Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
http://www.math.unl.edu
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar
Evaluation of the Gaussian Density Integral
Rating
Mathematically Mature: may contain mathematics beyond calculus with
proofs.
1
Section Starter Question
Is it possible to evaluate
Z ∞
x2
1
√
e− 2 dx
2π −∞
directly by standard integration techniques?
Key Concepts
1.
1
√
2π
Z
∞
x2
e− 2 dx = 1.
−∞
2. Several proofs use double integration in clever ways.
3. Another proof uses interpolation between two standard integrals.
4. Yet another proof uses an approximation which can be evaluated using
standard integration techniques. The approximate evaluations lead to
a sequence with a limit related to Wallis’ Formula.
Vocabulary
1. The standard Gaussian density function is
x2
1
√ e− 2 ,
2π
−∞ < x < ∞.
2
Mathematical Ideas
The Gaussian density integral
The standard Gaussian density function is
x2
1
√ e− 2 ,
−∞ < x < ∞.
2π
In order to claim this is truly a probability density function, we must show
that the total area under this function is 1.
Theorem 1.
Z ∞
x2
1
√
e− 2 dx = 1.
2π −∞
We provide several proofs of this evaluation.
According to Peter Lee, [2], the first complete evaluation of this integral
was by P. S. Laplace in 1774 starting from a more complicated integral,
using some changes of variables and limits, and finally reaching a special
case which can be reduced to the Gaussian integral. Lee [2] gives 8 methods
of evaluation, including 3 cited here along with other methods using contour
integration and the Gamma Function.
Evaluation with double integration
The first proof is the usual proof, relying on integration in polar coordinates.
R∞
R∞
y2
x2
Proof. Set J = −∞ e− 2 dx = −∞ e− 2 dy. Then we find that
Z ∞Z ∞
(x2 +y 2 )
2
J =
e− 2 dx dy
−∞ −∞
Z 2π Z ∞
r2
=
e− 2 r dr dθ
0
0
= 2π.
3
Then J =
√
2π.
According to Peter Lee, [2], this method is due to S. D. Poisson and was
popularized by J. K. F. Sturm.
A second proof using double integrals is also due to P. S. Laplace, [2].
Proof. Note that
Z
∞
K=
e
−z 2 /2
∞
Z
e−(xy)
dz =
0
2 /2
y dx
0
for any y by setting z = xy. Putting z in place of y, it follows that for any z
Z ∞
2
e−(zx) /2 z dx
K=
0
so that
Z
2
∞
e
K =
Z
0
∞
Z
∞
e
dz
−(zx)2 /2
z dx
0
∞
Z
e−(x
=
0
−z 2 /2
2 +1)z 2 /2
z dz dx .
0
Now set (1 + x2 )z 2 = 2t so that z dz = dt /(1 + x2 ) to get
Z ∞Z ∞
dt
2
e−t
K =
dx
(1 + x2 )
Z ∞
0Z ∞0
dx
−t
e dt
=
(1 + x2 )
0
0
∞
= −e−t 0 (arctan x|∞
0 )
π
=
2
p
and hence K = π/2. Finally double the integral using symmetry
Z ∞
√
2
e−y /2 dy = 2π.
−∞
A third double-integration method due to S. P. Eveson in 2005 is cited
by P. S. Lee [2].
4
Proof. Consider the volume under the surface z = e−(x
by
Z ∞Z ∞
2
2
V =
e−(x +y ) dx dy
−∞ −∞
∞
−x2
e
, which is given
2
Z
=
2 +y 2 )
dx
.
−∞
However V can also be thought of as a volume of revolution created by
2
revolving e−x about the z axis. Use the disk method
√ of evaluation for a
2
volume of revolution. Since z = e−x we have x(z) = − log z is the radius
of a disk at level z.
Z 1
Z 1
2
V =π
x dz = π
− log z dz = π[z − z log z]10 = π.
0
0
and hence
Z
∞
2
e−x dx =
√
π.
−∞
√
Changing variables x = y/ 2,
Z ∞
e−y
2 /2
dy =
√
2π.
−∞
Evaluation by interpolation between two integrals
Another proof uses an interpolation between two integrals.
Proof. Define
Z
H(t) =
t
−x2
e
2
dx
Z
+
0
0
1
2
2
e−t (x +1)
dx
x2 + 1
for t > 0. Compute the derivative
Z t
Z 1 −t2 (x2 +1)
e
0
−x2
−t2
H (t) = 2
e
dx · e +
· −2t(x2 + 1) dx
2
x +1
0
Z0 t
Z 1
2
2
2 2
=2
e−x dx · e−t +
(−2t)e−t (x +1) dx
0
0
5
Change variables in the first integral x = ty, so the new limits of integration
are 0 and 1 and dx = t dy. After this change of variables
Z 1
Z 1
2 2
0
−t2 y 2
−t2
H (t) = 2t
e
dy · e +
(−2t)e−t (x +1) dx
0
Z0 1
Z 1
2 2
−t2 x2
−t2
−t2
= 2t
e
dx · e − 2te
e−t x dx
0
0
= 0.
