THE GAUSSIAN INTEGRAL
KEITH CONRAD
Let
Z
∞
− 21 x2
Z
∞
dx, and K =
0
−∞
√
√
These numbers are positive, and J = I/(2 2) and K = I/ 2π.
e
I=
e
Z
−x2
dx, J =
Theorem. With notation as above, I =
√
∞
2
e−πx dx.
−∞
2π, or equivalently J =
√
π/2, or equivalently K = 1.
We will give multiple proofs of this result. (Other lists of proofs are in [3] and [8].) The theorem
1 2
2
2
is
subtle because there is no simple antiderivative for e− 2 x (or e−x or e−πx ). For comparison,
Z ∞
1 2
1 2
xe− 2 x dx can be computed using the antiderivative −e− 2 x : this integral is 1.
0
1. First Proof: Polar coordinates
The most widely known proof uses multivariable calculus: express J 2 as a double integral and
then pass to polar coordinates:
Z ∞
Z ∞Z ∞
Z ∞
2
2
2
2
e−x dx
e−y dy =
e−(x +y ) dx dy.
J2 =
0
0
0
0
This is a double integral over the first quadrant, which we will compute by using polar coordinates.
The first quadrant is {(r, θ) : r ≥ 0 and 0 ≤ θ ≤ π/2}. Writing x2 + y 2 as r2 and dx dy as r dr dθ,
Z π/2 Z ∞
2
2
J =
e−r r dr dθ
0
Z
=
0
∞
−r2
re
Z
π/2
dr ·
dθ
0
0
1 −r2 ∞ π
= − e
·2
2
0
1 π
=
·
2 2
π
=
.
4
√
Taking square roots, J = π/2. This method is due to Poisson [8, p. 3].
2. Second Proof: Another change of variables
Our next proof uses another change of variables to compute J 2 , but this will only rely on singlevariable calculus. As before, we have
Z ∞Z ∞
Z ∞ Z ∞
−(x2 +y 2 )
2
−(x2 +y 2 )
e
dx dy,
J =
e
dx dy =
0
0
0
1
0
2
KEITH CONRAD
but instead of using polar coordinates we make a change of variables x = yt on the inside integral,
with dx = y dt, so
Z ∞ Z ∞
Z ∞ Z ∞
−y 2 (t2 +1)
−y 2 (t2 +1)
2
ye
dy dt.
e
y dt dy =
J =
0
Z
Since
∞
2
ye−ay dy =
0
0
0
0
1
for a > 0, we have
2a
Z ∞
dt
1 π
π
2
J =
= · = ,
2
2(t + 1)
2 2
4
0
√
so J = π/2. This approach is due to Laplace [6, pp. 94–96] and historically precedes the more
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
For t > 0, set
Z
A(t) =
t
e
−x2
2
.
dx
0
2
J and
The integral we want to calculate is A(∞) =
then take a square root.
Differentiating A(t) with respect to t and using the Fundamental Theorem of Calculus,
Z t
Z t
2
−x2
−t2
−t2
0
e
dx · e
= 2e
e−x dx.
A (t) = 2
0
0
Let x = ty, so
2
A0 (t) = 2e−t
1
Z
2 y2
te−t
1
Z
dy =
0
2te−(1+y
2 )t2
dy.
0
The function under the integral sign is easily antidifferentiated with respect to t:
Z 1
Z
2 2
2 2
∂ e−(1+y )t
d 1 e−(1+y )t
0
A (t) =
−
dy = −
dy.
∂t 1 + y 2
dt 0
1 + y2
0
Letting
2
2
e−t (1+x )
dx,
1 + x2
0
we have A0 (t) = −B 0 (t) for all t > 0, so there is a constant C such that
Z
1
B(t) =
A(t) = −B(t) + C
(3.1)
Z
0+
0
−x2
2
e
for all t > 0. To find C, we let t →
in (3.1). The left side tends to
dx
= 0 while
0
Z 1
the right side tends to −
dx/(1 + x2 ) + C = −π/4 + C. Thus C = π/4, so (3.1) becomes
0
2
2
2
e−t (1+x )
e
dx
dx.
1 + x2
0
0
√
Letting t → ∞ in this equation, we obtain J 2 = π/4, so J = π/2.
A comparison of this proof with the first proof is in [18].
