Orthogonal polynomials arising in the numerical evaluation of inverse Laplace transforms

In finding f ( t ) f(t) , the inverse Laplace transform of F ( p ) F(p) , where (1) f ( t ) = ( 1 / 2 π j ) ∫ c − j ∞ c + j ∞ e p t F ( p ) d p f(t) = (1/2\pi j)\int _{c - j\infty }^{c + j\infty }{e^{{p^t}}}F(p)dp , the function F ( p ) F(p) may be either known only numerically or too complicated for evaluating f ( t ) f(t) by Cauchy’s theorem. When F ( p ) F(p) behaves like a polynomial without a constant term, in the variable 1 / p 1/p , along ( c − j ∞ , c + j ∞ c - j\infty ,\;c + j\infty ), one may find f ( t ) f(t) numerically using new quadrature formulas (analogous to those employing the zeros of the Laguerre polynomials in the direct Laplace transform). Suitable choice of p i {p_i} yields an n n -point quadrature formula that is exact when ρ 2 n {\rho _{2n}} is any arbitrary polynomial of the ( 2 n 2n )th degree in x ≡ 1 / p x \equiv 1/p without a constant term, namely: (2) ( 1 / 2 π j ) ∫ c − j ∞ c + j ∞ e p ρ 2 n ( 1 / p ) d p = ∑ i = 1 n A i ( n ) ρ 2 n ( 1 / p i ) (1/2\pi j)\int _{c - j\infty }^{c + j\infty } {{e^p}} {\rho _{2n}}(1/p)dp = {\sum \limits _{i = 1}^n {{A_i}} ^{(n)}}{\rho _{2n}}(1/{p_i}) . In (2), x i ≡ 1 / p i {x_i} \equiv 1/{p_i} are the zeros of the orthogonal polynomials P n ( x ) ≡ ∏ i = 1 n ( x − x i ) {P_n}(x) \equiv \prod \limits _{i = 1}^n {(x - {x_i})} where (3) ( 1 / 2 π j ) ∫ c − j ∞ c + j ∞ e p ( 1 / p ) p n ( 1 / p ) ( 1 / p ) i d p = 0 , i = 0 , 1 , ⋯ , n − 1 (1/2\pi j)\int _{c - j\infty }^{c + j\infty } {{e^p}} (1/p){p_n}(1/p){(1/p)^i}dp = 0,\;i = 0,1, \cdots ,n - 1 and A i ( n ) {A_i}^{(n)} correspond to the Christoffel numbers. The normalization P n ( 1 / p ) ≡ ( 4 n − 2 ) ( 4 n − 6 ) ⋯ 6 p n ( 1 / p ) , n ≥ 2 {P_n}(1/p) \equiv (4n - 2)(4n - 6) \cdots 6{p_n}(1/p),n \geq 2 , produces all integral coefficients. P n ( 1 / p ) {P_n}(1/p) is proven to be ( − 1 ) n e − p p n d n ( e p / p n ) / d p n {( - 1)^n}{e^{ - p}}{p^n}{d^n}({e^p}/{p^n})/d{p^n} . The normalization factor is proved, in three different ways, to be given by (4) ( 1 / 2 π j ) ∫ c − j ∞ c + j ∞ e p ( 1 / p ) [ P n ( 1 / p ) ] 2 d p = 1 2 ( − 1 ) n (1/2\pi j)\int _{c - j\infty }^{c + j\infty } {{e^p}} (1/p){[{P_n}(1/p)]^2}dp = \tfrac {1}{2}{( - 1)^n} . Proofs are given for the recurrence formula (5) ( 2 n − 3 ) P n ( x ) = [ ( 4 n − 2 ) ( 2 n − 3 ) x + 2 ] P n − 1 ( x ) + ( 2 n − 1 ) P n − 2 ( x ) (2n - 3){P_n}(x) = [(4n - 2)(2n - 3)x + 2]{P_{n - 1}}(x) + (2n - 1){P_{n - 2}}(x) , for n ≥ 3 n \geq 3 , and the differential equation (6) x 2 P n ( x ) + ( x − 1 ) P n ′ ( x ) − n 2 P n ( x ) = 0 {x^2}{P_n}^{}(x) + (x - 1){P_n}’(x) - {n^2}{P_n}(x) = 0 . The quantities p i ( n ) , 1 / p i ( n ) {p_i}^{(n)},1/{p_i}^{(n)} and A i ( n ) {A_i}^{(n)} were computed, mostly to 6S – 8S, for i = 1 ( 1 ) n , n = 1 ( 1 ) 8 i = 1(1)n,n = 1(1)8 .