Beyond the Gaussian
In this paper we present a non-Gaussian integral based on a cubic polynomial, instead of a quadratic, and give a fundamental formula in terms of its discriminant. It gives a mathematical reinforcement to the recent result by Morozov and Shakirov. We also present some related results. This is simply...
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nasplib_isofts_kiev_ua-123456789-1467912025-02-09T16:44:28Z Beyond the Gaussian Fujii, K. In this paper we present a non-Gaussian integral based on a cubic polynomial, instead of a quadratic, and give a fundamental formula in terms of its discriminant. It gives a mathematical reinforcement to the recent result by Morozov and Shakirov. We also present some related results. This is simply one modest step to go beyond the Gaussian but it already reveals many obstacles related with the big challenge of going further beyond the Gaussian. 2011 Article Beyond the Gaussian / K. Fujii // Symmetry, Integrability and Geometry: Methods and Applications. — 2011. — Т. 7. — Бібліогр.: 7 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 11D25; 11R29; 26B20; 81Q99 DOI:10.3842/SIGMA.2011.022 https://nasplib.isofts.kiev.ua/handle/123456789/146791 en Symmetry, Integrability and Geometry: Methods and Applications application/pdf Інститут математики НАН України |
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In this paper we present a non-Gaussian integral based on a cubic polynomial, instead of a quadratic, and give a fundamental formula in terms of its discriminant. It gives a mathematical reinforcement to the recent result by Morozov and Shakirov. We also present some related results. This is simply one modest step to go beyond the Gaussian but it already reveals many obstacles related with the big challenge of going further beyond the Gaussian. |
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Fujii, K. |
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Fujii, K. Beyond the Gaussian Symmetry, Integrability and Geometry: Methods and Applications |
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Fujii, K. |
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Fujii, K. |
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Beyond the Gaussian |
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Beyond the Gaussian |
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Beyond the Gaussian |
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Beyond the Gaussian |
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Beyond the Gaussian |
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beyond the gaussian |
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Інститут математики НАН України |
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2011 |
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https://nasplib.isofts.kiev.ua/handle/123456789/146791 |
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Beyond the Gaussian / K. Fujii // Symmetry, Integrability and Geometry: Methods and Applications. — 2011. — Т. 7. — Бібліогр.: 7 назв. — англ. |
| series |
Symmetry, Integrability and Geometry: Methods and Applications |
| work_keys_str_mv |
AT fujiik beyondthegaussian |
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2025-11-28T02:25:05Z |
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2025-11-28T02:25:05Z |
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1849999188543143936 |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 7 (2011), 022, 12 pages
Beyond the Gaussian
Kazuyuki FUJII
Department of Mathematical Sciences, Yokohama City University, Yokohama, 236-0027 Japan
E-mail: fujii@yokohama-cu.ac.jp
Received January 12, 2011, in final form February 28, 2011; Published online March 04, 2011
doi:10.3842/SIGMA.2011.022
Abstract. In this paper we present a non-Gaussian integral based on a cubic polynomial,
instead of a quadratic, and give a fundamental formula in terms of its discriminant. It
gives a mathematical reinforcement to the recent result by Morozov and Shakirov. We also
present some related results. This is simply one modest step to go beyond the Gaussian but
it already reveals many obstacles related with the big challenge of going further beyond the
Gaussian.
Key words: non-Gaussian integral; renormalized integral; discriminant; cubic equation
2010 Mathematics Subject Classification: 11D25; 11R29; 26B20; 81Q99
1 Introduction
The Gaussian is an abbreviation of all subjects related to the Gauss function e−(px
2+qx+r) like
the Gaussian beam, Gaussian process, Gaussian noise, etc. It plays a fundamental role in math-
ematics, statistics, physics and related disciplines. It is generally conceived that any attempts to
generalize the Gaussian results would meet formidable difficulties. Hoping to overcome this high
wall of difficulties of going beyond the Gaussian in the near future, a first step was introduced
in [1]. This paper is its polished version.
In the paper [2] the following “formula” is reported:∫∫
e−(ax3+bx2y+cxy2+dy3)dxdy =
1
6
√
−D
, (1)
where D is the discriminant of the cubic equation
ax3 + bx2 + cx+ d = 0,
and it is given by
D = b2c2 + 18abcd− 4ac3 − 4b3d− 27a2d2. (2)
The formula (1) is of course non-Gaussian. However, if we consider it in the framework of the
real category then (1) is not correct because the left hand side diverges. In this paper we treat
only the real category, and so a, b, c, d, x, y are real numbers.
