Numerical integration based on the hyperfunction theory

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Numerical Integration based on the Hyperfunction Theory
The 6th China-Japan-Korea Joint Conference on Numerical Mathematics
∗Hidenori Ogata (The University of Electro-Communications, Japan)
joint work with Hiroshi Hirayama (Kanaga Institute of Technology, Japan)
23 August, 2016

Contents
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An application of the hyperfunction theory to numerical analysis

Contents
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hyperfunction theory✓ ✏
generalized function theory proposed by M. Sato (1958)
based on the complex function theory
function with singularities
pole
discontinuity
delta function, ...
←−
complex analytic functions
familiar in
numerical computations
✒ ✑

Contents
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hyperfunction theory✓ ✏
generalized function theory proposed by M. Sato (1958)
based on the complex function theory
function with singularities
pole
discontinuity
delta function, ...
←−
complex analytic functions
familiar in
numerical computations
✒ ✑
In this speech, we show a numerical integration method based on
the hyperfunction theory, which is expected to be efﬁcient for integrals with
singularities.

Contents
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1. Hyperfunction theory
2. Numerical integration over a ﬁnite interval
3. Numerical integration over an inﬁnite interval
4. Numerical examples
5. Summary

Contents
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5. Summary

1. Hyperfunction theory (an example)
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b
a
φ(x)δ(x)dx = φ(0).
1
2πi C
φ(z)
z
dz = φ(0).
( a < 0 < b )
Dirac delta function Cauchy integral formula
O
C
D
φ(z) : analytic in D

5 / 29
b
a
φ(x)δ(x)dx = φ(0).
1
2πi C
φ(z)
z
dz = φ(0).
( a < 0 < b )
O
C
D
Oa b
1
2πi C
φ(z)
z
dz = −
1
2πi
b
a
φ(x)
1
x + i0
−
1
x − i0
dx.

5 / 29
b
a
φ(x)δ(x)dx = φ(0).
1
2πi C
φ(z)
z
dz = φ(0).
( a < 0 < b )
O
C
D
Oa b
b
a
φ(x)δ(x)dx =
1
2πi C
φ(z)
z
dz = −
1
2πi
b
a
φ(x)
1
x + i0
−
1
x − i0
dx.
∴ δ(x) = −
1
2πi
1
x + i0
−
1
x − i0
.

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Hyperfunction theory (M. Sato, 1958)✓ ✏
• A hyperfunction f(x) on an interval I is the difference of the boundary
values of an analytic function F(z).
f(x) = [F(z)] ≡ F(x + i0) − F(x − i0).
F(z) : the deﬁning function of the hyperfunction f(x)
analytic in D I,
where D is a complex neighborhood of the interval I.
✒ ✑
D
I
R

1. Hyperfunction theory: examples
7 / 29
• Dirac delta function
δ(x) = −
1
2πiz
= −
1
2πi
1
x + i0
−
1
x − i0
.
• Heaviside step function
H(x) =
1 ( x > 0 )
0 ( x < 0 )
= −
1
2πi
{log(−(x + i0)) − log(−(x − i0))} .
∗ log z is the principal value s.t. −π ≦ arg z < π
• Non-integral powers
x+
α
=
xα ( x > 0 )
0 ( x < 0 )
= −
(−(x + i0))α − (−(x − i0))α
2i sin(πα)
(α ∈ Z const.).
∗ zα
is the principal value s.t. −π ≦ arg z < π.

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Heaviside step function
H(x) =
1 ( x > 0 )
0 ( x < 0 )
= F(x+i0)−F(x−i0), F(z) = −
1
2πi
log(−z).
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1Re z
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Im z
-1
-0.5
0
0.5
1
Re F(z)
The real part of the deﬁning function F(z) = −
1
2πi
log(−z).

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Heaviside step function
H(x) =
1 ( x > 0 )
0 ( x < 0 )
= F(x+i0)−F(x−i0), F(z) = −
1
2πi
log(−z).
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1Re z
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Im z
-1
-0.5
0
0.5
1
Re F(z)
Many functions with singularities are expressed by analytic functions
in the hyperfunction theory.

1. Hyperfunction theory: integral
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Integral of a hyperfunction f(x) = F(x + i0) − F(x − i0)✓ ✏
I
f(x)dx ≡ −
C
F(z)dz
C : closed path which encircles I in the positive sense
and is included in D (F(z) is analytic in D I).
✒ ✑
D
C
I
• The integral in independent of the choise of C by the Cauchy integral theorem.

