About the Gamma FunctionNotes for Honors Calculus II, Originally Prepared in Spring 1995Basic Facts about the Gamma FunctionThe Gamma function is defined by the improper integral
(x) _{0}^{}t^{x} e^{t} dtt .
The integral is absolutely convergent for x 1 since
t^{x1} e^{t} e^{t2} , t 1
and _{0}^{}e^{t2} 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} _{r}^{1}t^{x1} e^{t} dt
exists for x 0 since
t^{x1} e^{t} t^{x1}
for t 0, and, therefore, the limiting value of the preceding integral
is no larger than that of
lim_{r 0} _{r}^{1}t^{x1} dt 1x .
Hence, (x) is defined by the
first formula above for all values x 0 If one integrates by parts the integral
(x 1) _{0}^{}t^{x} e^{t} dt ,
writing
_{0}^{}udv u()v() u(0)v(0) _{0}^{}vdu ,
with dv e^{t}dt and u t^{x}, one obtains the functional
equation
(x1) x (x) , x 0 . Obviously, (1) _{0}^{}e^{t} dt 1 , and, therefore,
(2) 1 (1) 1,
(3) 2 (2) 2,
(4) 3 (3) 3, , and, finally,
(n1) n
for each integer n 0Thus, the gamma function provides a way of giving a meaning to the
factorial of any positive real numberAnother reason for interest in the gamma function is its relation
to integrals that arise in the study of probability The graph of the
function defined by
(x) e^{x2}
is the famous bellshaped curve of probability theory It can be
shown that the antiderivatives of are not expressible in
terms of elementary functions. On the other hand,
(x) _{}^{x}(t) dt
is, by the fundamental theorem of calculus, an antiderivative of
, and information about its values is useful
One finds that
() _{}^{}e^{t2} dt (12)
by observing that
_{}^{}e^{t2} dt 2 _{0}^{}e^{t2} dt ,
and that upon making the substitution tu^{12} in the latter
integral, one obtains (12)To have some idea of the size of (12), it will be useful
to consider the qualitative nature of the graph of (x)
For that one wants to know the derivative of By definition (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 (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 seriesSince
ddx t^{x} t^{x}(log t) ,
one finds
ddx (x) _{0}^{}t^{x} (log t) e^{t} dtt ,
and, differentiating again,
d^{2}dx^{2} (x)
_{0}^{}t^{x} (log t)^{2} e^{t} dtt .
One observes that in the integrals for both and the second
derivative the integrand is always positive Consequently,
one has (x) 0 and (x) 0 for all x 0 This
means that the derivative of is a strictly increasing
function; one would like to know where it becomes positiveIf one differentiates the functional equation
(x1) x (x) , x 0 ,
one finds
(x1) 1x (x) , x 0 ,
where
(x) ddx log(x) (x)(x) ,
and, consequently,
(n1) (1) _{k0}^{n}1k .
Since the harmonic series diverges, its partial sum in the foregoing
line approaches as x Inasmuch as
(x) (x)(x), it is clear that approaches
as x since is steadily increasing
and its integer values (n1)(n) approach
Because 2 (3) 1 (2),
it follows that cannot be negative everywhere in the interval
2 x 3, and, therefore, since is increasing,
must be always positive for x 3 As a result, must be
increasing for x 3, and, since (n 1) n, one sees
that (x) approaches as x It is also the case that (x) approaches as
x 0 To see the convergence one observes that the
integral from
0 to defining (x) is greater than the integral from
0 to 1 of the same integrand Since e^{t} 1e for
0 t 1, one has
(x)_{0}^{1}(1e)t^{x1}dt
(1e)t^{x}x_{t0}^{t1}1ex.
It then follows from the mean value theorem combined with the fact that
always increases that (x) approaches
as x 0Hence, there is a unique number c 0 for which (c) 0,
and decreases steadily from to the minimum value
(c) as x varies from 0 to c and then increases to
as x varies from c to Since
(1) 1 (2), the number c must lie in the interval
from 1 to 2 and the minimum value (c) must be less than 1grmplgamma Figure1: Graph of the Gamma FunctionThus, the graph of (see Figure1) is concave upward
and lies entirely in the first quadrant of the plane It has the
yaxis as a vertical asymptote It falls steadily for 0 x c
to a postive minimum value (c) 1 For x c the graph
rises rapidly
It will be recalled, as one may show using lHopitals Rule,
that
e^{t} n lim 1tn^{n} .
