gaussian integral; Possion integral; Gauss积分; Possion积分; 高斯积分; 泊松积分

height z as r(z), we have A(z) = πr(z)2 and we need to find r(z). An equation of the surface at
height
z
is
z
=
e−
1 2
r(z)2
,
so
r(z)2
=
−2 log z.
Therefore
A(z)
=
−2πlog z,
00
0
∞
ye−y2(t2+1) dy dt.
0
Since
∞ 0
ye−ay2
dy
=
1 2a
for
a > 0,
we
have
J2 =
∞ dt
1π π
0
2(t2 + 1)
=
2
·
2
=
, 4
√
so J = π/2. This approach is due to Laplace [6, pp. 94–96] and historically precedes the more
0
1
e−nx2 dx ≤
0
1 dx 0 (1 + x2)n .
√
Under the changes of variables x = sin θ on the left, x = y/ n in the middle, and x = tan θ on the
right, (6.2)
√
π/2
(cos θ)2n+1 dθ
x2
(or e−x2
or e−πx2 ).
For comparison,
∞ 0
xe−
1 2
x2
dx
can
be
computed
using
the
antiderivative
−e−
1 2
x2
:
this
integral
is
1.
1. First Proof: Polar coordinates
The most widely known proof uses multivariable calculus: express J2 as a double integral and
=
π/4,
so
(3.1)
becomes
t
e−x2 dx
0
2π =− 4
1 e−t2(1+x2) 0 1 + x2 dx.
Letting
t
→
∞
in
this
equation,
we
obtain
J2
=
π/4,
so
J
=
√ π/2.
A comparison of this proof with the first proof is in [17].
familiar technique in the first proof above. We will see in our seventh proof that this was not
Laplace’s first method.
3. Third Proof: Differentiating under the integral sign
I = 2π.
5. Fifth Proof: The Γ-function
For any integer n ≥ 0, we have n! =
∞ 0
tne−t
dt.
For
x
>
0
we
define
Γ(x) =
∞
txe−t
dt ,
0
t
so Γ(n) = (n − 1)! when n ≥ 1. Using integration by parts, Γ(x + 1) = xΓ(x). One of the basic
We will give multiple proofs of this result. (Other lists of proofs are in [3] and [8].) The theorem
is
subtle
because
there
is
no
simple
antiderivative
for
e−
1 2
properties of the Γ-function [13, pp. 193–194] is
(5.1)
Γ(x)Γ(y) =
1
tx−1(1 − t)y−1 dt.
Γ(x + y) 0
Set x = y = 1/2:
12
1 dt
Γ
=
.
2
0 t(1 − t)
Note
1 Γ
=
∞√ te−t
dt
=
∞ e−t √ dt =
THE GAUSSIAN INTEGRAL
3
4. Fourth Proof: A volume integral
Our next proof is due to T. P. Jameson [4] (and rediscovered by A. L. Delgado [2]). Revolve the
so
1
1
V = −2π log z dz = −2π (z log z − z) = −2π(−1 − lim z log z).
0
0
z→0+
By L’Hopital’s rule limz→0+ z log z = 0, so V = 2π. (A calculation of V with cylinders instead of discs is in [10].)
and dx dy = r dr dθ,
J2 =
π/2 ∞
e−r2 r dr dθ
0
0
∞
π/2
=
re−r2 dr ·
dθ
0
0
=
− 1 e−r2
∞π ·
2
02
1π =·
22 π =. 4 √ Taking square roots, J = π/2. This method is due to Poisson [8, p. 3].
0
0
Let x = ty, so
1
1
A (t) = 2e−t2 te−t2y2 dy = 2te−(1+y2)t2 dy.
0பைடு நூலகம்
0
The function under the integral sign is easily antidifferentiated with respect to t:
1 ∂ e−(1+y2)t2
For a second method of computing V , let’s use slices in planes y = constant. We get V =
∞ −∞
A(y)
dy,
where
A(y0)
is
the
area
under
the
surface
lying
in
the
plane
of
d 1 e−(1+y2)t2
A (t) = − 0 ∂t
1 + y2
dy = − dt 0
1 + y2 dy.
