### 1.  Basic Facts about the Gamma Function

The Gamma function is defined by the improper integral

 Gamma(x)  =  INT_{0}^{INFTY}[t^{x} e^{-t} {dt}/{t} ]  .

The integral is absolutely convergent for x >= 1 since

 t^{x-1} e^{-t}  <=  e^{-t/2}  ,     t ≫ 1

and INT_{0}^{INFTY}[e^{-t/2} dt ] is convergent. The preceding inequality is valid, in fact, for all x. But for x < 1 the integrand becomes infinitely large as t approaches 0 through positive values. Nonetheless, the limit

 lim_{r  -->  0+} INT_{r}^{1}[t^{x-1} e^{-t} dt ]

exists for x > 0 since

 t^{x-1} e^{-t}  <=  t^{x-1}

for t > 0, and, therefore, the limiting value of the preceding integral is no larger than that of

 lim_{r  -->  0+} INT_{r}^{1}[t^{x-1} dt ]  =  {1}/{x}  .

Hence, Gamma(x) is defined by the first formula above for all values x > 0 .

If one integrates by parts the integral

 Gamma(x + 1)  =  INT_{0}^{INFTY}[t^{x} e^{-t} dt ]  ,

writing

 INT_{0}^{INFTY}[udv ]  =  u(INFTY)v(INFTY) - u(0)v(0) - INT_{0}^{INFTY}[vdu ]  ,

with dv = e^{-t}dt and u = t^{x}, one obtains the functional equation

 Gamma(x+1)  =  x Gamma(x)  ,   x > 0   .

Obviously, Gamma(1) = INT_{0}^{INFTY}[e^{-t} dt ] = 1 , and, therefore, Gamma(2) = 1 · Gamma(1) = 1, Gamma(3) = 2 · Gamma(2) = 2!, Gamma(4) = 3 Gamma(3) = 3!, …, and, finally,

 Gamma(n+1)  =  n!

for each integer n > 0.

Thus, the gamma function provides a way of giving a meaning to the “factorial” of any positive real number.

Another reason for interest in the gamma function is its relation to integrals that arise in the study of probability. The graph of the function varphi defined by

 varphi(x)  =  e^{-x^{2}}

is the famous “bell-shaped curve” of probability theory. It can be shown that the anti-derivatives of varphi are not expressible in terms of elementary functions. On the other hand,

 Phi(x)  =  INT_{-INFTY}^{x}[varphi(t) dt ]

is, by the fundamental theorem of calculus, an anti-derivative of varphi, and information about its values is useful. One finds that

 Phi(INFTY)  =  INT_{-INFTY}^{INFTY}[e^{-t^{2}} dt ]  =  Gamma(1/2)

by observing that

 INT_{-INFTY}^{INFTY}[e^{-t^{2}} dt ]  =  2 · INT_{0}^{INFTY}[e^{-t^{2}} dt ]  ,

and that upon making the substitution t = u^{1/2} in the latter integral, one obtains Gamma(1/2).

To have some idea of the size of Gamma(1/2), it will be useful to consider the qualitative nature of the graph of Gamma(x). For that one wants to know the derivative of Gamma.

By definition Gamma(x) is an integral (a definite integral with respect to the dummy variable t) of a function of x and t. Intuition suggests that one ought to be able to find the derivative of Gamma(x) by taking the integral (with respect to t) of the derivative with respect to x of the integrand. Unfortunately, there are examples where this fails to be correct; on the other hand, it is correct in most situations where one is inclined to do it. The methods required to justify “differentiation under the integral sign” will be regarded as slightly beyond the scope of this course. A similar stance will be adopted also for differentiation of the sum of a convergent infinite series.

Since

 {d}/{dx} t^{x}  =  t^{x}(log t)  ,

one finds

 {d}/{dx} Gamma(x)  =  INT_{0}^{INFTY}[t^{x} (log t) e^{-t} {dt}/{t} ] ,

and, differentiating again,

 {d^{2}}/{dx^{2}} Gamma(x)  =   INT_{0}^{INFTY}[t^{x} (log t)^{2} e^{-t} {dt}/{t} ]  .

One observes that in the integrals for both Gamma and the second derivative Gamma^{″} the integrand is always positive. Consequently, one has Gamma(x) > 0 and Gamma^{″}(x) > 0 for all x > 0. This means that the derivative Gamma^{′} of Gamma is a strictly increasing function; one would like to know where it becomes positive.

If one differentiates the functional equation

 Gamma(x+1)  =  x Gamma(x)  ,   x > 0  ,

one finds

 psi(x+1)  =  {1}/{x} + psi(x)  ,   x > 0  ,

where

 psi(x)  =  {d}/{dx} logGamma(x)  =  {Gamma^{′}(x)}/{Gamma(x)}  ,

and, consequently,

 psi(n+1)  =  psi(1) + SUM_{k = 0}^{n}[{1}/{k} ] .