Formally, by performing the change of variable x = tρ, y = ρ for (1) we have
l.h.s. of (1) =
∫∫
e−ρ
3(at3+bt2+ct+d)|ρ|dtdρ =
∫ {∫
e−(at3+bt2+ct+d)ρ3 |ρ|dρ
}
dt
=
∫
|σ|e−σ3
dσ
∫
1∣∣ 3
√
(at3 + bt2 + ct+ d)
∣∣ 3
√
(at3 + bt2 + ct+ d)
dt
by the change of variable σ = 3
√
at3 + bt2 + ct+ dρ.
mailto:fujii@yokohama-cu.ac.jp
http://dx.doi.org/10.3842/SIGMA.2011.022
2 K. Fujii
The divergence comes from∫
|σ|e−σ3
dσ,
while the main part is∫
1∣∣ 3
√
(ax3 + bx2 + cx+ d)
∣∣ 3
√
(ax3 + bx2 + cx+ d)
dx
under the change t→ x. As a kind of renormalization the integral may be defined like
‡
∫∫
R2
e−(ax
3+bx2y+cxy2+dy3)dxdy ‡ =
∫
R
1∣∣ 3
√
(ax3+ bx2+ cx+ d)
∣∣ 3
√
(ax3+ bx2+ cx+ d)
dx.
However, the right hand side lacks proper symmetry. If we set
F (a, b, c, d) =
∫∫
DR
e−(ax3+bx2y+cxy2+dy3)dxdy,
where DR = [−R,R]× [−R,R], then it is easy to see
F (−a,−b,−c,−d) = F (a, b, c, d).
Namely, F is invariant under Z2-action. This symmetry is important and must be kept even
in the renormalization process. The right hand side in the “definition” above is clearly not
invariant. Therefore, by modifying it slightly we reach the renormalized integral
Definition 1.
‡
∫∫
R2
e−(ax
3+bx2y+cxy2+dy3)dxdy ‡ =
∫
R
1
3
√
(ax3 + bx2 + cx+ d)2
dx. (3)
We believe that the definition is not so bad (see the Section 4).
In the paper we calculate the right hand side of (3) directly, which will give some interesting
results and a new perspective. The result gives a mathematical reinforcement to the result [2]
by Morozov and Shakirov.
2 Main result
Before stating the result let us make some preparations. The Gamma function Γ(p) is defined
by
Γ(p) =
∫ ∞
0
e−xxp−1dx (p > 0) (4)
and the Beta function B(p, q) is
B(p, q) =
∫ 1
0
xp−1(1− x)q−1dx (p, q > 0).
Note that the Beta function is rewritten as
B(p, q) =
∫ ∞
0
xp−1
(1 + x)p+q
dx.
See [3] for more detail. Now we are in a position to state the result.
Beyond the Gaussian 3
Fundamental formula.
(I) For D < 0∫
R
1
3
√
(ax3 + bx2 + cx+ d)2
dx =
C−
6
√
−D
, (5)
where
C− =
3
√
2B
(
1
2
,
1
6
)
.
(II) For D > 0∫
R
1
3
√
(ax3 + bx2 + cx+ d)2
dx =
C+
6
√
D
, (6)
where
C+ = 3B
(
1
3
,
1
3
)
.
(III) C− and C+ are related by C+ =
√
3C− through the identity
√
3B
(
1
3
,
1
3
)
=
3
√
2B
(
1
2
,
1
6
)
. (7)
Our result shows that the integral depends on the sign of D, and so our question is as follows.
Problem. Can the result be derived from the method developed in [2]?
A comment is in order. If we treat the Gaussian case (e−(ax
2+bxy+cy2)) then the integral is
reduced to∫
R
1
ax2 + bx+ c
dx =
2π√
−D
(8)
if a > 0 and D = b2 − 4ac < 0. Noting
π =
√
π
√
π
1
=
Γ(12)Γ(12)
Γ(1)
= B
(
1
2
,
1
2
)
(8) should be read as∫
R
1
ax2 + bx+ c
dx =
2B
(
1
2 ,
1
2
)
√
−D
.
3 Proof of the formula
The proof is delicate. In order to prevent possible misunderstanding we present a detailed proof
in this section.
Proof of (I). We prove (5) in case of D < 0.
First we consider the special case where a = 0 in the cubic equation ax3 + bx2 + cx + d.
Namely, we calculate the integral∫
R
1
3
√
(bx2 + cx+ d)2
dx.
