I've been studying probability on my own using an old textbook
(~1958).  One of the problems reduces to showing that the result of an
integral equals the gaussian density function.  I can use various
computational methods to prove that the integral does in fact equal
the guassian density function, but I have no idea how to evaluate the
integral using techniques available in 1958.  Out of curiosity, I'd be
very happy if someone could show me how.

Here is the integral:

Int[abs(k) to inf] ((x*E^(-(1/2)*x^2)) / (Pi*Sqrt[x^2 - k^2])) dx

This does in fact equal the gaussian density function of k:

E^(-(1/2)*k^2) / Sqrt[2*Pi]

Can you prove this using only techniques available to a college student in 1958?

Thanks!

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Request for Question Clarification by mathtalk-ga on 18 Aug 2004 04:41 PDT

Hi, missionpeak-ga:

Are you looking for hints on how to do this or a detailed solution?

There's a u-substitution that gets rid of the dependence on k, and
leaves you with a definite integral:

INTEGRAL exp(-(1/2)t^2) dt FROM -oo to +oo

for which there's a "trick" to integrating using a bit of
multidimensional calculus that you may already know.

regards, mathtalk-ga

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Clarification of Question by missionpeak-ga on 20 Aug 2004 21:11 PDT

Hi kprema,

Wow!  Thanks for a very complete and easy to follow explanation. 

Mathtalk,

While I probably could have figured it out with a few hints, I was
very happy to see the complete explanation from kprema.  My
programming ability exceeded my math ability quite some time ago and I
enjoy learning from folks who obviously know what they're doing.

Thanks again to both of you.

- Dave

Hi: Let us first look at the case when k >= 0.  The integral then is

(1)  I = \int_{x=k}^{inf} x*exp(-x^2/2)/(pi*sqrt(x^2-k^2)) dx.

Use the substitution u^2 = x^2 - k^2 --> u du = x dx.  Then we have

(2)  I = exp(-k^2/2)/pi * \int_{u=0}^{inf} exp(-u^2/2) du.

But, it is well established that (check a handbook on integrals) 

(3)  \int_{u=0}^{inf} exp(-a^2*u^2) du = sqrt(pi)/(2*a), for a > 0.

When a = 1/sqrt(2), this reduces to

(4)  \int_{u=0}^{inf} exp(-u^2/2) du = sqrt(pi/2).

This is exactly the integral that appears in (2).  Hence, we have 

(5)  I = exp(-k^2/2)/sqrt(2*pi),

which is what you wanted I think.

Now, let us look at the case when k < 0.  The integral then reduces to
an expression that is identical to (1) except that the integration
limits are from
-k to inf (instead of k to inf).  But, when the substitution u^2 = x^2
- k^2 is made, one is left with (2).  Hence, the result is the same.

I hope this helps.

kprema

Hi, kprema:

The original integral uses |k| as the lower limit of integration (and
only k^2 appears within the integral).  So you can consider only k >=
0 and know that the result depends only on |k|, i.e. an even function
of k.

Your u-substitution (1) is a bit different than the one I had in mind
(see above), and I like it.  There is a way to evaluate (4) without
appealing to a handbook, but the "trick" requires stepping up to two
dimensions.

regards, mathtalk-ga

Hi mathtalk-ga:

You are right, and may I also thank you for the inspiring discussions.
 I was actually about to show how to do the integral when I realized
that you had posted a comment.  In any case, let me include the
derivation (at laest for the sake of completeness):

First, let

(1) I = \int_{x=0}^{inf} exp(-x^2) dx.

Then

(2) I^2 = \int_{y=0}^{inf} \inf_{x=0}^{inf} exp(-(x^2+y^2)) dx dy.

Use the following "polar" coordinate substitution:

(3) x^2+y^2 = r^2, x = r*cos(theta), y=r*sin(theta).

This yields

(4) I^2 = \int_{theta=0}^{pi/2} \int_{r=0}^{inf} exp(-r^2)*r dr dtheta,

where the "extra" r term on the right-hand-side is generated from the
corresponding Jacobian, i.e.,

(5) det([dx/dr dx/dtheta; dy/dr dy/dtheta]) = r.

So, let us get back to (4).  Integrate out theta to get 

(6) I^2 = (pi/2) \int_{r=0}^{inf} r*exp(-r^2) dr.

Substitute r^2 = x --> 2*r dr = dx:

(7) I^2 = pi/4 --> I = sqrt(pi)/2.

Remember that, what we wanted was an expression for 

(8) J = int_{x=0}^{inf} exp(-a^2*x^2) dx.

Well, with (7) in hand, this is quite easy when one uses the substitution 

(9) a*x = y with a > 0.

Note: It is essential that a > 0 to keep the integration sensible.

kprema