Z
t
−x2
π
= −
4
Z
1
THE GAUSSIAN INTEGRAL
3
4. Fourth Proof: Another differentiation under the integral sign
Here is a second approach to finding J with differentiation under the integral sign. I learned it
from Michael Rozman [13], who modified an idea on math.stackexchange [20].
For t ∈ R, set
Z ∞ −t2 (1+x2 )
e
F (t) =
dx.
1 + x2
0
R∞
Then F (0) = 0 dx/(1 + x2 ) = π/2 and F (∞) = 0. Differentiating under the integral sign,
Z ∞
Z ∞
2
0
−t2 (1+x2 )
−t2
F (t) =
−2te
dx = −2te
e−(tx) dx.
0
0
Make the substitution y = tx, with dy = t dx, so
Z ∞
2
2
0
−t2
e−y dy = −2Je−t .
F (t) = −2e
0
For b > 0, integrate both sides from 0 to b and use the Fundamental Theorem of Calculus:
Z b
Z b
Z b
2
2
F 0 (t) dt = −2J
e−t dt =⇒ F (b) − F (0) = −2J
e−t dt.
0
0
Letting b → ∞,
0−
0
√
π
π
π
= −2J 2 =⇒ J 2 = =⇒ J =
.
2
4
2
5. Fifth Proof: A volume integral
Our next proof is due to T. P. Jameson [4] and it was rediscovered by A. L. Delgado [2]. Revolve
1 2
1
2
2
the curve z = e− 2 x in the xz-plane around the z-axis to produce the “bell surface” z = e− 2 (x +y ) .
See below, where the z-axis is vertical and passes through the top point, the x-axis lies just under
the surface through the point 0 in front, and the y-axis lies just under the surface through the
point 0 on the left. We will compute the volume V below the surface and above the xy-plane in
two ways.
Z
1
First we compute V by horizontal slices, which are discs: V =
A(z) dz where A(z) is the area
0
of the disc formed by slicing the surface at height z. Writing the radius of the disc at height z as
1 2
r(z), A(z) = πr(z)2 . To compute r(z), the surface cuts the xz-plane at a pair of points (x, e− 2 x )
1 2
where the height is z, so e− 2 x = z. Thus x2 = −2 ln z. Since x is the distance of these points from
the z-axis, r(z)2 = x2 = −2 ln z, so A(z) = πr(z)2 = −2πln z. Therefore
1
Z 1
V =
−2π ln z dz = −2π (z ln z − z) = −2π(−1 − lim z ln z).
0
z→0+
0
By L’Hospital’s rule, limz→0+ z ln z = 0, so V = 2π. (A calculation of V by shells is in [10].)
Next we compute the volume by vertical slices in planes x = constant. Vertical slices are scaled
bell curves: look at the black contour lines in the picture. The equation of the bell curve along the
1
2
2
top ofZ the vertical slice with x-coordinate x is z = e− 2 (x +y ) , where y varies and x is fixed. Then
∞
V =
A(x) dx, where A(x) is the area of the x-slice:
−∞
Z
∞
A(x) =
− 12 (x2 +y 2 )
e
−∞
− 12 x2
Z
∞
dy = e
−∞
1 2
1 2
e− 2 y dy = e− 2 x I.
4
KEITH CONRAD
Z
∞
Z
∞
− 12 x2
∞
Z
1 2
e− 2 x dx = I 2 .
−∞
−∞
−∞
√
Comparing the two formulas for V , we have 2π = I 2 , so I = 2π.
Thus V =
A(x) dx =
e
I dx = I
6. Sixth Proof: The Γ-function
Z ∞
For any integer n ≥ 0, we have n! =
tn e−t dt. For x > 0 we define
0
Z
Γ(x) =
0
∞
tx e−t
dt
,
t
so Γ(n) = (n − 1)! when n ≥ 1. Using integration by parts, Γ(x + 1) = xΓ(x). One of the basic
properties of the Γ-function [14, pp. 193–194] is
Z 1
Γ(x)Γ(y)
(6.1)
=
tx−1 (1 − t)y−1 dt.
Γ(x + y)
0
Set x = y = 1/2:
2 Z 1
1
dt
p
Γ
=
.