Evaluating H at 0,
1
Z
H(0) =
x2
0
1
π
dx = .
+1
4
As t → ∞,
Z
∞
e
lim H(t) =
t→∞
−x2
2
dx
0
Putting these all together,
Z
∞
√
−x2
e
dx =
0
π
.
2
√
Changing variables again, x = y/ 2,
√
Z ∞
π
−y 2 /2
e
dy = √ .
2
0
Finally double the integral using symmetry
Z ∞
√
2
e−y /2 dy = 2π.
−∞
Evaluation using Wallis’ Formula
Lemma 2. For n > 1 and all x ∈ [0, n],
0≤e
−x
x n e−1
− 1−
≤
.
n
n
6
.
Proof. To prove the left inequality is equivalent to showing
n
x n
1−
≤ e−x/n
n
for x ∈ [0, n]. It suffices to show
x
0≤ 1−
≤ e−x/n
n
or
0 ≤ (1 − t) ≤ e−t
for t ∈ [0, 1]. This follows immediately since 1 − t is the expression for the
tangent line to e−t at the point (0, 1) and e−t is concave up.
To prove the right inequality, let F (x) = e−x −(1− nx )n . Note that F (0) =
0, F (n) = e−n , and F 0 (0) = −e−n . The function F attains a maximum at
some point x0 of the interval [0, n]. The maximum is positive, since F (n) >
F (0) = 0. The maximum does not occur at n since F 0 (n) < 0. At the
maximum
x0 n−1
.
e−x0 = 1 −
n
Then using this we can rewrite
x0 x0 e−x0
=
F (x0 ) = e−x0 − e−x0 1 −
n
n
The function xe−x has a maximum value of e−1 at x = 1, so
F (x) ≤ F (x0 ) ≤
e−1
.
n
Now we use this lemma to approximate the Gaussian density integral.
2
−x
Proof. Approximate the integrand
in the Gaussian density integral with
√ e
2
n
(1 − x /n) on the interval [0, n].
n Z √n Z √n −1
x2
e
−x2
e
− 1−
0≤
dx ≤
dx .
n
n
0
0
Rewrite this as
√
Z
0≤
√
n
e
0
−x2
Z
n
dx −
0
7
x2
1−
n
n
e−1
dx ≤ √ .
n
Therefore, taking the limit as n → ∞,
Z
Z ∞
−x2
e
dx = lim
n→∞
0
0
√
n
n
x2
1−
dx .
n
Now evaluate the integral on the right side using the substitution x =
n
Z √n Z π/2
√
x2
1−
cos2n+1 t dt .
dx = n
n
0
0
√
n sin t
R π/2
Consider Jk = 0 cosk (x) dx. Then integrating by parts just as in
Wallis Formula we get kJk = (k − 1)Jk−2 . Now J1 = 1 so recursively J3 = 32 ,
and inductively
J5 = 2·4
3·5
J2n+1 =
Hence,
∞
Z
2 · 4 · · · (2n − 2) · (2n)
.
1 · 3 · · · (2n − 1) · (2n + 1)
2
e−x dx = lim
n→∞
0
√
n
2 · 4 · · · (2n − 2) · (2n)
.
1 · 3 · · · (2n − 1) · (2n + 1)
Now use Wallis’ Formula
2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)
π
lim
= .
n→∞
1 · 3 · 3 · 5 · 5 . . . (2n − 1) · (2n + 1)
2
Then
Z
0
∞
2
e−x dx
s
√
n
2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)
= lim √
n→∞
2n + 1 1 · 3 · 3 · 5 · 5 . . . (2n − 1) · (2n + 1)
r
π
1
=√
.
2 2
Finally changing variables and using symmetry as before
Z ∞
y2
1
√
e− 2 dy = 1.
2π −∞
8
Sources
The first double-integral proof is standard and occurs in many sources. The
interpolation proof is from a footnote in the article by Michel, [4], where it
is credited to exercises in Apostol, [1]. The proof using Wallis’ Formula is
from the short article by Levrie and Daems, [3].
Problems to Work for Understanding
1.
2.
3.
4.
Reading Suggestion:
References
[1] Tom M. Apostol. Mathematical Analysis. Addison-Wesley, second edition,
1974.
[2] Peter M. Lee. The probability integral. http://www.york.ac.uk/depts/
maths/histstat/normal_history.pdf. [Online; accessed 15-October2011].
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[3] Paul Levrie and Walter Daems. Evaluating the probability integral using Wallis product formula for π. American Mathematical Monthly,
116(6):538–541, June-July 2009.
[4] Reinhard Michel. On Stirling’s formula. American Mathematical Monthly,
109(4):388–390, April 2002.
Outside Readings and Links:
1.
2.
3.
4.
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Steve Dunbar’s Home Page, http://www.math.unl.edu/~sdunbar1
Email to Steve Dunbar, sdunbar1 at unl dot edu
Last modified: Processed from LATEX source on October 22, 2011
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