4 K. Fujii
Noting −D = b2(4bd− c2) > 0 we obtain∫
R
1
3
√
(bx2 + cx+ d)2
dx =
∫
R
1
3
√
b2 3
√
(x2 + c
bx+ d
b )2
dx
=
1
3
√
b2
∫
R
1
3
√(
(x+ c
2b)
2 + d
b −
c2
4b2
)2dx =
1
3
√
b2
∫
R
1
3
√(
x2 + 4bd−c2
4b2
)2dx
T 2≡ 4bd−c2
4b2
>0
=
1
3
√
b2
∫
R
1
3
√
(x2 + T 2)2
dx
x=Ty
=
1
3
√
b2
∫
R
T
3
√
T 4 3
√
(y2 + 1)2
dy
=
2
3
√
b2T
∫ ∞
0
1
3
√
(y2 + 1)2
dy
y=
√
x
=
2
3
√
b2T
∫ ∞
0
1
3
√
(x+ 1)2
dx
2
√
x
=
1
3
√
b2T
∫ ∞
0
x−
1
2
(x+ 1)
2
3
dx =
B
(
1
2 ,
1
6
)
3
√
b2T
=
B
(
1
2 ,
1
6
)
6
√
b4T 2
=
3
√
2B
(
1
2 ,
1
6
)
6
√
b2(4bd− c2)
=
3
√
2B
(
1
2 ,
1
6
)
6
√
−D
. (9)
Now we consider the general case of a 6= 0. From the condition D < 0 there is (only) one real
root of the cubic equation ax3 + bx2 + cx+ d = 0. Let us denote it by α. From the equation
ax3 + bx2 + cx+ d = (x− α)(ax2 + kx+ l), aα3 + bα2 + cα+ d = 0
we have easily
b = k − aα, c = l − kα, d = −lα. (10)
First we assume α = 0. In this case d = 0 and
ax3 + bx2 + cx+ d = x
(
ax2 + bx+ c
)
.
Then ∫
R
1
3
√
x2(ax2 + bx+ c)2
dx =
∫ ∞
0
1
3
√
x2(ax2 + bx+ c)2
dx+
∫ 0
−∞
1
3
√
x2(ax2 + bx+ c)2
dx
x= 1
y
=
∫ 0
∞
1
3
√
1
y2
(
a
y2
+ b
y + c
)2
(
−dy
y2
)
+
∫ −∞
0
1
3
√
1
y2
(
a
y2
+ b
y + c
)2
(
−dy
y2
)
=
∫ ∞
0
1
3
√
(cy2 + by + a)2
dy +
∫ 0
−∞
1
3
√
(cy2 + by + a)2
dy
=
∫
R
1
3
√
(cy2 + by + a)2
dy
(9) (b→c; c→b; d→a)
=
3
√
2B
(
1
2 ,
1
6
)
6
√
c2(4ac− b2)
=
3
√
2B
(
1
2 ,
1
6
)
6
√
−D
.
Next, let us calculate the case α 6= 0:∫
R
1
3
√
(x− α)2(ax2 + kx+ l)2
dx
x=y+α
=
∫
R
1
3
√
y2{a(y + α)2 + k(y + α) + l}2
dy
=
∫
R
1
3
√
y2{ay2 + (2aα+ k)y + (aα2 + kα+ l)}2
dy
(10)
=
3
√
2B
(
1
2 ,
1
6
)
6
√
(aα2 + kα+ l)2{4a(aα2 + kα+ l)− (2aα+ k)2}
=
3
√
2B
(
1
2 ,
1
6
)
6
√
(aα2 + kα+ l)2(4al − k2)
. (11)
Beyond the Gaussian 5
Key Lemma. From (10) the following equation holds(
aα2 + kα+ l
)2(
4al − k2
)
= 27a2d2 + 4ac3 − 18abcd− b2c2 + 4b3d = −D. (12)
The proof is straightforward but tedious.
Therefore, from both (11) and (12) we obtain the formula∫
R
1
3
√
(x− α)2(ax2 + kx+ l)2
dx =
3
√
2B
(
1
2 ,
1
6
)
6
√
−D
. �
Proof of (II). We prove (6) in case of D > 0. Let us start with the evaluation of the following
integral∫
R
1
3
√
x2(x− α)2
dx
for α > 0. Then∫
R
1
3
√
x2(x− α)2
dx =
∫ 0
−∞
1
3
√
x2(x− α)2
dx+
∫ ∞
0
1
3
√
x2(x− α)2
dx
=
∫ ∞
0
1
3
√
x2(x+ α)2
dx+
∫ ∞
0
1
3
√
x2(x− α)2
dx, (13)
where the change of variable x→ −x for the first term of the right hand side was made.