1. Hyperfunction theory: integral
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Integral of a hyperfunction f(x) = F(x + i0) − F(x − i0)✓ ✏
I
f(x)dx ≡ −
C
F(z)dz
C : closed path which encircles I in the positive sense
and is included in D (F(z) is analytic in D I).
✒ ✑
D
I
I
f(x)dx =
I
{F(x + i0) − F(x − i0)} dx.

Contents
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5. Summary

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We consider the evaluation of an integral
I
f(x)w(x)dx,
f(x) : analytic in a domain D s.t.
I ⊂ D ⊂ C,
w(x) : weight function.
D
I
R

11 / 29
We consider the evaluation of an integral
I
f(x)w(x)dx,
I ⊂ D ⊂ C,
D
I
R
We regard the integrand as a hyperfunction.
✓ ✏
f(x)w(x)χI(x) = −
1
2πi
{f(x + i0)Ψ(x + i0) − f(x − i0)Ψ(x − i0)}
with χI(x) =
1 (x ∈ I)
0 (x ∈ I)
, Ψ(z) =
I
w(x)
z − x
dx
Hilbert
transform
✒ ✑

12 / 29
From the deﬁnition of hyperfunction integrals, we have
✓ ✏
I
f(x)w(x)dx =
1
2πi C
f(z)Ψ(z)dz
=
1
2πi
uperiod
0
f(ϕ(u))Ψ(ϕ(u))ϕ′
(u)du,
C : z = ϕ(u) ( 0 ≦ u ≦ uperiod ) periodic function of period uperiod.
✒ ✑
D
C : z = ϕ(u)
I
Approximating the r.h.s. by the trapezoidal rule, we have ...

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Hyperfunction method✓ ✏
I
f(x)w(x)dx ≃
h
2πi
N−1
k=0
f(ϕ(kh))Ψ(ϕ(kh))ϕ′
(kh),
with Ψ(z) =
b
a
w(x)
z − x
dx and h =
uperiod
N
.
✒ ✑
D
C : z = ϕ(u), 0 ≦ u ≦ uperiod
I

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I
f(x)w(x)dx ≃
h
2πi
N−1
k=0
(kh),
with Ψ(z) =
b
a
w(x)
z − x
dx and h =
uperiod
N
.
✒ ✑
Ψ(z) for typical weight functions w(x)
I w(x) Ψ(z)
(a, b) 1 log
z − a
z − b
∗
(0, 1) xα−1(1 − x)β−1 B(α, β)z−1F(α, 1; α + β; z−1)∗∗
( α, β > 0 )
∗ log z is the principal value s.t. −π ≦ arg z < π.
∗∗ F(α, 1; α + β; z−1
) can be easily evaluated by the continued fraction.

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I
f(x)w(x)dx ≃
h
2πi
N−1
k=0
(kh),
with Ψ(z) =
b
a
w(x)
z − x
dx and h =
uperiod
N
.
✒ ✑
In the hyperfunction method, we numerically evaluate the complex integral
which deﬁnes the desired integral as a hyperfunction integral.

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If f(x) is real valued for real x, we can reduce the number of sampling points
N by half by the reﬂection principle.
I
f(x)w(x)dx ≃
h
π
Im
1
2
f(ϕ(0))Ψ(ϕ(0))ϕ′
(0) +
1
2
f(ϕ(Nh))Ψ(ϕ(Nh))ϕ′
(Nh)
+
N′
k=1
(kh) ,
where z = ϕ(u) is a parameterization of C s.t.
ϕ(−u) = ϕ(u) and h =
uperiod
2N′
.
D
C : z = ϕ(u)
I

2. Numerical integration over a ﬁnite interval (error estimate)
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The hyperfunction method is very accurate.
∵ The trapezoidal rule is efﬁcient for the integrals of periodic analytic functions.