2
t(1 − t)
0
THE GAUSSIAN INTEGRAL
5
Note
Z ∞
Z ∞ −x2
Z ∞
√ −t dt Z ∞ e−t
1
e
2
√ dt =
Γ
=
te
=
2x dx = 2
e−x dx = 2J,
2
t
x
t
0
0
0
0
p
R
1
2
2
so 4J = 0 dt/ t(1 − t). With the substitution t = sin θ,
Z π/2
2 sin θ cos θ dθ
π
2
4J =
= 2 = π,
sin θ cos θ
2
0
√
√
√
so J = π/2. Equivalently,
Γ(1/2) = π. Any method that proves Γ(1/2) = π is also a method
Z ∞
2
e−x dx.
that calculates
0
7. Seventh Proof: Asymptotic estimates
√
We will show J = π/2 by a technique whose steps are based on [15, p. 371].
For x ≥ 0, power series expansions show 1 + x ≤ ex ≤ 1/(1 − x). Reciprocating and replacing x
with x2 , we get
1
2
(7.1)
1 − x2 ≤ e−x ≤
.
1 + x2
for all x ∈ R.
For any positive integer n, raise the terms in (7.1) to the nth power and integrate from 0 to 1:
Z 1
Z 1
Z 1
dx
2 n
−nx2
(1 − x ) dx ≤
e
dx ≤
.
2 n
0
0
0 (1 + x )
√
Under the changes of variables x = sin θ on the left, x = y/ n in the middle, and x = tan θ on the
right,
Z √n
Z π/2
Z π/4
1
2
2n+1
(7.2)
(cos θ)
dθ ≤ √
e−y dy ≤
(cos θ)2n−2 dθ.
n
0
0
0
R π/2
k
Set Ik = 0 (cos θ) dθ, so I0 = π/2, I1 = 1, and (7.2) implies
Z √n
√
√
2
(7.3)
nI2n+1 ≤
e−y dy ≤ nI2n−2 .
0
We will show that as k → ∞,
√
kIk2
nI2n+1
→ π/2. Then
r
√
√
n √
1
π
π
=√
2n + 1I2n+1 → √
=
2
2n + 1
2 2
and
√
nI2n−2 = √
√
Z
so by (7.3)
n
2
e−y dy →
√
√
n √
1
2n − 2I2n−2 → √
2n − 2
2
π/2. Thus J =
0
√
r
√
π
π
=
,
2
2
π/2.
To show kIk2 → π/2, first we compute several values of Ik explicitly by a recursion. Using
integration by parts,
Z π/2
Z π/2
Ik =
(cos θ)k dθ =
(cos θ)k−1 cos θ dθ = (k − 1)(Ik−2 − Ik ),
0
0
6
KEITH CONRAD
so
k−1
Ik−2 .
k
Using (7.4) and the initial values I0 = π/2 and I1 = 1, the first few values of Ik are computed and
listed in Table 1.
(7.4)
Ik =
Ik
k
Ik
k
0
π/2
1
1
3
2/3
2 (1/2)(π/2)
4 (3/8)(π/2)
5 8/15
6 (15/48)(π/2) 7 48/105
Table 1.
From Table 1 we see that
1 π
2n + 1 2
for 0 ≤ n ≤ 3, and this can be proved for all n by induction using (7.4). Since 0 ≤ cos θ ≤ 1 for
k
θ ∈ [0, π/2], we have Ik ≤ Ik−1 ≤ Ik−2 = k−1
Ik by (7.4), so Ik−1 ∼ Ik as k → ∞. Therefore (7.5)
implies
1 π
π
2
2
I2n
∼
=⇒ (2n)I2n
→
2n 2
2
as n → ∞. Then
π
2
2
(2n + 1)I2n+1
∼ (2n)I2n
→
2
√
as n → ∞, so kIk2 → π/2 as k → ∞. This completes our proof that J = π/2.