Since the harmonic series diverges, its partial sum in the foregoing line approaches INFTY as x --> INFTY. Inasmuch as Gamma^{′}(x) = psi(x)Gamma(x), it is clear that Gamma^{′} approaches INFTY as x --> INFTY since Gamma^{′} is steadily increasing and its integer values (n-1)!psi(n) approach INFTY. Because 2 = Gamma(3) > 1 = Gamma(2), it follows that Gamma^{′} cannot be negative everywhere in the interval 2 <= x <= 3, and, therefore, since Gamma^{′} is increasing, Gamma^{′} must be always positive for x >= 3. As a result, Gamma must be increasing for x >= 3, and, since Gamma(n + 1) = n!, one sees that Gamma(x) approaches INFTY as x --> INFTY.

It is also the case that Gamma(x) approaches INFTY as x --> 0. To see the convergence one observes that the integral from 0 to INFTY defining Gamma(x) is greater than the integral from 0 to 1 of the same integrand. Since e^{-t} >= 1/e for 0 <= t <= 1, one has

Gamma(x)>INT_{0}^{1}[(1/e)t^{x-1}dt]  =   (1/e)
 [ {t^{x}}/{x} ]
_{t = 0}^{t = 1} = {1}/{ex} .

It then follows from the mean value theorem combined with the fact that Gamma^{′} always increases that Gamma^{′}(x) approaches -INFTY as x --> 0.

Hence, there is a unique number c > 0 for which Gamma^{′}(c) = 0, and Gamma decreases steadily from INFTY to the minimum value Gamma(c) as x varies from 0 to c and then increases to INFTY as x varies from c to INFTY. Since Gamma(1) = 1 = Gamma(2), the number c must lie in the interval from 1 to 2 and the minimum value Gamma(c) must be less than 1.

 Figure 1: Graph of the Gamma Function

Thus, the graph of Gamma (see Figure 1) is concave upward and lies entirely in the first quadrant of the plane. It has the y-axis as a vertical asymptote. It falls steadily for 0 < x < c to a postive minimum value Gamma(c) < 1. For x > c the graph rises rapidly.

### 2.  Product Formulas

It will be recalled, as one may show using l'Hôpital's Rule, that

e^{-t}  =  \underset{n  -->  INFTY}{lim}
 ( 1-{t}/{n} )
^{n}  .

From the original formula for Gamma(x), using an interchange of limits that in a more careful exposition would receive further comment, one has

 Gamma(x)  =  \underset{n  -->  INFTY}{lim}  Gamma(x,n)  ,

where Gamma(x,n) is defined by

Gamma(x,n) = INT_{0}^{n}[t^{x-1}
 ( 1-{t}/{n} )
^{n} dt] , n >=  1 .

The substitution in which t is replaced by nt leads to the formula

 Gamma(x,n)  =  n^{x} INT_{0}^{1}[t^{x-1} (1 - t)^{n} dt]  .

This integral for Gamma(x,n) is amenable to integration by parts. One finds thereby:

Gamma(x,n) = {1}/{x}
 ( {n}/{n-1} )
^{x+1}Gamma(x+1,n-1) ,n  >=  2  .

For the smallest value of n, n = 1 , integration by parts yields:

 Gamma(x,1)  =  {1}/{x(x+1)}  .

Iterating n-1 times, one obtains:

 Gamma(x,n)  =  n^{x} {n!}/{x(x+1)(x+2)⋯(x+n)}  ,  n  >=  1  .

Thus, one arrives at the formula

 Gamma(x)  =  \underset{n  -->  INFTY}{lim}  n^{x} {n!}/{x(x+1)(x+2)⋯(x+n)}  .

This last formula is not exactly in the form of an infinite product

 PROD_{k = 1}^{INFTY}[p_{k} ]  =  \underset{n  -->  INFTY}{lim} {PROD_{k = 1}^{n}[p_{k}]}  .

But a simple trick enables one to maneuver it into such an infinite product. One writes n as a “collapsing product”:

 n+1 = {n+1}/{n}·{n}/{n-1}·  ⋯ ·{3}/{2}·{2}/{1}

or

n+1  =  PROD_{k = 1}^{n}[
 ( 1 + {1}/{k} )
]  ,

and, taking the xth power, one has

(n+1)^{x}  =  PROD_{k = 1}^{n}[
 ( 1 + {1}/{k} )
^{x}]    .

Since

 lim_{n  -->  INFTY}{n^{x}}/{(n+1)^{x}}  =  1  ,

one may replace the factor n^{x} by (n+1)^{x} in the last expression above for Gamma(x) to obtain

Gamma(x)  =  {1}/{x}lim_{n  -->  INFTY}{PROD_{k = 1}^{n}[{
 ( 1 + {1}/{k} )
^{x}}/{
 ( 1 + {x}/{k} )
}]}  ,

or

Gamma(x) = {1}/{x}{PROD_{k = 1}^{INFTY}[{
 ( 1 + {1}/{k} )
^{x}}/{
 ( 1 + {x}/{k} )
}]} .

The convergence of this infinite product for Gamma(x) when x > 0 is a consequence, through the various maneuvers performed, of the convergence of the original improper integral defining Gamma(x) for x > 0.