2. Numerical integration over a ﬁnite interval (error estimate)
15 / 29
The hyperfunction method is very accurate.
∵ The trapezoidal rule is efﬁcient for the integrals of periodic analytic functions.
Theoretical error estimate✓ ✏
If f(ϕ(w)) and ϕ(w) are analytic in | Im w| < d0, the error of
the hyperfunction method (with N reduction by half) is bounded
by the inequality
|error| ≦ 2uperiod max
Im w=±d
|f(ϕ(w))Ψ(ϕ(w))ϕ′
(w)|
×
exp(−(4πd/uperiod)N)
1 − exp(−(4πd/uperiod)N)
( 0 < ∀d < d0 ).
. . . Geometric convergence.
✒ ✑

Contents
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5. Summary

17 / 29
We consider the evaluation of an integral over the inﬁnite interval (0, +∞)
∞
0
f(x)w(x)dx,
[0, +∞) ⊂ D ⊂ C,
D
R
O

17 / 29
We consider the evaluation of an integral over the inﬁnite interval (0, +∞)
∞
0
f(x)w(x)dx,
[0, +∞) ⊂ D ⊂ C,
D
R
O
We regard the integrand as a hyperfunction.
✓ ✏
f(x)w(x)H(x) = −
1
2πi
{f(x + i0)Ψ(x + i0) − f(x − i0)Ψ(x − i0)}
with H(x) =
1 (x > 0)
0 (x < 0)
(Heaviside’s step function),
w(x) xα−1 ( α > 0, α ∈ Z ) 1
Ψ(z) −πzα−1/ sin(πα) − log(−z)
✒ ✑

18 / 29
From the deﬁnition of hyperfunction integrals, we have
✓ ✏
∞
0
f(x)w(x)dx =
1
2πi C
f(z)Ψ(z)dz
=
1
2πi
+∞
−∞
f(ϕ(u))Ψ(ϕ(u))ϕ′
(u)du,
C : z = ϕ(u) ( −∞ < u < +∞ ).
✒ ✑
R
D
C : z = ϕ(u)
O
Approximating the r.h.s. by the trapezoidal rule, we have ...

19 / 29
∞
0
f(x)w(x)dx ≃
h
2πi
∞
k=−∞
(kh)
≃
h
2πi
N1
k=−N2
(kh).
✒ ✑
R
D
C : z = ϕ(u)
O
∗ Actually, we use the DE transform
to make the convergence of the inﬁnite sum fast.

19 / 29
∞
0
f(x)w(x)dx ≃
h
2πi
∞
k=−∞
(kh)
≃
h
2πi
N1
k=−N2
(kh).
✒ ✑
R
D
C : z = ϕ(u)
O
∗ We can reduce the number of sampling points N by half
by the reﬂection principle also in the cases of inﬁnite intervals.

Contents
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5. Summary

4. Example 1: numerical integration over a ﬁnite interval
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1
0
ex
xα−1
(1 − x)β−1
dx = B(α, β)F(α; α + β; 1) ( α, β > 0 ).
We evaluated the integral by
• the hyperfunction method (with N reduction by half)
• the DE formula (efﬁcient for integrals with end-point singularities)
and compared the errors of the two methods.
• C++ programs, double precision.
• integral path for the hyperfunction method
z =
1
2
+
1
4
ρ +
1
ρ
cos u +
i
4
ρ −
1
ρ
sin u ( ρ = 10 )
= 0.5 + 2.575 cos u + i2.425 sin u.

22 / 29
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
log10(relativeerror)
N
hyperfunction rule
DE rule
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
hyperfunction rule
DE rule
α = 0.5 α = 10−4 (very strong singularity)
The errors of the hyperfunction method and the DE rule.
hyperfunction method DE rule
α = 0.5 O(0.025N ) O(0.36N )
α = 10−4 O(0.025N ) —

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-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
hyperfunction rule
DE rule
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
hyperfunction rule
DE rule
α = 0.5 O(0.025N ) O(0.36N )
α = 10−4 O(0.025N ) —
The DE rule does not work if the end-point singularities are very strong.

22 / 29
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
hyperfunction rule
DE rule
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
hyperfunction rule
DE rule
α = 0.5 O(0.025N ) O(0.36N )
α = 10−4 O(0.025N ) —
The convergence of the hyperfunction method is not affected
by the end-point singularities.

23 / 29
1
0
xα−1(1 − x)β−1
1 + x2
dx = B(α, β) Re{F(α, 1; α + β; i)} ( α, β > 0 ).
• integral path for the hyperfunction method
z =
1
2
+
1
4
ρ +
1
ρ
cos u +
i
4
ρ −
1
ρ
sin u ( ρ = 2 )
= 0.5 + 0.625 cos u + i0.375 sin u.

