Evaluation of the non-elementary integral e

Noname manuscript No. will be inserted by the editor Evaluation of the non-elementary integral e λx dx,, and related integrals Victor Nijimbere Received: date / Accepted: date Abstract A formula for the non-elementary integral e λx dx, where λ is complex and is real and greater or equal two, is obtained in terms of the confluent hypergeometric function F. This result is verified by directly evaluating the area under the Gaussian Bell curve, corresponding to =, using the asymptotic expression for the confluent hypergeometric function and the Fundamental Theorem of Calculus FTC. Two different but equivalent expressions, one in terms of the confluent hypergeometric function F and another one in terms of the hypergeometric function F, are obtained for each of these integrals, coshλx dx, sinhλx dx, cosλx dx and sinλx dx, λ C,. And the hypergeometric function F is expressed in terms of the confluent hypergeometric function F. Keywords Non-elementary integral Hypergeometric function Confluent hypergeometric function Asymptotic evaluation Fundamental theorem of calculus Mathematics Subject Classification 6A36 33C5 3E5 Introduction Definition An elementary function is a function of one variable built up using that variable and constants, together with a finite number of repeated algebraic operations and the taking of exponentials and logarithms [6]. Victor Nijimbbere School of mathematics and Statistics Carleton University Ottawa ON KS 5B6, Canada Tel.: +-63-5-6 ext. 37 Fax: +-63-5-3536 E-mail: nijimberevictor@cmail.carleton.ca

Victor Nijimbere Definition An indefinite integral is non-elementary if it can not be expressed in terms of elementary functions [6]. In 835, Joseph Liouville established conditions in his theorem, known as Liouville 835 s Theorem [,6], which can be used to determine whether or not an indefinite integral is elementary or non-elementary. Using Liouville 835 s Theorem, one can show that the indefinite integral e λx dx,, is noelementary [], and to my knowledge, there does not exist a formula for this non-elementary integral. For instance, if =, λ = β <, where β is a real constant, the area under the Gaussian Bell curve can be calculated using double integration and then polar coordinates to obtain e β x dx = π β. Is that possible to evaluate by directly using the Fundamental Theorem of Calculus FTC as in equation? e β x dx = t t e β x dx + t t e β x dx. The Central it Theorem CLT in Probability theory [] states that the probability that a random variable X does not exceed some observed value z is P X < z = π z e x dx. 3 If the antiderivative of the function gx = e λx is known, then one may choose to use the FTC to compute the cumulative probability P X < z in 3 if the value of z is given or is known, instead of using numerical integration. There are many other examples where the antiderivative of gx = e λx, can be useful. For instance, using the FTC, formulas for integrals such as x e tn+ dt, x < ; x e tn+ dt, x > ; x t n e t dt, x, where n is a positive integer, can be obtained if the antiderivative of gx = e λx, is known. In this paper, the antiderivative of gx = e λx,, is expressed in terms of a special function, the confluent hypergeometric F []. And the confluent hypergeometric F is an entire function [3] and its properties are well known [,5].

Evaluation of the non-elementary integral e λx dx,, and related integrals 3 The main goal here is to consider the most general case with λ complex λ C, evaluate the non-elementary integral e λx dx,, and thus make possible the use of the FTC to compute the definite integral B A e λx dx, λ C,, 5 for any A and B. And once 5 is evaluated, then integrals such as,, 3, can also be evaluated. Using the hyperbolic and Euler identities, coshλx = e λx + e λx /, sinhλx = e λx e λx /, cosλx = e iλx + e iλx / and sinλx = e iλx e iλx /i, the integrals coshλx dx, sinhλx dx, sinλx dx and cosλx dx, where λ C and, are evaluated in terms of F. They are also expressed in terms of the hypergeometric F. And some expressions of the hypergeometric function F in terms of the confluent hypergeometric function F are therefore obtained. Evaluation of B e λx dx A Proposition The function Gx = x F ; + ; λx, where F is a confluent hypergeometric function [], λ is an arbitrarily constant and, is the antiderivative of the function gx = e λx. Thus, e λx dx = x F ; + ; λx + C. 6 Proof We expand gx = e λx as a Taylor series and integrate the series term by term. We also use the Pochhammer s notation [] for the gamma function, + n = n, where n = + + n, and the property of the gamma function + = []. For example, n + + = n + n + for any real n. We then obtain λ n gxdx = e λx dx = x n dx = x n + λx n n + + + C = x n λx n + + C n = x F ; + ; λx + C = Gx + C. 7