(7.5)
I2n I2n+1 =
Remark 7.1. This proof is closely related to the fifth proof using the Γ-function. Indeed, by (6.1)
Z 1
1
Γ( k+1
2 )Γ( 2 )
=
t(k+1)/2+1 (1 − t)1/2−1 dt,
1
Γ( k+1
+
)
0
2
2
and with the change of variables t = (cos θ)2 for 0 ≤ θ ≤ π/2, the integral on the right is equal to
R π/2
2 0 (cos θ)k dθ = 2Ik , so (7.5) is the same as
π
= I2n I2n+1
2(2n + 1)
=
1
2n+2
1
Γ( 2n+1
2 )Γ( 2 ) Γ( 2 )Γ( 2 )
2Γ( 2n+2
2Γ( 2n+3
2 )
2 )
=
1 2
Γ( 2n+1
2 )Γ( 2 )
4Γ( 2n+1
2 + 1)
=
1 2
Γ( 2n+1
2 )Γ( 2 )
2n+1
4 2n+1
2 Γ( 2 )
=
Γ( 12 )2
,
2(2n + 1)
or equivalently Γ(1/2)2 = π. We saw in the fifth proof that Γ(1/2) =
√
π if and only if J =
√
π/2.
THE GAUSSIAN INTEGRAL
7
8. Eighth Proof: The original proof
√
The original proof that J = π/2 is due to Laplace [7] in 1774. (An English translation of
Laplace’s article is mentioned in the bibliographic citation for [7], with preliminary comments on
that article in [16].) He wanted to compute
Z 1
dx
√
.
(8.1)
− log x
0
R∞
√
√
2
Setting y = − log x, this integral is 2 0 e−y dy = 2J, so we expect (8.1) to be π.
Laplace’s starting point for evaluating (8.1) was a formula of Euler:
Z 1
Z 1 s+r
xr dx
x
dx
1
π
√
√
=
(8.2)
s(r + 1) 2
1 − x2s 0
1 − x2s
0
for positive r and s. (Laplace himself said this formula held “whatever be” r or s, but if s < 0 then
the number under the square root is negative.) Accepting (8.2), let r → 0 in it to get
Z 1
Z 1
dx
xs dx
1π
√
√
(8.3)
.
=
2s
2s
s2
1−x
1−x
0
0
Now let s → 0 in (8.3). Then 1 − x2s ∼ −2s log x by L’Hopital’s rule, so (8.3) becomes
Z 1
2
dx
√
= π.
− log x
0
√
Thus (8.1) is π.
Euler’s formula (8.2) looks mysterious, but we have met it before. In the formula let xs = cos θ
with 0 ≤ θ ≤ π/2. Then x = (cos θ)1/s , and after some calculations (8.2) turns into
Z π/2
Z π/2
π
1
(r+1)/s−1
.
(8.4)
(cos θ)
dθ
(cos θ)(r+1)/s dθ =
(r + 1)/s 2
0
0
R π/2
We used the integral Ik = 0 (cos θ)k dθ before when k is a nonnegative integer. This notation
1 π
makes sense when k is any positive real number, and then (8.4) assumes the form Iα Iα+1 = α+1
2 for
α = (r +1)/s − 1, which is (7.5) with a possibly nonintegral index. Letting r = 0 and s = 1/(2n + 1)
in (8.4) recovers (7.5). Letting s → 0 in (8.3) corresponds to letting n → ∞ in (7.5), so the 6th
proof is in essence a more detailed version of Laplace’s 1774 argument.
9. Ninth Proof: Contour Integration
Z
We will calculate
∞
e−x
2 /2
dx using contour integrals and the residue theorem. However, we
−∞
2
e−z /2 ,
can’t just integrate
as this function has no poles. For a long time nobody knew how to
handle this integral using contour integration. For instance, in 1914 Watson
Z [17, p. 79] wrote
∞
“Cauchy’s theorem cannot be employed to evaluate all definite integrals; thus
0
2
e−x dx has not
been evaluated except by other methods.” In the 1940s several contour integral solutions were
published using awkward contours such as parallelograms [9], [11, Sect. 5] (see [1, Exer. 9, p. 113]
for a recent appearance). Our approach will follow Kneser [5, p. 121] (see also [12, pp. 413–414] or
[19]), using a rectangular contour and the function
2
e−z /2
√
.
1 − e− π(1+i)z
8
KEITH CONRAD
This function comes out of nowhere, so our first task is to motivate the introduction of this function.
We seek a meromorphic function f (z) to integrate around the rectangular contour γR in the
figure below, with vertices at −R, R, R + ib, and −R + ib, where b will be fixed and we let R → ∞.