Each term can be evaluated elementarily:∫ ∞
0
1
3
√
x2(x+ α)2
dx
x=αt
=
1
3
√
α
∫ ∞
0
1
3
√
t2(t+ 1)2
dt = α−
1
3
∫ ∞
0
t−
2
3
(t+ 1)
2
3
dt = α−
1
3B
(
1
3
,
1
3
)
,
while ∫ ∞
0
1
3
√
x2(x− α)2
dx
x=αt
= α−
1
3
∫ ∞
0
1
3
√
t2(t− 1)2
dt
= α−
1
3
{∫ 1
0
1
3
√
t2(t− 1)2
dt+
∫ ∞
1
1
3
√
t2(t− 1)2
dt
}
= α−
1
3
{∫ 1
0
1
3
√
t2(1− t)2
dt+
∫ ∞
1
1
3
√
t2(t− 1)2
dt
}
= 2α−
1
3
∫ 1
0
1
3
√
t2(1− t)2
dt = 2α−
1
3
∫ 1
0
t−
2
3 (1− t)−
2
3dt
= 2α−
1
3B
(
1
3
,
1
3
)
,
where we have used∫ ∞
1
1
3
√
t2(t− 1)2
dt
t= 1
s=
∫ 0
1
1
3
√
1
s2
(1−s)2
s2
(
−ds
s2
)
=
∫ 1
0
1
3
√
s2(1− s)2
ds =
∫ 1
0
1
3
√
t2(1− t)2
dt.
From (13) we have∫
R
1
3
√
x2(x− α)2
dx = 3α−
1
3B
(
1
3
,
1
3
)
=
3B
(
1
3 ,
1
3
)
3
√
α
.
6 K. Fujii
Now we consider the special case a = 0 in the cubic equation ax3 + bx2 + cx + d. Then by
D = b2(c2 − 4bd) > 0 we obtain∫
R
1
3
√
(bx2 + cx+ d)2
dx =
∫ ∞
−∞
1
3
√{
b(x+ c
2b)
2 − c2−4bd
4b
}2
dx =
∫ ∞
−∞
1
3
√
(bx2 − c2−4bd
4b )2
dx
α2= c2−4bd
4b2
>0
=
1
3
√
b2
∫ ∞
−∞
1
3
√
(x2 − α2)2
dx =
1
3
√
b2
∫ ∞
−∞
1
3
√
(x− α)2(x+ α)2
dx
y=x+α
=
1
3
√
b2
∫ ∞
−∞
1
3
√
y2(y − 2α)2
dy
(13)
=
1
3
√
b2
3B
(
1
3 ,
1
3
)
3
√
2α
=
3B
(
1
3 ,
1
3
)
3
√
2αb2
=
3B
(
1
3 ,
1
3
)
6
√
4α2b4
=
3B
(
1
3 ,
1
3
)
6
√
b2(c2 − 4bd)
=
3B
(
1
3 ,
1
3
)
6
√
D
. (14)
Next we consider the remaining general case of a 6= 0. From the condition D > 0 there
are three real solutions in the equation ax3 + bx2 + cx + d = 0. We denote one of them by α.
Remember the relations b = k − aα, c = l − kα, d = −lα from the equation
ax3 + bx2 + cx+ d = (x− α)
(
ax2 + kx+ l
)
, aα3 + bα2 + cα+ d = 0.