Victor Nijimbere Example One can now evaluate x n e λx dx for any positive integer n. Using integration by parts, x n e λx dx = xn λ eλx n x n e λx dx. 8 λ. For n =, x e λx dx = x λ eλx λ. For n =, x e λx dx = x3 λ eλx 3 x e λx dx λ = x3 λ eλx 3x λ eλx + 3x λ F e λx dx = x λ eλx x λ F ; 3 ; λx +C. 9 ; 3 ; λx + C. Example Using the method of integrating factors, the first-order ordinary differential equation, y x y =, has solution yx = e x e x dx + C = xe x F ; 3 ; x + Ce x. It is also important to note that the function erfx, erfx = π x e t dt = x π F ; 3 ; x, 3 is a particular solution of the first-order initial-value problem πy = e x, x, y =. It only takes positive values of x and corresponds to a particular case of Proposition with λ =, = and the constant C =. Lemma Assume that an antiderivative of gx = e λx, λ C and, exists and is Gx.. If the real part of λ < and is even, then the its Gx and x Gx are finite constants. And thus the Lebesgue integral e λx dx < x.. If λ is real, then the point, G =, is an inflection point of the curve Y = Gx, x R.

Evaluation of the non-elementary integral e λx dx,, and related integrals 5 3. And if λ < and real, and is even, then the its Gx and x Gx are finite. And there exists a positive real constant θ > such x that x Gx = θ and Gx = θ. x Proof. If λ is complex, then the antiderivative Gx of the function gx is an entire function on the whole complex plane C. And if the real part of λ < and is even, then x ± G x = gx = x ± x ± eλx =. By Liouville theorem, Gx has to be constant as x ± section 3..3 in [3]. Let λ r and λ i be the real and imaginary parts of λ respectively. Then, the Lebesgue integral, e λx dx = e λ r+iλ i x dx = e λ rx e iλ ix dx = e λ rx dx < since Gx is constant as x ±. Otherwise, if the real part of λ < and is odd, then and so does the integral gx diverges x e λx dx. And if the real part of λ >, then gx diverges and so does e λx dx regardless of the value of. x. At x =, g =. And so, around x =, the antiderivative Gx x because G = g =. This gives, G =,. Since, G x = g x = λx e λx for, then G =. Hence, by the second derivative test, if λ is real, then the point, G =, is an inflection point of the curve Y = Gx, x R. 3. If λ is a real constant then Gx is analytic on R. For λ <, gx = x gx =. Therefore, by Liouville theorem, Gx has to be constant as x x ± since if λ C and its real part is negative, then e λx dx <. In addition, the fact that G x < if x < and G x > if x implies that Gx is concave upward on the interval, while is concave downward on the interval,. Moreover, the fact that gx = G x is symmetric about the y-axis even implies that Gx has to be antisymmetric about the y-axis odd. Hence there exists a positive constant θ > such that proof. x Gx = θ while x Gx = θ. This completes the If λ = and =, then e x dx = x F ; 3 ; x + C. 5 According to 5, the antiderivative of gx = e x is Gx = x F ; 3 ; x. Its graph as a function of x, sketched using MATLAB, is shown in Figure,

6 Victor Nijimbere it is in agreement with Lemma. It is actually seen in Figure that, is an inflection point and that Gx reaches some constants as x ± as predicted by Lemma..8 θ = π.6. Gx....6 θ = π.8 5 5 5 5 x Fig. Plot of the antiderivative of gx = e x given by 5. Lemma Consider Gx in Proposition.. Then for x, Gx = x F ; + ; λx + e i π λ + e i π λ x x + eλx λx, + eλx λx, if is even if is odd 6. Assume that is even and λ <, and set λ = β, where β is a real number, preferably positive. Then, and G = Gx x = x x F ; + ; β x = β + 7 G = Gx x = x x F ; + ; β x = β +. 8