Letting
1 e−t2(1+x2)
B(t) =
0
1 + x2 dx,
we have A (t) = −B (t) for all t > 0, so there is a constant C such that
For t > 0, set
A(t) =
t
2
e−x2 dx .
0
The integral we want to calculate is A(∞) = J2 and then take a square root.
Differentiating A(t) with respect to t,
t
t
A (t) = 2 e−x2 dx · e−t2 = 2e−t2 e−x2 dx.
points
y
=
y0.
Since
∞
∞
∞
A(y) =
z(y) dx =
e−
1 2
(x2
+y2
)
dx
=
e−
1 2
y2
e−
1 2
x2
dx
=
e−
1 2
y2
I
,
−∞
−∞
−∞
we h√ave V =
∞ −∞
e−
1 2
y2 I
dy
=
I2.
Comparing the two formulas for V , we have 2π = I2, so
≤
√1
n
π/4
e−y2 dy ≤
(cos θ)2n−2 dθ.
0
n0
0
Set Ik = 0π/2(cos θ)k dθ, so I0 = π/2, I1 = 1, and (6.2) implies
∞∞
J2 =
e−(x2+y2) dx dy,
00 1
2
KEITH CONRAD
but instead of using polar coordinates we make a change of variables x = yt with dx = y dt, so
∞∞
∞
J2 =
e−y2(t2+1)y dt dy =
(6.1)
1
−
x2
≤
e−x2
≤
1
1 + x2 .
for all x ∈ R. For any positive integer n, raise the terms in (6.1) to the nth power and integrate from 0 to 1:
1
(1 − x2)n dx ≤
then pass to polar coordinates:
∞
∞
∞∞
J2 =
e−x2 dx
e−y2 dy =
e−(x2+y2) dx dy.
0
0
00
This is a double integral over the first quadrant, which we will compute by using polar coordinates. In polar coordinates, the first quadrant is {(r, θ) : r ≥ 0 and 0 ≤ θ ≤ π/2}. Writing x2 + y2 = r2
∞ e−x2 2x dx = 2
∞
e−x2 dx = 2J,
2
0
t
0t
0x
0
so 4J 2 =
1 0
dt/
t(1 − t). With the substitution t = sin2 θ,
4J 2 =
π/2 2 sin θ cos θ dθ π = 2 = π,
0
sin θ cos θ
2
√
√
√
so J = π/2. Equivalently, Γ(1/2) = π. Any method that proves Γ(1/2) = π is also a method
that calculates
∞ 0
e−x2
dx.
4
KEITH CONRAD
6. Sixth Proof: Asymptotic Estimates √ We will show J = π/2 by a technique whose steps are based on [14, p. 371]. For x ≥ 0, power series expansions show 1 + x ≤ ex ≤ 1/(1 − x). Reciprocating and replacing x with x2, we get
(3.1)
A(t) = −B(t) + C
for
all
t
>
0.
To
find
C,
we
let
t
→ 0+
in
(3.1).
The
left
side
tends
to
(
0 0
e−x2
dx)2
=
0
while
the
right side tends to −
1 0
dx/(1 + x2) + C
=
−π/4 + C.
Thus
C
ways.
First we compute the volume by horizontal slices, which are discs: V =
1 0
A(z) dz
where
A(z)
is the area of the disc formed by slicing the surface at height z. Writing the radius of this disc at
THE GAUSSIAN INTEGRAL
KEITH CONRAD
Let
∞
∞
I=
e−
1 2
x2
dx,
J
=
e−x2 dx, and K =
−∞
√0
√
These numbers are positive, and J = I/(2 2) and K = I/ 2π.
∞
e−πx2 dx.
−∞
√
√
Theorem. With notation as above, I = 2π, or equivalently J = π/2, or equivalently K = 1.
2. Second Proof: Another change of variables
Our next proof uses another change of variables to compute J2, but this will only rely on single-
variable calculus. As before, we have
curve
z
=
e−
1 2
x2
in
the
xz-plane
around
the
z-axis
to
produce
the
“bell
surface”
z
=
e−
1 2
(x2
+y2
)
in
R3. We will compute the volume V of the region below this surface and above the xy-plane in two