23 / 29
1
0
xα−1(1 − x)β−1
1 + x2
dx = B(α, β) Re{F(α, 1; α + β; i)} ( α, β > 0 ).
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
hyperfunction method
DE rule
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
N
DE rule

23 / 29
1
0
xα−1(1 − x)β−1
1 + x2
dx = B(α, β) Re{F(α, 1; α + β; i)} ( α, β > 0 ).
α = 0.5 O(0.18N ) O(0.54N )
α = 10−4 O(0.25N ) —
• The DE rule does not work if the end-point singularities are very strong.
• The convergence of the hyperfunction method is not affected

4. Example: Why the hyperfunction method works well?
24 / 29
integrand
e z
• (DE rule) The sampling points accumulate at the singularities.
• (hyperfunction method) The sampling points are distributed on a curve
in the complex plane where the integrand varies slowly.

4. Example: Why the hyperfunction method works well?
24 / 29
integrand
e z
• (DE rule) The sampling points accumulate at the singularities.
• (hyperfunction method) The sampling points are distributed on a curve
in the complex plane where the integrand varies slowly.
Thus, the hyperfunction method is not affected by the end-point
singularities.

4. Example 3: Numerical integration over an inﬁnite interval
25 / 29
∞
0
xα−1
1 + x2
dx =
π/2
sin(πα/2)
We computed it by the hyperfunction method (with N reduction by half)
and the DE rule.
• C++ program & double precision
• integral path
z = ϕ(u) =
w(u)
iπ
log
1 + iw(u)
1 − iw(u)
,
w = sinh(sinh u)
DE transform
+0.5i.
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
Imz Re z

26 / 29
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100
N
DE rule
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100
N
DE rule
The error of the hyperfunction method and the DE rule.
α = 0.5 O(0.51N ) O(0.34N )
α = 10−4 O(0.46N ) O(0.57N )

26 / 29
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100
N
DE rule
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100
N
DE rule
The error of the hyperfunction method and the DE rule.
α = 0.5 O(0.51N ) O(0.34N )
α = 10−4 O(0.46N ) O(0.57N )
The convergence rate of the hyperfunction method is not affected
by the end-point singularity.

27 / 29
∞
0
xα−1
e−x
dx = Γ(α) ( α > 0 ).
• integral path for the hyperfunction
method
z = ϕ(u) =
w(u)
iπ
log
1 + iw(u)
1 − iw(u)
,
w = sinh u
DE transform
+0.5i.
-3
-2
-1
0
1
2
3
-3 -2 -1 0 1 2 3
Imz
Re z

27 / 29
∞
0
xα−1
e−x
dx = Γ(α) ( α > 0 ).
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100
log10(error)
N
alpha=0.5
alpha=0.1
alpha=0.01
alpha=1.0e-4
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100
log10(error)
N
d=0.5
d=0.1
d=0.01
d=1.0e-4
The erros of the hyperfunction method and the DE rule.

27 / 29
∞
0
xα−1
e−x
dx = Γ(α) ( α > 0 ).
α 0.5 0.1 0.01 10−4
hyperfunction rule O(0.46N ) O(0.46N ) O(0.46N ) O(0.46N )
error
DE rule O(0.39N ) O(0.47N ) O(0.53N ) O(0.55N )
The convergence of the hyperfunction method is not affected

Contents
28 / 29
5. Summary

5. Summary
29 / 29
• The hyperfunction theory is a generalized function theory based on the
complex function theory.
• The hyperfunction method approximately computes desired integral by
evaluating the complex integrals which deﬁne them as hyperfunction
integrals
• Numerical examples show that the hyperfunction method is efﬁcient for
integral with end-point singularities.

5. Summary
29 / 29
integrals
functions with singularities
(poles, discontinuities,
delta functions, ...)
←−←−←−
hyperfunction
analytic
functions

5. Summary
29 / 29
integrals
←−←−←−
hyperfunction
analytic
functions
Hyperfunctions connect singular functions with analytic functions.
We expect that we can apply the hyperfunction theory to a wide range of
scientiﬁc computations.

5. Summary
29 / 29
integrals
←−←−←−
hyperfunction
analytic
functions
Hyperfunctions connect singular functions with analytic functions.
We expect that we can apply the hyperfunction theory to a wide range of
scientiﬁc computations.
Thank you!

Numerical integration based on the hyperfunction theory

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