Evaluation of the non-elementary integral e λx dx,, and related integrals 7 3. And by the FTC, e β x dx = G G = β = β + β + +. 9 Proof. To prove 6, we use the asymptotic series for the confluent hypergeometric function that is valid for z [], formula 3.5., F a; b; z b = e±iπa z a b a + ez z a b a { R { S a n + a b n z n + O z R } b a n a n z n + O z S }, where a and b are constants, and the upper sign being taken if π < argz < 3π and the lower sign if 3π < argz π. We set z = λx, a = and b = +, and obtain F ; { + ; λx + = ei π R } n λx n + O λx R λx { + S } eλx n n λx λx n + O λx S. Then, for x, e i π λx while eλx λx { R { S } n λx n + O λx R n n e i π λ e i π λ x, λx n + O λx S } Therefore for x, F ; + ; λx + e i π λ + e i π x + eλx λx, λ x + eλx if is even x, if is odd eλx λx. 3 if is even λx, if is odd

8 Victor Nijimbere Hence for x, Gx = x F ; + ; λx + e i π λ + e i π λ x x + eλx λx, + eλx λx, if is even if is odd 5. Setting λ = β, where β is real and positive and using 6, then for even, Gx = x F ; + ; β x Hence, while G = G = 3. By the FTC, e β x dx = x β + x β x e β, 6 x x F x ; + ; β x = β +, 7 x F x ; + ; β x = β +. 8 y y = y y F e β x dx + y y ; + ; β y e β x dx y F y ; + ; β y = G G = β + β + = β +. 9 We now verify wether 9 is correct or not by double integration. We first observe that 9 is valid for all even. And so, if 9 is verified for =, we are done since 9 is valid for all even. Considering =,

Evaluation of the non-elementary integral e λx dx,, and related integrals 9 and setting = in 9 gives e β x dx = y y On the other hand, e β x dx + y y e β x dx = y F y ; 3 ; β y y F y ; 3 ; β y = G G = 3 β = π π = β β. 3 In polar coordinate, Hence, e β x dx = = e β x +y dydx = π e β x dx = as before. This completes the proof. e β x dx e β y dy 3 e β x +y dydx. 3 e β r rdrdθ = β π e x +y π dydx = β Example 3 Setting λ = and = in Lemma gives Gx = x F x x ; 3 π ; x = and dθ = π β. 33 3 35 Gx = x F x x ; 3 π ; x =. 36 This implies θ = π in Lemma. And this is exactly the value of Gx as x in Figure. And also Gx = θ = π x as in Figure. The FTC gives π π e x dx = G G = = π, 37

Victor Nijimbere and e x dx = G G = π π = = π e x dx = G G = = e x dx 38 π. 39 Example In this example, the integral x e tn+ dt, x <, where n is a positive integer, is evaluated using Proposition and the asymptotic expression 6. Setting λ = and = n+ in Proposition, and using 6 gives x e tn+ dt = x y y e tn+ dt = x F n + ; n + n + ; xn+ = x F n + ; n + n + ; xn+ One can also obtain, y F y n + n + n + ; n + n + ; yn+, x <. x e tn+ dt = y y x e tn+ dt = y F y n + ; n + n + ; yn+ n + = n + x F x F n + ; n + n + ; xn+ n + ; n + n + ; xn+, x >. Theorem For any A and B, the FTC gives B A e λx dx = GB GA, 3 where G is the antiderivative of the function gx = e λx and is given in Proposition. And λ is any complex or real constant, and.

Evaluation of the non-elementary integral e λx dx,, and related integrals Proof Gx = x F ; + ; λx, where λ is any constant, is the antiderivative of gx = e λx, by Proposition, Lemma and Lemma. And since the FTC works for A = and B = in 37, A = and B = in 38 and A = and B = in 39 by Lemma if λ = and =, and for all λ < and all even, then it has to work for other values of A, B R and for any λ C and. This completes the proof. Example 5 In this example, we apply Theorem to the Central Limit Theorem in Probability theory []. The normal zero-one distribution of a random variable X is the measure µdx = g X xdx, where dx is the Lebesgue measure and the function g X x is the probability density function p.d.f of the normal zero-one distribution [], and is g X x = π e x, < x <. A comparison with the function gx in Proposition and Lemma gives λ = β = / and =. By Theorem, the cumulative probability, P X < z, is then given by P X < z = µ{, z} = z = π g X xdx z e x dx = + z F π ; 3 ; z. 5 One may also use Theorem to compute the probabilities, P < X < = µ{, } =.77.77 =.95, P < X < = µ{, } =.77.33 =.885 and so on. 3 Other related non-elementary integrals Proposition For, the function Gx = x F ;, + ; λ x, where F is a hypergeometric function [] and λ is an arbitrarily constant, is the antiderivative of the hyperbolic function gx = cosh λx. Thus, cosh λx dx = x F ;, + ; λ x + C. 6