First we assume α = 0. Then
ax3 + bx2 + cx+ d = x(ax2 + bx+ c)
and from D = c2(b2 − 4ac) we have∫
R
1
3
√
x2(ax2 + bx+ c)2
dx =
∫ ∞
0
1
3
√
x2(ax2 + bx+ c)2
dx+
∫ 0
−∞
1
3
√
x2(ax2 + bx+ c)2
dx
x= 1
y
=
∫ 0
∞
1
3
√
1
y2
(
a
y2
+ b
y + c
)2
(
−dy
y2
)
+
∫ −∞
0
1
3
√
1
y2
(
a
y2
+ b
y + c
)2
(
−dy
y2
)
=
∫ ∞
0
1
3
√
(cy2 + by + a)2
dy +
∫ 0
−∞
1
3
√
(cy2 + by + a)2
dy =
∫
R
1
3
√
(cy2 + by + a)2
dy
(14) (c→b; b→c; a→d)
=
3B
(
1
3 ,
1
3
)
6
√
c2(b2 − 4ac)
=
3B
(
1
3 ,
1
3
)
6
√
D
. (15)
For the case α 6= 0 we obtain the formula∫
R
1
3
√
(x− α)2(ax2 + kx+ l)2
dx
x=y+α
=
∫
R
1
3
√
y2{a(y + α)2 + k(y + α) + l}2
dy
=
∫
R
1
3
√
y2{ay2 + (2aα+ k)y + (aα2 + kα+ l)}2
dy
(15)
=
3B
(
1
3 ,
1
3
)
6
√
(aα2 + kα+ l)2{(2aα+ k)2 − 4a(aα2 + kα+ l)}
=
3B
(
1
3 ,
1
3
)
6
√
(aα2 + kα+ l)2(k2 − 4al)
(12)
=
3B
(
1
3 ,
1
3
)
6
√
D
. �
Proof of (III). We prove the relation (7). Let us make some preparations. For the Gamma
function (4) there are well-known formulas (see for example [3])
B(x, y) =
Γ(x)Γ(y)
Γ(x+ y)
(x, y > 0), (16)
Beyond the Gaussian 7
Γ(x)Γ(1− x) =
π
sin(πx)
(0 < x < 1), (17)
Γ
(x
2
)
Γ
(
x+ 1
2
)
=
√
π
2x−1
Γ(x) = 21−xΓ
(
1
2
)
Γ(x). (18)
(18) is called the Legendre’s relation. In the formula we set x = 2/3, then
Γ
(
1
3
)
Γ
(
5
6
)
=
3
√
2Γ
(
1
2
)
Γ
(
2
3
)
.
Multiplying both sides by Γ(1/6) gives
Γ
(
1
3
)
Γ
(
5
6
)
Γ
(
1
6
)
=
3
√
2Γ
(
1
2
)
Γ
(
2
3
)
Γ
(
1
6
)
⇐⇒ Γ
(
1
3
)
π
sin(π6 )
=
3
√
2Γ
(
1
2
)
Γ
(
1
6
)
Γ
(
2
3
)
⇐⇒ 2πΓ
(
1
3
)
=
3
√
2Γ
(
1
2
)
Γ
(
1
6
)
Γ
(
2
3
)
⇐⇒ 2π
Γ
(
1
3
)
Γ
(
2
3
)2 =
3
√
2
Γ
(
1
2
)
Γ
(
1
6
)
Γ
(
2
3
)
⇐⇒ 2π
Γ
(
1
3
)
Γ
(
1
3
)
Γ
(
2
3
)
Γ
(
1
3
)
Γ
(
2
3
) =
3
√
2B
(
1
2
,
1
6
)
⇐⇒ 2π
Γ
(
1
3
)2
π
sin(π
3
)Γ
(
2
3
) =
3
√
2B
(
1
2
,
1
6
)
⇐⇒
√
3B
(
1
3
,
1
3
)
=
3
√
2B
(
1
2
,
1
6
)
,
where we have used formulas (16) and (17) several times.
The proof of (7) is now complete. �
4 Renormalized integral revisited
In this section let us check whether the renormalized integral (3) is reasonable or not by making
use of the results in the Section 3.
In the introduction we introduced the following integral defined on DR = [−R,R]× [−R,R]
F (a, b, c, d) =
∫∫
DR
e−(ax3+bx2y+cxy2+dy3)dxdy.
For this it is easy to see(
∂
∂a
∂
∂d
− ∂
∂b
∂
∂c
)
F (a, b, c, d)
=
∫∫
DR
(
x3 · y3 − x2y · xy2
)
e−(ax3+bx2y+cxy2+dy3)dxdy = 0,(
∂
∂b
∂
∂b
− ∂
∂a
∂
∂c
)
F (a, b, c, d)
=
∫∫
DR
(
x2y · x2y − x3 · xy2
)
e−(ax3+bx2y+cxy2+dy3)dxdy = 0,
8 K. Fujii(
∂
∂c
∂
∂c
− ∂
∂b
∂
∂d
)
F (a, b, c, d)
=
∫∫
DR
(
xy2 · xy2 − x2y · y3
)
e−(ax3+bx2y+cxy2+dy3)dxdy = 0. (19)
On the other hand, if we set
F(a, b, c, d) =
∫
R
1
3
√
(ax3 + bx2 + cx+ d)2
dx =
C±
6
√
±D
,
then we can also verify the same relations:(
∂
∂a
∂
∂d
− ∂
∂b
∂
∂c
)
F(a, b, c, d) = 0,
(
∂
∂b
∂
∂b
− ∂
∂a
∂
∂c
)
F(a, b, c, d) = 0,(
∂
∂c
∂
∂c
− ∂
∂b
∂
∂d
)
F(a, b, c, d) = 0. (20)
Verification by hand is rather tough, but it can be done easily by use of MATHEMATICA1.