It is now possible to represent logGamma(x) as the sum of an infinite series by taking the logarithm of the infinite product formula. But first it must be noted that

 {(1+t)^{r}}/{1 + rt} > 0   for  t > 0  ,   r > 0  .

Hence, the logarithm of each term in the preceding infinite product is defined when x > 0.

Taking the logarithm of the infinite product one finds:

 log Gamma(x)  =  - log x + SUM_{k = 1}^{INFTY}[u_{k}(x)]  ,

where

u_{k}(x)  =  xlog
 ( 1 + {1}/{k} )
-log
 ( 1 + {x}/{k} )
.

It is, in fact, almost true that this series converges absolutely for all real values of x. The only problem with non-positive values of x lies in the fact that log(x) is meaningful only for x > 0, and, therefore, log(1+x/k) is meaningful only for k > |x|. For fixed x, if one excludes the finite set of terms u_{k}(x) for which k <= |x|, then the remaining “tail” of the series is meaningful and is absolutely convergent. To see this one applies the “ratio comparison test” which says that an infinite series converges absolutely if the ratio of the absolute value of its general term to the general term of a convergent positive series exists and is finite. For this one may take as the “test series”, the series

 SUM_{k = 1}^{INFTY}[{1}/{k^{2}}]  .

Now as k approaches INFTY, t = 1/k approaches 0, and so

 lim_{k  -->  INFTY}{u_{k}(x)}/{1/k^{2}}
 =
 lim_{t  -->  0}{xlog(1+t)-log(1+xt)}/{t^{2}}
 =
 lim_{t  -->  0}{{x}/{1+t}-{x}/{1+xt}}/{2t}
 =
 lim_{t  -->  0}{x[(1+xt)-(1+t)]}/{2t(1+t)(1+xt)}
 =
 {x(x-1)}/{2}    .

Hence, the limit of |u_{k}(x)/k^{-2}| is |x(x-1)/2|, and the series SUM[u_{k}(x)] is absolutely convergent for all real x. The absolute convergence of this series foreshadows the possibility of defining Gamma(x) for all real values of x other than non-positive integers. This may be done, for example, by using the functional equation

 Gamma(x+1)  =  x Gamma(x)

or

 Gamma(x)  =  {1}/{x} Gamma(x + 1)

to define Gamma(x) for -1 < x < 0 and from there to -2 < x < -1, etc.

Taking the derivative of the series for logGamma(x) term-by-term – once again a step that would receive justification in a more careful treatment – and recalling the previous notation psi(x) for the derivative of logGamma(x), one obtains

 psi(x) + {1}/{x}
 =
SUM_{k = 1}^{INFTY}[
{log
 ( 1 + {1}/{k} )
-{{1}/{k}}/{
 ( 1 + {x}/{k} )
}}
]
 =
\underset{n  -->  INFTY}{lim} SUM_{k = 1}^{n}[
 { log{k+1}/{k}-{1}/{x+k} }
]
 =
\underset{n  -->  INFTY}{lim}
 { log(n+1)-SUM_{k = 1}^{n}[{1}/{x+k}] }
 =
\underset{n  -->  INFTY}{lim}
{log(n+1)-SUM_{k = 1}^{n}[{1}/{k}]  +SUM_{k = 1}^{n}[
 ( {1}/{k}-{1}/{x+k} )
]}
 =
\underset{n  -->  INFTY}{lim}
 { log(n+1)-SUM_{k = 1}^{n}[{1}/{k}]  +xSUM_{k = 1}^{n}[{1}/{k(x+k)}] }
 =
 -gamma + xSUM_{k = 1}^{INFTY}[{1}/{k(x+k)}]  ,

where gamma denotes Euler's constant

gamma  =  \underset{n  -->  INFTY}{lim}
 ( SUM_{k = 1}^{n}[{1}/{k}] - log n )
.

When x = 1 one has

 psi(1)  =  -1 - gamma + SUM_{k = 1}^{INFTY}[{1}/{k(k+1)}]  ,

and since

 {1}/{k(k+1)}  =  {1}/{k} - {1}/{k+1}  ,

this series collapses and, therefore, is easily seen to sum to 1. Hence,

 psi(1)  =  - gamma  ,   psi(2)  =  psi(1) + 1/1  =  1 - gamma  .

Since Gamma^{′}(x) = psi(x)Gamma(x), one finds:

 Gamma^{′}(1)  =  - gamma  ,

and

 Gamma^{′}(2)  =  1 - gamma  .

These course notes were prepared while consulting standard references in the subject, which included those that follow.

### REFERENCES

[1]   R. Courant, Differential and Integral Calculus (2 volumes), English translation by E. J. McShane, Interscience Publishers, New York, 1961.
[2]   E. T. Whittaker & G. N. Watson, A Course of Modern Analysis, 4th edition, Cambridge University Press, 1969.
[3]   David Widder, Advanced Calculus, 2nd edition, Prentice Hall, 1961.