Proof. Using Theorem 8, we obtain

sin (Ax^) 1 f eiAx^ — e iAx^

dx = — / -dx

2i / x"

1 Ax/3-0+1 2ft-a+1

a 1 a 1

? + ? + + i a 1 a 1

? + ? + + 5

„( a 1 a 1 , «\

+ 2F2 —-g + — + 1; 2, —— + — + 2; —iXx^J

+ C.

Hence, comparing (2.7) with (5.32) gives (5.30). Or on the other hand,

„ f sinh (Ax^) , f eAx^ — e-Ax^ Ax^-"+1

2 / -dx = / -dx = —-x

J x" J x" ft — a + 1

a 1 a 1 a 1 a 1 , Ai ^

2ii (l, -- + - + 1; 2, -- + - + 2; Ax") +2 (l, -- + - + 1; 2, -- + - + 2; -Ax")] + <7.

(5.33)

Hence, comparing (2.9) with (5.33) gives (5.31). □

Theorem 10. For any constants a, ft and A,

¿x2^-"+1 ( a 1 a 1 3 A2 x2^J x^-"+1 2ft-« + l2F3(v1'-2ft + 2ft+1;-2ft + 2ft+2'2'2;--—)- P-a + lX

„( a 1 a 1 , ^ / a 1 a 1 1

fi (l, ■-J - + 1; 2. -0 + J + 2; /Ax ) - 2F2 (l, -- + - + 1; 2, -- + - + 2; -Ax")].

(5.34)

"

x

Or,

x

2S-"+1

2ft — a + 1

2F3

x

S-"+1

ft—a+1

a 1 a 13 A2x2g

1___I___Li-___I___|_ 2 - 2- _

' 2/3 2/3 ' 2/3 2/3 ' 2' ' 4

„( a 1 a 1 , a

2F2 1, -- - - + 1; 2, -- + - + 2; Xx'3

ftft

ftft

(5.35)

We prove Theorem 10 as Theorem 9 using Theorems 4 and 8.

6. Conclusion

Sis," = /[sin (Axa)/(Ax")]dx, ft > 1, a < ft + 1, and Cia," = /[cos (Axs)/(Ax")]dx, ft > 1, a < 2ft + 1, were expressed in terms of the hypergeometric functions 1F2 and 2F3 respectively, and their asymptotic expressions for |x| » 1 were obtained (see Theorems 1,2, 3, 4 and 5). Once derived, formulas for the hyperbolic sine and hyperbolic cosine integrals were readily deduced from those of the sine and cosine integrals.

On the other hand, the exponential integral Eig," = f (eAx^/x")dx, ft > 1, a < ft + 1, and the logarithmic integral / dx/ ln x were expressed in terms of the hypergeometric function 2F2, and their asymptotic expressions for |x| » 1 were also obtained (see Theorems 6, 7 and 8). Therefore, their corresponding definite integrals can now be evaluated using the FTC rather than using numerical integration.

Using the Euler and hyperbolic identities Sis," and Cig," were expressed in terms of Eig,". And hence, some expressions of the hypergeometric functions 1F2 and 2F3 in terms of 2F2 were derived (see Theorems 9 and 10).

The evaluation of the logarithmic integral / dx/ ln x in terms of the function 2F2 and its asymptotic expression 2F2 for |x| » 1 allowed us to add the term ln (ln x/ln ¡) — 2 — lni 2F2(1,1; 2, 2; ln ¡), I > 1, to the known asymptotic expression of the logarithmic integral, which is Li = J2x dt/ ln t ~ x/lnx [1, 9], so that it is given by Li = JJ dt/lnt ~ x/lnx+ln (lnx/ln¡) — 2 — ln12F2(1,1; 2, 2; ln ¡) in Theorem 7. Beside, this leads to Corollary 2 which is an improvement of the prime number Theorem [3].

In addition, other non-elementary integrals which can be written in terms of Eig;1 and f dx/ ln x and then evaluated were given as examples. For instance, using substitution, the f eAe^xdx was written in terms of Eig,1 and therefore evaluated in terms of 2F2, and using integration by parts, the non-elementary integral J" ln(ln x)dx was written in terms of J" dx/ ln x and therefore evaluated in terms of 2F2.

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