From (19) and (20) we can conclude that our renormalized integral (3) is reasonable enough.
5 Discriminant
In this section we make some comments on the discriminant (2). See [4] for more details ([4] is
strongly recommended).
For the equations
f(x) = ax3 + bx2 + cx+ d, f ′(x) = 3ax2 + 2bx+ c (21)
the resultant R(f, f ′) of f and f ′ is given by
R(f, f ′) =
∣∣∣∣∣∣∣∣∣∣
a b c d 0
0 a b c d
3a 2b c 0 0
0 3a 2b c 0
0 0 3a 2b c
∣∣∣∣∣∣∣∣∣∣
. (22)
It is easy to calculate (22) and the result becomes
1
a
R(f, f ′) = 27a2d2 + 4ac3 − 18abcd− b2c2 + 4b3d = −D.
On the other hand, if α, β, γ are three solutions of f(x) = 0 in (21), then the following
relations are well–known
α+ β + γ = − b
a
, αβ + αγ + βγ =
c
a
, αβγ = −d
a
.
From these it is easy to see
α+ β + γ = − b
a
, α2 + β2 + γ2 =
b2 − 2ac
a2
, α3 + β3 + γ3 = −b
3 + 3a2d− 3abc
a3
,
α4 + β4 + γ4 =
b4 + 4a2bd+ 2a2c2 − 4ab2c
a4
.
1The author owes the calculation to Hiroshi Oike.
Beyond the Gaussian 9
If we set
∆ = (α− β)(α− γ)(β − γ)
the discriminant D is given by
D = a4∆2.
Let us calculate ∆2 directly. For the Vandermonde matrix
V =
1 1 1
α β γ
α2 β2 γ2
=⇒ |V | = −∆
we obtain by some manipulations of determinant
∆2 = (−|V |)2 = |V ||V T | = |V V T | =
∣∣∣∣∣∣
3 α+ β + γ α2 + β2 + γ2
α+ β + γ α2 + β2 + γ2 α3 + β3 + γ3
α2 + β2 + γ2 α3 + β3 + γ3 α4 + β4 + γ4
∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣∣∣
3 − b
a
b2 − 2ac
a2
− b
a
b2 − 2ac
a2
−b
3 + 3a2d− 3abc
a3
b2 − 2ac
a2
−b
3 + 3a2d− 3abc
a3
b4 + 4a2bd+ 2a2c2 − 4ab2c
a4
∣∣∣∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣∣∣
3 − b
a
b2 − 2ac
a2
2b
a
−2c
a
−3ad− bc
a2
−2b2 + 2ac
a2
−3ad− 3bc
a2
4abd+ 2ac2 − 2b2c
a3
∣∣∣∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣∣∣
3 − b
a
b2 − 2ac
a2
2b
a
−2c
a
−3ad− bc
a2
−2c
a
−3ad− bc
a2
abd+ 2ac2 − b2c
a3
∣∣∣∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣∣∣
3 − b
a
−2c
a
2b
a
−2c
a
−3ad+ bc
a2
−2c
a
−3ad− bc
a2
−2bd+ 2c2
a2
∣∣∣∣∣∣∣∣∣∣∣∣
=
∣∣∣∣∣∣∣∣∣∣∣∣
3 − b
a
−2c
a
0 2
b2 − 3ac
3a2
bc− 9ad
3a2
0
bc− 9ad
3a2
2
c2 − 3bd
3a2
∣∣∣∣∣∣∣∣∣∣∣∣
= 3
∣∣∣∣∣∣∣∣
2
b2 − 3ac
3a2
bc− 9ad
3a2
bc− 9ad
3a2
2
c2 − 3bd
3a2
∣∣∣∣∣∣∣∣
=
1
a4
−1
3
{
(bc− 9ad)2 − 4(b2 − 3ac)(c2 − 3bd)
}
.
This result is very suggestive. In fact, from the cubic equation
ax3 + bx2 + cx+ d = 0
we have three data
A = b2 − 3ac, B = bc− 9ad, C = c2 − 3bd,
10 K. Fujii
and so if we consider the quadratic equation
AX2 +BX + C = 0
then the discriminant is just B2 − 4AC. This is very interesting.
Problem. Clarify the above connection.
As a result we have
D =
−1
3
{
(bc− 9ad)2 − 4(b2 − 3ac)(c2 − 3bd)
}
= b2c2 + 18abcd− 4ac3 − 4b3d− 27a2d2.
6 Some calculations
In this section we calculate some quantities coming from the integral.