Suppose f (z) → 0 along the right and left sides of γR uniformly as R → ∞. Then by applying
the residue theorem and letting R → ∞, we would obtain (if the integrals converge)
Z −∞
Z ∞
X
f (x + ib) dx = 2πi
Resz=a f (z),
f (x) dx +
∞
−∞
a
where the sum is over poles of f (z) with imaginary part between 0 and b. This is equivalent to
Z ∞
X
(f (x) − f (x + ib)) dx = 2πi
Resz=a f (z).
−∞
a
Therefore we want f (z) to satisfy
(9.1)
f (z) − f (z + ib) = e−z
2 /2
,
where f (z) and b need to be determined.
2
Let’s try f (z) = e−z /2 /d(z), for an unknown denominator d(z) whose zeros are poles of f (z).
We want f (z) to satisfy
(9.2)
f (z) − f (z + τ ) = e−z
2 /2
for some τ (which will not be purely imaginary, so (9.1) doesn’t quite work, but (9.1) is only
2
motivation). Substituting e−z /2 /d(z) for f (z) in (9.2) gives us
!
2
1
e−τ z−τ /2
2
−z 2 /2
(9.3)
e
−
= e−z /2 .
d(z)
d(z + τ )
Suppose d(z + τ ) = d(z). Then (9.3) implies
d(z) = 1 − e−τ z−τ
2 /2
,
2
and with this definition of d(z), f (z) satisfies (9.2) if and only if eτ = 1, or equivalently τ 2 ∈ 2πiZ.
√
2
The simplest nonzero solution is τ = π(1 + i). From now on this is the value of τ , so e−τ /2 =
e−iπ = −1 and then
2
2
e−z /2
e−z /2
f (z) =
=
,
d(z)
1 + e−τ z
which is Kneser’s function mentioned earlier. This function satisfies (9.2) and we henceforth ignore
the motivation (9.1). Poles of f (z) are at odd integral multiples of τ /2.
We will integrate this f (z) around the rectangular contour γR below, whose height is Im(τ ).
THE GAUSSIAN INTEGRAL
9
The poles of f (z) nearest the origin
√ are plotted in the figure; they lie along the line y = x. The
only pole of f (z) inside γR (for R > π/2) is at τ /2, so by the residue theorem
Z
2
2
√
e−τ /8
2πie3τ /8
√
f (z) dz = 2πiResz=τ /2 f (z) = 2πi
=
=
2π.
2
− π(1 + i)
(−τ )e−τ /2
γR
As R → ∞, the value of |f (z)| tends to 0 uniformly along the left and right sides of γR , so
Z ∞
Z −∞+i√π
√
2π =
f (x) dx +
f (z) dz
√
−∞
∞+i π
Z ∞
Z ∞
√
=
f (x) dx −
f (x + i π) dx.
−∞
−∞
√
In the second integral, write i π as τ − π and use (real) translation invariance of dx to obtain
Z ∞
Z ∞
√
2π =
f (x) dx −
f (x + τ ) dx
−∞
−∞
Z ∞
(f (x) − f (x + τ )) dx
=
−∞
Z ∞
2
=
e−x /2 dx by (9.2).
−∞
10. Tenth Proof: Stirling’s Formula
Z ∞
√
1 2
Besides the integral formula
e− 2 x dx = 2π that we have been discussing, another place
−∞
√
in mathematics where 2π appears is in Stirling’s formula:
nn √
2πn as n → ∞.
en
√
In 1730 De Moivre proved n! ∼ C(nn /en ) n for some
√ positive number C without being able to
determine C. Stirling soon thereafter showed C = 2π and wound up having the whole formula
√
named after him. We will show that
determining
that
the
constant
C
in
Stirling’s
formula
is
2π
√
√
is equivalent to showing that J = π/2 (or, equivalently, that I = 2π).
n! ∼
10
KEITH CONRAD
Applying (7.4) repeatedly,
I2n =
=
2n − 1
I2n−2
2n
(2n − 1)(2n − 3)
I2n−4
(2n)(2n − 2)
..
.
=
(2n − 1)(2n − 3)(2n − 5) · · · (5)(3)(1)
I0 .