From the original formula for (x), using an interchange of limits
that in a more careful exposition would receive further comment, one has
(x) n lim (x,n) ,
where (x,n) is defined by
(x,n)_{0}^{n}t^{x1}1tn^{n} dt,n 1.
The substitution in which t is replaced by nt leads to
the formula
(x,n) n^{x} _{0}^{1}t^{x1} (1 t)^{n} dt .
This integral for (x,n) is amenable to integration by parts
One finds thereby:
(x,n)1xnn1^{x1}(x1,n1),n 2 .
For the smallest value of n, n 1 , integration by parts yields:
(x,1) 1x(x1) .
Iterating n1 times, one obtains:
(x,n) n^{x} nx(x1)(x2)(xn) , n 1 .
Thus, one arrives at the formula
(x) n lim n^{x} nx(x1)(x2)(xn) . This last formula is not exactly in the form of an infinite product
_{k1}^{}p_{k} n lim_{k1}^{n}p_{k} .
But a simple trick enables one to maneuver it into such an infinite
product One writes n as a collapsing product:
n1n1nnn1
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or
n1 _{k1}^{n}1 1k
and, taking the xth power, one has
(n1)^{x} _{k1}^{n}1 1k^{x}
Since
lim_{n }n^{x}(n1)^{x} 1
one may replace the factor n^{x} by (n1)^{x} in the last expression above
for (x) to obtain
(x) 1xlim_{n }_{k1}^{n}1 1k^{x}1 xk ,
or
(x)1x_{k1}^{}1 1k^{x}1 xk.The convergence of this infinite product for (x) when x 0
is a consequence, through the various maneuvers performed, of the
convergence of the original improper integral defining (x) for
x 0It is now possible to represent log(x) as the sum of an infinite
series by taking the logarithm of the infinite product formula But first
it must be noted that
(1t)^{r}1 rt 0 for t 0 , r 0 .
Hence, the logarithm of each term in the preceding infinite product is
defined when x 0Taking the logarithm of the infinite product one finds:
log (x) log x _{k1}^{}u_{k}(x) ,
where
u_{k}(x) xlog1 1klog1 xk .
It is, in fact, almost true that this series converges absolutely for
all real values of x The only problem with nonpositive
values of x lies in the fact that log(x) is meaningful only for
x 0, and, therefore, log(1xk) 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
_{k1}^{}1k^{2} .
Now as k approaches , t 1k approaches 0,
and so
lim_{k }u_{k}(x)1k^{2} lim_{t 0}xlog(1t)log(1xt)t^{2} lim_{t 0}x1tx1xt2t lim_{t 0}x[(1xt)(1t)]2t(1t)(1xt) x(x1)2
Hence, the limit of u_{k}(x)k^{2} is x(x1)2, and the series
u_{k}(x) is absolutely convergent for all real x The absolute
convergence of this series foreshadows the possibility of defining
(x) for all real values of x other than nonpositive integers
This may be done, for example, by using the functional equation
(x1) x (x)
or
(x) 1x (x 1)
to define (x) for 1 x 0 and from there
to 2 x 1, etcTaking the derivative of the series for log(x) termbyterm
once again a step that would receive justification in a more careful
treatment and recalling the previous notation (x) for the
derivative of log(x), one obtains
(x) 1x
_{k1}^{}log1 1k
1k1 xk n lim_{k1}^{n}logk1k1xk n limlog(n1)_{k1}^{n}1xk n limlog(n1)_{k1}^{n}1k
_{k1}^{n}1k1xk n limlog(n1)_{k1}^{n}1k
x_{k1}^{n}1k(xk) x_{k1}^{}1k(xk) ,
where denotes Eulers constant
n lim_{k1}^{n}1k log n . When x 1 one has
(1) 1 _{k1}^{}1k(k1) ,
and since
1k(k1) 1k 1k1 ,
this series collapses and, therefore, is easily seen to sum to 1
Hence,
(1) , (2) (1) 11 1 .
Since (x) (x)(x), one finds:
(1) ,
and
(2) 1 .
These course notes were prepared while consulting standard references
in the subject, which included those that follow
courant R.Courant, Differential and Integral
Calculus (2 volumes), English translation by E.J. McShane,
Interscience Publishers, New York, 1961
whittakerWatson E.T. Whittaker G.N. Watson,
ACourse of Modern Analysis, 4th edition, Cambridge University
Press, 1969
widder David Widder, Advanced Calculus, 2nd edition,
Prentice Hall, 1961