Victor Nijimbere Proof gxdx = cosh λx dx = = x = x = x λx n dx λ n x n n + + C n + n + n + + x n + C λ n λ x n + C n + n = x F ;, + ; λ x + C = Gx + C. 7 Proposition 3 For, the function Gx = λx+ + F + ; 3, + 3 ; λ x, where F is a hypergeometric function [] and λ is an arbitrarily constant, is the antiderivative of the hyperbolic function gx = sinh λx. Thus, sinh λx dx = λx+ + F + ; 3, + 3 ; λ x + C. 8 Proof gxdx = = sinh λx dx = λx+ = λx+ = λx+ + λx n+ n +! dx λ n x n n +! n + + n + + n + n + 3 = λx+ + F + n n + 3 n + C + 3 n λ x λ x n + C + C + ; 3, + 3 ; λ x + C = Gx + C. 9

Evaluation of the non-elementary integral e λx dx,, and related integrals 3 We also can show as above that cos λx dx = x F ;, + ; x λ + C 5 and sin λx dx = λx+ + F + ; 3, + 3 ; x λ + C, 5 Theorem For any constants and λ,. F ;, + ; λ x = [F ; + ; λx + F ; ] + ; λx. 5. And F ;, + ; x λ = [F ; + ; iλx + F ; ] + ; iλx. 53 Proof. Using Proposition, we obtain e coshλx λx + e λx dx = dx = x [F ; + ; λx + F ; ] + ; λx + C, 5 Hence, comparing 7 with 5 gives 5.. On the other hand, e cosλx iλx + e iλx dx = dx = x [F ; + ; iλx + F ; ] + ; iλx + C, 55 Hence, comparing 5 with 55 gives 53. Theorem 3 For any constants and λ,

Victor Nijimbere. λx + F + ; 3, + 3 ; λ x = [F ; + ; λx F ; ] + ; λx. 56. And λx + F + ; 3, + 3 ; x λ = [F i ; + ; iλx F ; ] + ; iλx. 57 Proof. Using Proposition, we obtain e sinh λx λx e λx dx = dx = x [F ; + ; λx F ; ] + ; λx. 58 Hence, comparing 9 with 58 gives 56.. On the other hand, e sin λx iλx e iλx dx = dx = x [F i ; + ; iλx F ; ] + ; iλx. 59 Hence, comparing 5 with 59 gives 57. Conclusion The non-elementary e λx dx, where λ is an arbitrary constant and, was expressed in term of the confluent hypergeometric function F. And using the properties of the confluent hypergeometric function F, the asymptotic expression for x of this integral was derived too. As established in Theorem, the definite integral 5 can now be computed using the FTC. For example, one can evaluate the area under the Gaussian Bell curve using the FTC rather than using double integration and then polar coordinates. One can also choose to use Theorem to compute the cumulative probability for the normal distribution.

Evaluation of the non-elementary integral e λx dx,, and related integrals 5 On one hand, the integrals coshλx dx, sinhλx dx and sinλx dx and cosλx dx, λ C, were evaluated in terms of F, while on another hand, they were expressed in terms of the hypergeometric F. This allowed to express the hypergeometric function F in terms of the confluent hypergeometric function F. References. Abramowitz M., Stegun I. A., Handbook of mathematical functions with formulas, graphs and mathematical tables, National Bureau of Standards 96.. Billingsley P., Probability and measure, Wiley series in Probability and Mathematical Statistics 986. 3. Krantz S. G., Handbook of Complex variables, Boston, MA Birkäusser 999.. Marchisotto E. A., Zakeri G.-A.: An invitation to integration in finite terms, College Mathematical Journal 5 pp. 95 38 99. 5. NIST Digital Library of Mathematical Functions, http://dlmf.nist.gov/. 6. Rosenlicht M., Integration in finite terms, American Mathematical Monthly 799 pp. 963 97 97.