DEMONSTRATIO MATHEMATICA Vol. XLVI No 3 2013 M. A. Pathan, O. A. Daman LAPLACE TRANSFORMS OF THE LOGARITHMIC FUNCTIONS AND THEIR APPLICATIONS Abstract. This paper deals with theorems and formulas using the technique of Laplace and Steiltjes transforms expressed in terms of interesting alternative logarith- mic and related integral representations. The advantage of the proposed technique is illustrated by logarithms of integrals of importance in certain physical and statistical problems. 1. Introduction The aim of this paper is to obtain some theorems and formulas for the evaluation of finite and infinite integrals for logarithmic and related func- tions using technique of Laplace transform. Basic properties of Laplace and Steiltjes transforms and Parseval type relations are explicitly used in combi- nation with rules and theorems of operational calculus. Some of the integrals obtained here are related to stochastic calculus [6] and common mathemati- cal objects, such as the logarithmic potential [3], logarithmic growth [2] and Whittaker functions [2, 3, 4, 6, 7] which are of importance in certain physical and statistical applications, in particular in energies, entropies [3, 5, 7 (22)], intermediate moment problem [2] and quantum electrodynamics. The ad- vantage of the proposed technique is illustrated by the explicit computation of a number of different types of logarithmic integrals. We recall here the definition of the Laplace transform 1 −st (1) L ff (t)g = L[f(t); s] = \ e f (t) dt: 0 Closely related to the Laplace transform is the generalized Stieltjes trans- form 2010 Mathematics Subject Classification: 33C05, 33C90, 44A10. Key words and phrases: Laplace and Steiltjes transforms, logarithmic functions and integrals, stochastic calculus and Whittaker functions. 534 M. A. Pathan, O. A. Daman 1 f(t)dt (2) S ff(t)g = = G(a; ρ) ρ \ (a + t)ρ 0 which, for ρ = 1, gives the Stieltjes transform 1 f(t)dt (3) Sff(t)g = = G(a): \ (a + t) 0 After a change of integration variable, (2) is transfomed into 1 f(−a ln x) (4) dx = aρ−1G(a; ρ): \ x(1 − ln x)ρ 0 Special cases of (2) and (4), when f(t) = tλ−1e−t, are generalizations of gamma function given by Kobayashi [7] 1 λ−1 −ρ (5) Γρ(λ, a) = \(− ln x) (a − ln x) dx; R(λ) > 0: 0 Kobayashi [7] applied this generalized gamma function integral in diffraction theory. 2. Theorems In this section, we state and prove some theorems in the study of integral transforms, and briefly discuss some apparent, known and new special cases of these theorems. We will apply systematically the rules and theorems of the operational calculus assuming the existence of the Laplace transforms of the functions involved and the permissibility of performed mathematical operations. Theorem 1. If (6) Lff(t)g = '(s) and (7) Lfh(t)g = g(s) then 1 − ln xn − ln x − ln x g f dx \ s s s 0 1 n (8) = s \ '(t)h (t − s)dt s 1 n (9) = s \ '(s + x)h (x)dx 0 Laplace transforms of the logarithmic functions and their applications 535 1 1 hn(t) (10) L[fhn(t − a)H(t − a)g; s]ds = Γ(n + 1) dt \ \ (t + a)n+1 0 0 and 1 φ( − ln x ) 1 (11) u dx = L[φ(t − a)H(t − a); x]dx \ (au − ln x) \ 0 u provided that, f(t) 2 L2(0; 1), e−sttng(t) 2 L2(0; 1), h(t) 2 L2(0; 1) and hn(t) denotes the nth differential coefficient of h(t) such that h0(0) = h00(0) = ··· = hn−1(0) = 0. H(t) is the Heaviside’s unit function and integrals in (8) to (11) are convergent. Proof. Consider the Laplace transform 1 −st (12) Lff(t)g = L[f(t); s] = \ e f(t)dt = '(s): 0 Recall a well known property of the Laplace transform [4, p. 129] that is, if (13) Lfh(t)g = g(s) then (14) Lfhn(t)g = sng(s); h0(0) = h00(0) = ··· = hn−1(0) = 