Vistas of Special Functions by Shigeru Kanemitsu, Haruo Tsukada
By Shigeru Kanemitsu, Haruo Tsukada
It is a distinct ebook for learning certain services via zeta-functions. Many vital formulation of specific features scattered through the literature can be found of their right positions and readers get enlightened entry to them during this ebook. The parts lined comprise: Bernoulli polynomials, the gamma functionality (the beta and the digamma function), the zeta-functions (the Hurwitz, the Lerch, and the Epstein zeta-function), Bessel capabilities, an creation to Fourier research, finite Fourier sequence, Dirichlet L-functions, the rudiments of advanced services and summation formulation. The Fourier sequence for the (first) periodic Bernoulli polynomial is successfully used, familiarizing the reader with the connection among designated services and zeta-functions.
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4 Determine the value of the probability integral Γ x by considering two functions f (x) 1 0 = 2 2 e−t dt and g(x) 0 2 2 1 e−x (t +1) dt. t2 + 1 Solution Recalling the fundamental theorem of calculus, we obtain f (x) = 2 e−x x 2 2 e−t dt. 0 We may differentiate g(x) under the integral sign to get 1 g (x) = 0 t2 1 d −x2 (t2 +1) e dt + 1 dx 1 = −2x e−x 2 (t2 +1) 0 by the change of variable u = xt. dt = −2 e−x x 2 0 2 e−u du 1 2 = March 27, 2007 17:14 WSPC/Book Trim Size for 9in x 6in vista 35 The theory of the gamma and related functions Hence we conclude that f (x) + g (x) = 0, whence by the Newton-Leibniz principle (cf.
R+1 Hence LHS = −m−n+r n n (−Br ) − m−n Br r r or m−n (mr − 1)br = −(m−n − m−n+r ) whence br = n Br , r n Br , r March 27, 2007 17:14 WSPC/Book Trim Size for 9in x 6in 20 vista Vistas of Special Functions completing the proof. 5). Let F (x, t) = t ext et − 1 and expand it into the Taylor series in t: F (x, t) = ∞ 1 Gn (x)tn , n! 29) with Gn (x) a polynomial of degree n, (n) Gn (x) = a0 xn + · · · + a(n) n , say. 29), we get ∞ 1 n 1 yt et = y Gn eyt − 1 n! y n=0 1 y and t by ty in tn , which leads, as y → 0, to t e = (since y n Gn (n) Hence a0 Now, 1 m 1 y ∞ 1 (n) n a0 t n!
U + 1 − l) x 1 (x + a)u+1 + ζ(−u, a), u = −1, + u+1 log(x + a) − ψ(a), u = −1.