The expectation value 〈x3〉 is formally given by
〈x3〉 =
∫∫
x3e−(ax3+bx2y+cxy2+dy3)dxdy∫∫
e−(ax3+bx2y+cxy2+dy3)dxdy
= − ∂
∂a
log
{∫∫
e−(ax3+bx2y+cxy2+dy3)dxdy
}
,
so renormalized expectation values 〈x3〉RN, 〈x2y〉RN, 〈xy2〉RN, 〈y3〉RN are defined as
Definition 2.
〈x3〉RN = − ∂
∂a
log
{
‡
∫∫
R2
e−(ax
3+bx2y+cxy2+dy3)dxdy ‡
}
,
〈x2y〉RN = − ∂
∂b
log
{
‡
∫∫
R2
e−(ax
3+bx2y+cxy2+dy3)dxdy ‡
}
,
〈xy2〉RN = − ∂
∂c
log
{
‡
∫∫
R2
e−(ax
3+bx2y+cxy2+dy3)dxdy ‡
}
,
〈y3〉RN = − ∂
∂d
log
{
‡
∫∫
R2
e−(ax
3+bx2y+cxy2+dy3)dxdy ‡
}
.
From the integral forms (5) and (6) it is easy to calculate the above. Namely, we have
〈x3〉RN =
18bcd− 4c3 − 54ad2
6D
, 〈x2y〉RN =
2bc2 + 18acd− 12b2d
6D
,
〈xy2〉RN =
2b2c+ 18abd− 12ac2
6D
, 〈y3〉RN =
18abc− 4b3 − 54a2d
6D
,
where D = b2c2 + 18abcd− 4ac3 − 4b3d− 27a2d2.
We can calculate other quantities like 〈x5y〉RN or 〈x4y2〉RN by use of these ones, which will
be left to readers.
7 Concluding remarks
In this paper we calculated the non-Gaussian integral (1) in a direct manner and, moreover,
calculated some renormalized expectation values. It is not clear at the present time whether
these results are useful enough or not. It would be desirable to accumulate many supporting
evidences. Some application(s) will be reported elsewhere [5].
Beyond the Gaussian 11
At this stage we can consider a further generalization. Namely, for the general degree n
polynomial
f(x) = a0x
n + a1x
n−1 + · · ·+ an−1x+ an
the (non-Gaussian) integral becomes∫
R
1
n
√
f(x)2
dx. (23)
The discriminant D of the equation f(x) = 0 is given by the resultant R(f, f ′) of f and f ′
like
1
a0
R(f, f ′) = (−1)
n(n−1)
2 D ⇐⇒ D = (−1)
n(n−1)
2 R(f, f ′)/a0
where
R(f, f ′) =
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
a0 a1 · · · an−1 an
a0 a1 · · · an−1 an
. . .
. . .
a0 a1 · · · an−1 an
na0 (n− 1)a1 · · · an−1
na0 (n− 1)a1 · · · an−1
. . .
. . .
na0 (n− 1)a1 · · · an−1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
,
see (21) and (22).
For example, for n = 4
R(f, f ′) =
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
a0 a1 a2 a3 a4 0 0
0 a0 a1 a2 a3 a4 0
0 0 a0 a1 a2 a3 a4
4a0 3a1 2a2 a3 0 0 0
0 4a0 3a1 2a2 a3 0 0
0 0 4a0 3a1 2a2 a3 0
0 0 0 4a0 3a1 2a2 a3
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
and
D = 256a30a
3
4 − 4a31a
3
3 − 27a20a
4
3 − 27a41a
2
4 − 128a20a
2
2a
2
4 + a21a
2
2a
2
3 + 16a0a
4
2a4
− 4a0a
3
2a
2
3 − 4a21a
3
2a4 + 144a20a2a
2
3a4 − 6a0a
2
1a
2
3a4 + 144a0a
2
1a2a
2
4 − 192a20a1a3a
2
4
+ 18a0a1a2a
3
3 + 18a31a2a3a4 − 80a0a1a
2
2a3a4,
and for n = 5
R(f, f ′) =
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
a0 a1 a2 a3 a4 a5 0 0 0
0 a0 a1 a2 a3 a4 a5 0 0
0 0 a0 a1 a2 a3 a4 a5 0
0 0 0 a0 a1 a2 a3 a4 a5
5a0 4a1 3a2 2a3 a4 0 0 0 0
0 5a0 4a1 3a2 2a3 a4 0 0 0
0 0 5a0 4a1 3a2 2a3 a4 0 0
0 0 0 5a0 4a1 3a2 2a3 a4 0
0 0 0 0 5a0 4a1 3a2 2a3 a4
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
12 K. Fujii
and
D = 3125a40a
4
5 − 2500a30a1a4a