0 and (15) Lfhn(t − a)H(t − a)g = e−assng(s); where H(t) is a Heaviside’s unit function. To prove (8) and (9), we use (12) and (15) in the Parseval theorem and then, by changing the integration variable x = e−st, we find 1 ln xn ln x ln x (16) − g − f − dx \ s s s 0 1 1 n n = s \ '(t)h (t − s)dt = s \ '(s + x)h (x)dx; s 0 where a is replaced by s. Now, we integrate both sides of (15) and use (14) to obtain 1 n (17) \ Lfh (t − a)H(t − a)gds 0 1 1 1 "1 # −as n −st n n n −as−st = \ e s \ e h (t)dtds = \ h (t) \ s e ds dt: 0 0 0 0 536 M. A. Pathan, O. A. Daman By evaluating the integral on the right hand side of (17), we obtain a Parseval relation for (13). Thus 1 1 hn(t) (18) Lfhn(t − a)gH(t − a)gdt = Γ(n + 1) dt: \ \ (t + a)n+1 0 0 1 1 h(t) For n = 0, (18) gives T0 Lfh(t − a)H(t − a)gdt = T0 (t+a) dt: For n = 0 and a = 0, (18) gives a known result [5, p. 110 (2.4)] 1 1 h(t) (19) Lfh(t)gds = dt: \ \ t 0 0 Now, set h(t) = e−utφ(t) and n = 0 in (19) and use shift property 1 −ut −ut−st (20) Lfh(t)g = Lfe φ(t)g = \ e φ(t)dt = L[φ(t); u + s]; 0 to get another Parseval-type relation 1 φ(t) (21) \ L[φ(t − a)H(t − a); x]dx = L ; u ; u a + t by making the change of variable u + s = x. On changing the integration variable x = e−ut, we get (11). For a = 0, (21) gives a known result [5, p. 110 (2.8)] 1 φ(t) (22) \ L[φ(t); x]dx = L ; u : u t If we take h(t) = tλe−at and use [4, p. 129] dn (23) Lftnf(t)g = (−1)n F (s); n = 1; 2; ::: dsn and binomial theorem in Theorem 1, we obtain Theorem 2. If L[f(t); s] = F (s) then 1 n a n − ln x n+1 d (24) x s (ln x) f dx = s F (s + a) \ s dsn 0 and 1 ln xn ln x−λ−1 ln x (25) − a − f − dx \ s s s 0 n ! X (−1)n−ran−r n = s φ(a; s) Γ(λ − r + 1) r=0 r Laplace transforms of the logarithmic functions and their applications 537 where 1 λ−r −ax (26) φ(a; s) = \ x e F (s + x)dx 0 provided that Laplace transform of jf(t)j exists, λ > n − 1; R(s + a) > 0 and the integrals in (25) and (26) are convergent. 1 In the simplest case f(t) = 1 and F (s) = , we have immediately from s (25) and [4, p. 294 (6)] 1 ln xn ln x−λ−1 (27) − a − dx \ s s 0 n n−r n−(λ+r+1)=2 ! as X (−1) a n = e a W (as); s(λ−r−1)=2 k;m r=0 r where k = (r−λ−1)=2; m = (λ−r)=2 and Wk;m(x) is Whittaker function [4]. Evidently, if we set f(t) = 1 in (24), we get 1 s + a−n−1 (28) xa=s(ln x)ndx = (−1)nn! : \ s 0 Another example is 1 1 s + 1 s L[f(t)] = L = − ; R(s) > 0 1 + e−t 2 s 2 which leads to 1 xa=s(ln x)n sn+1 a + s + 1 a + s (29) = (n) − (n) ; \ 1=s 2n+1 2 2 0 1 + x n = 1; 2;::: , R(s) > 0; where (ζ) is a psi-function and (n)(ζ) means the nth derivatives of psi-function [4]. On the other hand, the special cases of (24) and (25), for n = 0 and a = 0, yield known results [1, p. 241, equations (23), (26) and (28)]. Theorem 3. Let α > 0; β > 0; then 1 − ln x f( α+β ) p (30) \ p dx = πα(α + β)L[g(θ; t); α] 0 ln(1=x) where t e−βθf(θ) (31) g(θ; t) = \ p dθ: 0 π θ(t − θ) 538 M. A. Pathan, O. A. Daman Proof. Let g(θ; t) be defined by (31). Then 1 t −βθ −αt e f(θ) L[g(θ; t); α] = \ e \ p dθ dt 0 0 π θ(t − θ) " # 1 e−βθ 1 e−αt = \ p f(θ) \ p dt dθ 0 π θ θ t − θ " # 1 1 e−(α+β)θ 1 e−αs = p f(θ) p ds dθ π \ \ s 0 θ 0 where in the inner integral, we have changed the variable of integration by setting t − θ = s. It follows from 1 e−τt rπ (32) p dt = ; R(τ) > 0 \ τ 0 t −(α+β)θ that L[g(θ; t); α] = p1 1 e p f(θ)dθ. The uniqueness of Laplace trans- πα T0 θ forms and the substitution e−(α+β)θ = x implies the required result. It will be shown that, if we set f(θ) = 1 in the integral (31) and use (32), then Theorem 3 reduces to the following P. Levy’s Arc-Sine Law for occupation time of (0; 1) [6, p.