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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      .

Orthogonal polynomials arising in the numerical evaluation of inverse Laplace transforms