Orthogonal polynomials arising in the numerical evaluation of inverse Laplace transforms
Authors: Salzer Herbert E. 
Published: June 2010

In finding

f
(
t
)

f(t)

, the inverse Laplace transform of

F
(
p
)

F(p)

, where (1)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="f left-parenthesis t right-parenthesis equals left-parenthesis 1 slash 2 pi j right-parenthesis integral Subscript c minus j normal infinity Superscript c plus j normal infinity Baseline e Superscript p Super Superscript t Baseline upper F left-parenthesis p right-parenthesis d p">

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)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="left-parenthesis 1 slash 2 pi j right-parenthesis integral Subscript c minus j normal infinity Superscript c plus j normal infinity Baseline e Superscript p Baseline rho Subscript 2 n Baseline left-parenthesis 1 slash p right-parenthesis d p equals sigma-summation Underscript i equals 1 Overscript n Endscripts upper A Subscript i Baseline Superscript left-parenthesis n right-parenthesis Baseline rho Subscript 2 n Baseline left-parenthesis 1 slash p Subscript i Baseline right-parenthesis">

(
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)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="left-parenthesis 1 slash 2 pi j right-parenthesis integral Subscript c minus j normal infinity Superscript c plus j normal infinity Baseline e Superscript p Baseline left-parenthesis 1 slash p right-parenthesis p Subscript n Baseline left-parenthesis 1 slash p right-parenthesis left-parenthesis 1 slash p right-parenthesis Superscript i Baseline d p equals 0 comma i equals 0 comma 1 comma midline-horizontal-ellipsis comma n minus 1">

(
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
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="upper P Subscript n Baseline left-parenthesis 1 slash p right-parenthesis identical-to left-parenthesis 4 n minus 2 right-parenthesis left-parenthesis 4 n minus 6 right-parenthesis midline-horizontal-ellipsis 6 p Subscript n Baseline left-parenthesis 1 slash p right-parenthesis comma n greater-than-or-equal-to 2">

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
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="left-parenthesis negative 1 right-parenthesis Superscript n Baseline e Superscript negative p Baseline p Superscript n Baseline d Superscript n Baseline left-parenthesis e Superscript p Baseline slash p Superscript n Baseline right-parenthesis slash d p Superscript n">

(
−
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)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="left-parenthesis 1 slash 2 pi j right-parenthesis integral Subscript c minus j normal infinity Superscript c plus j normal infinity Baseline e Superscript p Baseline left-parenthesis 1 slash p right-parenthesis left-bracket upper P Subscript n Baseline left-parenthesis 1 slash p right-parenthesis right-bracket squared d p equals one half left-parenthesis negative 1 right-parenthesis Superscript n">

(
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)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="left-parenthesis 2 n minus 3 right-parenthesis upper P Subscript n Baseline left-parenthesis x right-parenthesis equals left-bracket left-parenthesis 4 n minus 2 right-parenthesis left-parenthesis 2 n minus 3 right-parenthesis x plus 2 right-bracket upper P Subscript n minus 1 Baseline left-parenthesis x right-parenthesis plus left-parenthesis 2 n minus 1 right-parenthesis upper P Subscript n minus 2 Baseline left-parenthesis x right-parenthesis">

(
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)
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="x squared upper P Subscript n Baseline Superscript Baseline left-parenthesis x right-parenthesis plus left-parenthesis x minus 1 right-parenthesis upper P prime Subscript n Baseline left-parenthesis x right-parenthesis minus n squared upper P Subscript n Baseline left-parenthesis x right-parenthesis equals 0">

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

.