Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
Voice: 402-472-3731
Fax: 402-472-8466

Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar


Evaluation of the Gaussian Density Integral


Note: To read these pages properly, you will need the latest version of the Mozilla Firefox browser, with the STIX fonts installed. In a few sections, you will also need the latest Java plug-in, and JavaScript must be enabled. If you use a browser other than Firefox, you should be able to access the pages and run the applets. However, mathematical expressions will probably not display correctly. Firefox is currently the only browser that supports all of the open standards.




Mathematically Mature: may contain mathematics beyond calculus with proofs.


Section Starter Question

Section Starter Question

Is it possible to evaluate

1 2πex2 2 dx

directly by standard integration techniques?


Key Concepts

Key Concepts

  1. 1 2πex2 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.




  1. The standard Gaussian density function is
    1 2πex2 2 , < x < .


Mathematical Ideas

Mathematical Ideas

The Gaussian density integral

The standard Gaussian density function is

1 2πex2 2 , < x < .

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.

1 2πex2 2 dx = 1.

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.

Proof. Set J =ex2 2 dx =ey2 2 dy. Then we find that

J2 =e(x2+y2) 2 dxdy =02π0er2 2 rdr dθ = 2π.

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

K =0ez22 dz =0e(xy)22ydx

for any y by setting z = xy. Putting z in place of y, it follows that for any z

K =0e(zx)22zdx

so that

K2 = 0ez22 dz 0e(zx)22zdx =00e(x2+1)z22zdz dx.

Now set (1 + x2)z2 = 2t so that zdz = dt(1 + x2) to get

K2 =00et dt (1 + x2) dx = 0et dt 0 dx (1 + x2) = et 0arctan x 0 = π 2

and hence K = π 2. Finally double the integral using symmetry

ey22 dy = 2π.

A third double-integration method due to S. P. Eveson in 2005 is cited by P. S. Lee [2].

Proof. Consider the volume under the surface z = e(x2+y2), which is given by

V =e(x2+y2) dxdy = ex2 dx2.

However V can also be thought of as a volume of revolution created by revolving ex2 about the z axis. Use the disk method of evaluation for a volume of revolution. Since z = ex2 we have x(z) = log z is the radius of a disk at level z.

V = π01x2 dz = π01 log zdz = π[z z log z] 01 = π.

and hence

ex2 dx = π.

Changing variables x = y2,

ey22 dy = 2π.

Evaluation by interpolation between two integrals

Another proof uses an interpolation between two integrals.

Proof. Define

H(t) = 0tex2 dx2 +01et2(x2+1) x2 + 1 dx

for t > 0. Compute the derivative

H(t) = 2 0tex2 dx et2 +01et2(x2+1) x2 + 1 2t(x2 + 1) dx = 2 0tex2 dx et2 +01(2t)et2(x2+1) dx

Change variables in the first integral x = ty, so the new limits of integration are 0 and 1 and dx = tdy. After this change of variables

H(t) = 2t 01et2y2 dy et2 +01(2t)et2(x2+1) dx = 2t 01et2x2 dx et2 2tet2 01et2x2 dx = 0.

Evaluating H at 0,

H(0) =01 1 x2 + 1 dx = π 4.

As t ,

lim tH(t) = 0ex2 dx2.

Putting these all together,

0ex2 dx = π 2 .

Changing variables again, x = y2,

0ey22 dy = π 2.

Finally double the integral using symmetry

ey22 dy = 2π.

Evaluation using Wallis’ Formula

Lemma 2. For n > 1 and all x [0,n],

0 ex 1 x nn e1 n .

Proof. To prove the left inequality is equivalent to showing

1 x nn exn n

for x [0,n]. It suffices to show

0 1 x n exn


0 1 t et

for t [0, 1]. This follows immediately since 1 t is the expression for the tangent line to et at the point (0, 1) and et is concave up.

To prove the right inequality, let F(x) = ex (1 x n)n. Note that F(0) = 0, F(n) = en, and F(0) = en. 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(n) < 0. At the maximum

ex0 = 1 x0 n n1.

Then using this we can rewrite

F(x0) = ex0 ex0 1 x0 n = x0ex0 n

The function xex has a maximum value of e1 at x = 1, so

F(x) F(x0) e1 n .

Now we use this lemma to approximate the Gaussian density integral.

Proof. Approximate the integrand ex2 in the Gaussian density integral with (1 x2n)n on the interval [0,n].

0 0n ex2 1 x2 n n dx 0ne1 n dx.

Rewrite this as

0 0nex2 dx0n 1 x2 n n dx e1 n.

Therefore, taking the limit as n ,

0ex2 dx = lim n0n 1 x2 n n dx.

Now evaluate the integral on the right side using the substitution x = n sin t

0n 1 x2 n n dx = n0π2 cos 2n+1tdt.

Consider Jk =0π2 cos k(x) dx. Then integrating by parts just as in Wallis Formula. we get kJk = (k 1)Jk2. Now J1 = 1 so recursively J3 = 2 3, J5 = 24 35 and inductively

J2n+1 = 2 4(2n 2) (2n) 1 3(2n 1) (2n + 1).


0ex2 dx = lim nn 2 4(2n 2) (2n) 1 3(2n 1) (2n + 1).

Now use Wallis’ Formula

lim n 2 2 4 4 6 6(2n) (2n) 1 3 3 5 5(2n 1) (2n + 1) = π 2.


0ex2 dx = lim n n 2n + 1 2 2 4 4 6 6 (2n) (2n) 1 3 3 5 5(2n 1) (2n + 1) = 1 2π 2.

Finally changing variables and using symmetry as before

1 2πey2 2 dy = 1.


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

Problems to Work for Understanding



Reading Suggestion:


[1]   Tom M. Apostol. Mathematical Analysis. Addison-Wesley, second edition, 1974.

[2]   Peter M. Lee. The probability integral. [Online; accessed 15-October-2011].

[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:


I check all the information on each page for correctness and typographical errors. Nevertheless, some errors may occur and I would be grateful if you would alert me to such errors. I make every reasonable effort to present current and accurate information for public use, however I do not guarantee the accuracy or timeliness of information on this website. Your use of the information from this website is strictly voluntary and at your risk.

I have checked the links to external sites for usefulness. Links to external websites are provided as a convenience. I do not endorse, control, monitor, or guarantee the information contained in any external website. I don’t guarantee that the links are active at all times. Use the links here with the same caution as you would all information on the Internet. This website reflects the thoughts, interests and opinions of its author. They do not explicitly represent official positions or policies of my employer.

Information on this website is subject to change without notice.

Steve Dunbar’s Home Page,

Email to Steve Dunbar, sdunbar1 at unl dot edu

Last modified: Processed from LATEX source on October 22, 2011