(2n)(2n − 2)(2n − 4) · · · (6)(4)(2)
Inserting (2n − 2)(2n − 4)(2n − 6) · · · (6)(4)(2) in the top and bottom,
(2n − 1)(2n − 2)(2n − 3)(2n − 4)(2n − 5) · · · (6)(5)(4)(3)(2)(1) π
(2n − 1)!
π
=
.
2
n−1
2
(2n)((2n − 2)(2n − 4) · · · (6)(4)(2))
2
2n(2
(n − 1)!) 2
√
Applying De Moivre’s asymptotic formula n! ∼ C(n/e)n n, ,
√
√
1
(2n − 1)2n 2n−1
2n − 1
π
π
C((2n − 1)/e)2n−1 2n − 1
√
=
I2n ∼
1
2(n−1)
2n
n−1
n−1
2
2n · 2
Ce(n − 1) (n−1)2 (n − 1) 2
2n(2
C((n − 1)/e)
n − 1) 2
I2n =
as n → ∞. For any a ∈ R, (1 + a/n)n → ea as n → ∞, so (n + a)n ∼ ea nn . Substituting this into
the above formula with a = −1 and n replaced by 2n,
(10.1)
Since Ik−1
Therefore
e−1 (2n)2n √12n
π
π
= √ .
2
2n ·
C 2n
√
√
∼ Ik , the outer terms in (7.3) are both asymptotic to nI2n ∼ π/(C 2) by (10.1).
I2n ∼
22(n−1) Ce(e−1 nn )2 n12 n
√
Z
n
π
2
e−y dy → √
C 2
0
√
√
√
as n → ∞, so J = π/(C 2). Therefore C = 2π if and only if J = π/2.
11. Eleventh Proof: Fourier transforms
For a continuous function f : R → C that is rapidly decreasing at ±∞, its Fourier transform is
the function Ff : R → C defined by
Z ∞
(Ff )(y) =
f (x)e−ixy dx.
−∞
R∞
For example, (Ff )(0) = −∞ f (x) dx.
Here are three properties of the Fourier transform.
• If f is differentiable, then after using differentiation under the integral sign on the Fourier
transform of f we obtain
Z ∞
(Ff )0 (y) =
−ixf (x)e−ixy dx = −i(F(xf (x)))(y).
−∞
• Using integration by parts on the Fourier transform of f , with u = f (x) and dv = e−ixy dx,
we obtain
F(f 0 )(y) = iy(Ff )(y).
THE GAUSSIAN INTEGRAL
11
• If we apply the Fourier transform twice then we recover the original function up to interior
and exterior scaling:
(F 2 f )(x) = 2πf (−x).
(11.1)
Let’s show the appearance of 2π in (11.1) is equivalent to the evaluation of I as
2
Fixing a > 0, set f (x) = e−ax , so
√
2π.
f 0 (x) = −2axf (x).
1
Applying the Fourier transform to both sides of this equation implies iy(Ff )(y) = −2a −i
(Ff )0 (y),
1
1
0
0
which simplifies to (Ff ) (y) = − 2a y(Ff )(y). The general solution of g (y) = − 2a yg(y) is g(y) =
2
Ce−y /(4a) , so
2
(Ff )(y) = Ce−y /(4a)
for some constant C. Letting a = 12 , so f (x) = e−x
2 /2
, we obtain
−y 2 /2
(Ff )(y) = Ce
= Cf (y).
R ∞ −x2 /2
dx = I, so I = Cf (0) = C.
Setting y = 0, the left side is (Ff )(0) = −∞ e
Applying the Fourier transform to both sides of the equation (Ff )(y)
√ = Cf (y), we get 2πf (−x) =
2
2
C(Ff
√)(x) = C f (x). At x = 0 this becomes 2π = C , so I = C = ± 2π. Since I > 0, the number
I is 2π. If we didn’t know the constant on the right side of (11.1) were 2π, whatever its value is
2
would
√ wind up being C , so saying 2π appears on the right side of (11.1) is equivalent to saying
I = 2π.
References
[1] C. A. Berenstein and R. Gay, Complex
Z ∞ Variables, Springer-Verlag, New York, 1991.
2
[2] A. L. Delgado, “A Calculation of
e−x dx”, The College Math. J. 34 (2003), 321–323.
0
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[19] http://math.stackexchange.com/questions/34767/int-infty-infty-e-x2-dx-with-complex-analysis.
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