3
5 − 3750a30a2a3a
3
5 + 2000a30a2a
2
4a
2
5 + 2250a30a
2
3a4a
2
5
− 1600a30a3a
3
4a5 + 256a30a
5
4 + 2000a20a
2
1a3a
3
5 − 50a20a
2
1a
2
4a
2
5 + 2250a20a1a
2
2a
3
5
− 2050a20a1a2a3a4a
2
5 + 160a20a1a2a
3
4a5 − 900a20a1a
3
3a
2
5 + 1020a20a1a
2
3a
2
4a5
− 192a20a1a3a
4
4 − 900a20a
3
2a4a
2
5 + 825a20a
2
2a
2
3a
2
5 + 560a20a
2
2a3a
2
4a5 − 128a20a
2
2a
4
4
− 630a20a2a
3
3a4a5 + 144a20a2a
2
3a
3
4 + 108a20a
5
3a5 − 27a20a
4
3a
2
4 − 1600a0a
3
1a2a
3
5
+ 160a0a
3
1a3a4a
2
5− 36a0a
3
1a
3
4a5 + 1020a0a
2
1a
2
2a4a
2
5+ 560a0a
2
1a2a
2
3a
2
5− 746a0a
2
1a2a3a
2
4a5
+ 144a0a
2
1a2a
4
4 + 24a0a
2
1a
3
3a4a5 − 6a0a
2
1a
2
3a
3
4 − 630a0a1a
3
2a3a
2
5 + 24a0a1a
3
2a
2
4a5
+ 356a0a1a
2
2a
2
3a4a5 − 80a0a1a
2
2a3a
3
4 − 72a0a1a2a
4
3a5 + 18a0a1a2a
3
3a
2
4 + 108a0a
5
2a
2
5
− 72a0a
4
2a3a4a5 + 16a0a
4
2a
3
4 + 16a0a
3
2a
3
3a5 − 4a0a
3
2a
2
3a
2
4 + 256a51a
3
5 − 192a41a2a4a
2
5
− 128a41a
2
3a
2
5 + 144a41a3a
2
4a5 − 27a41a
4
4 + 144a31a
2
2a3a
2
5 − 6a31a
2
2a
2
4a5 − 80a31a2a
2
3a4a5
+ 18a31a2a3a
3
4 + 16a31a
4
3a5 − 4a31a
3
3a
2
4 − 27a21a
4
2a
2
5 + 18a21a
3
2a3a4a5 − 4a21a
3
2a
3
4
− 4a21a
2
2a
3
3a5 + a21a
2
2a
2
3a
2
4.
However, to write down the general case explicitly is not easy (almost impossible).
Problem. Calculate (23) for n = 4 (and n = 5) directly.
The wall called Gaussian is very high and not easy to overcome, and therefore hard work will
be needed.
Recently the subsequent paper [6] by Morozov and Shakirov appeared. Our works are deeply
related to so-called non-linear algebras, so we will make some comments on this point in a near
future. As a general introduction to them see for example [7].
The author thanks referees and Hiroshi Oike, Ryu Sasaki for many useful suggestions and
comments.
References
[1] Fujii K., Beyond Gaussian: a comment, arXiv:0905.1363.
[2] Morozov A., Shakirov Sh., Introduction to integral discriminants, J. High Energy Phys. 2009 (2009), no. 12,
002, 39 pages, arXiv:0903.2595.
[3] Whittaker E.T., Watson G.N., A course of modern analysis, Cambridge University Press, Cambridge, 1996.
[4] Satake I., Linear algebra, Shokabo, Tokyo, 1989 (in Japanese).
[5] Fujii K., Beyond the Gaussian. II. Some applications, in progress.
[6] Morozov A., Shakirov Sh., New and old results in resultant theory, Theoret. and Math. Phys. 163 (2010),
587–617, arXiv:0911.5278.
[7] Dolotin V., Morozov A., Introduction to non-linear algebra, World Scientific Publishing Co. Pte. Ltd.,
Hackensack, NJ, 2007, hep-th/0609022.
http://arxiv.org/abs/0905.1363
http://dx.doi.org/10.1088/1126-6708/2009/12/002
http://arxiv.org/abs/0903.2595
http://dx.doi.org/10.1007/s11232-010-0044-0
http://arxiv.org/abs/0911.5278
http://arxiv.org/abs/hep-th/0609022
1 Introduction
2 Main result
3 Proof of the formula
4 Renormalized integral revisited
5 Discriminant
6 Some calculations
7 Concluding remarks
References
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