The Gamma Function and Its Derivative

Date: 05/26/98 at 12:39:33
From: Eric Geib
Subject: Derivative of n!

N! = the integral from 0 to infinity of (x ^ n)(e ^ -x) dx

Does this hold true for all real numbers and not just non-negative 
integers?

If so, does the derivative of n! with respect to n exist?

Date: 05/26/98 at 15:56:10
From: Doctor Rob
Subject: Re: Derivative of n!

The function given by the above integral is called the Gamma function.
Actually, you have written Gamma(n+1). It is true that for natural 
numbers n, Gamma(n+1) = n!. It is also true that Gamma is defined for 
all nonnegative real numbers. Furthermore, Gamma(n) is differentiable, 
so you could define the derivative of n! with respect to n to be the 
derivative of Gamma(n+1) with respect to n. 

Gamma can be extended to a function defined for all complex numbers 
except for 0 and negative integral values, at which points it has a 
pole. This function is also called the Gamma function, and it is also 
infinitely differentiable wherever it is defined.

-Doctor Rob,  The Math Forum
 http://mathforum.org/dr.math/   

Date: 05/26/98 at 16:39:47
From: Evan M Glazer
Subject: Re: Derivative of n!

Dr. Math,

Thanks for the quick response to my question. I have one more follow 
up. What is (and how would you determine) the derivative of the Gamma 
function?  Thanks!

-Eric Geib

Date: 05/27/98 at 17:02:42
From: Doctor Rob
Subject: Re: Derivative of n!

When the integral converges absolutely, as this one does, you can find 
the derivative by interchanging the integration and differentiation:

     Gamma(n) = Integral x^(n-1)*e^(-x) dx          from 0 to infinity

 dGamma(n)/dn = Integral d(x^[n-1]*e^[-x])/dn dx    from 0 to infinity                  
              = Integral d(e^[{n-1}*ln(x)-x])/dn dx from 0 to infinity
              = Integral x^(n-1)*e^(-x)*ln(x) dx    from 0 to infinity

In general, for any integer k > 0,

   d^k(Gamma[n])/dn^k = Integral x^(n-1)*e^(-x)*ln(x)^k dx  
                                                    from 0 to infinity

I am quite sure that this cannot be integrated in closed form in terms 
of the familiar functions of calculus, so it must be evaluated either 
by means of an infinite series or by numerical methods.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/