About the Gamma FunctionNotes for Honors Calculus II, Originally Prepared in Spring 19951Basic Facts about the Gamma FunctionThe Gamma function is defined by the improper integral
x0txetdtt.
The integral is absolutely convergent for x1 since
tx1etet2,t1
and 0et2dt is convergent The preceding
inequality is valid, in fact, for all x But for x1
the integrand becomes infinitely large as t approaches 0 through
positive values Nonetheless, the limit
limr0r1tx1etdt
exists for x0 since
tx1ettx1
for t0, and, therefore, the limiting value of the preceding integral
is no larger than that of
limr0r1tx1dt1x.
Hence, x is defined by the
first formula above for all values x0If one integrates by parts the integral
x10txetdt,
writing
0udvuvu0v00vdu,
with dvetdt and utx, one obtains the functional
equationx1xx,x0.Obviously, 10etdt1, and, therefore,
2111, 3222, 4333, , and, finally,
n1n
for each integer n0Thus, 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
xex2
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,
xxtdt
is, by the fundamental theorem of calculus, an antiderivative of
, and information about its values is useful
One finds that
et2dt12
by observing that
et2dt20et2dt,
and that upon making the substitution tu12 in the latter
integral, one obtains 12To 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
ddxtxtxlogt,
one finds
ddxx0txlogtetdtt,
and, differentiating again,
d2dx2x0txlogt2etdtt.
One observes that in the integrals for both and the second
derivative the integrand is always positive Consequently,
one has x0 and x0 for all x0 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
x1xx,x0,
one finds
x11xx,x0,
where
xddxlogxxx,
and, consequently,
n11k0n1k.
Since the harmonic series diverges, its partial sum in the foregoing
line approaches as x Inasmuch as
xxx, it is clear that approaches
as x since is steadily increasing
and its integer values n1n approach
Because 2312,
it follows that cannot be negative everywhere in the interval
2x3, and, therefore, since is increasing,
must be always positive for x3 As a result, must be
increasing for x3, and, since n1n, one sees
that x approaches as xIt is also the case that x approaches as
x0 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 et1e for
0t1, one has
x011etx1dt1etxxt0t11ex.
It then follows from the mean value theorem combined with the fact that
always increases that x approaches
as x0Hence, there is a unique number c0 for which c0,
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
112, 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 0xc
to a postive minimum value c1 For xc the graph
rises rapidly2Product FormulasIt will be recalled, as one may show using lHopitals Rule,
that
etnlim1tnn.
From the original formula for x, using an interchange of limits
that in a more careful exposition would receive further comment, one has
xnlimx,n,
where x,n is defined by
x,n0ntx11tnndt,n1.
The substitution in which t is replaced by ntleads to
the formula
x,nnx01tx11tndt.
This integral for x,n is amenable to integration by parts
One finds thereby:
x,n1xnn1x1x1,n1,n2.
For the smallest value of n, n1, integration by parts yields:
x,11xx1.
Iterating n1 times, one obtains:
x,nnxnxx1x2xn,n1.
Thus, one arrives at the formula
xnlimnxnxx1x2xn.This last formula is not exactly in the form of an infinite product
k1pknlimk1npk.
But a simple trick enables one to maneuver it into such an infinite
product One writes n as a collapsing product:
n1n1nnn13221
or
n1k1n11k
and, taking the xth power, one has
n1xk1n11kx
Since
limnnxn1x1
one may replace the factor nx by n1x in the last expression above
for x to obtain
x1xlimnk1n11kx1xk,
or
x1xk111kx1xk.The convergence of this infinite product for x when x0
is a consequence, through the various maneuvers performed, of the
convergence of the original improper integral defining x for
x0It is now possible to represent logx as the sum of an infinite
series by taking the logarithm of the infinite product formula But first
it must be noted that
1tr1rt0fort0,r0.
Hence, the logarithm of each term in the preceding infinite product is
defined when x0Taking the logarithm of the infinite product one finds:
logxlogxk1ukx,
where
ukxxlog11klog1xk.
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 logx is meaningful only for
x0, and, therefore, log1xk is meaningful only for
kx For fixed x, if one excludes the finite set of terms
ukx for which kx, 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
k11k2.
Now as kapproaches , t1kapproaches 0,
and so
limkukx1k2limt0xlog1tlog1xtt2limt0x1tx1xt2tlimt0x1xt1t2t1t1xtxx12
Hence, the limit of ukxk2 is xx12, and the series
ukx 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
x1xx
or
x1xx1
to define x for 1x0 and from there
to 2x1, etcTaking the derivative of the series for logx termbyterm
once again a step that would receive justification in a more careful
treatment and recalling the previous notation x for the
derivative of logx, one obtains
x1xk1log11k1k1xknlimk1nlogk1k1xknlimlogn1k1n1xknlimlogn1k1n1kk1n1k1xknlimlogn1k1n1kxk1n1kxkxk11kxk,
where denotes Eulers constant
nlimk1n1klogn.When x1 one has
11k11kk1,
and since
1kk11k1k1,
this series collapses and, therefore, is easily seen to sum to 1
Hence,
1,21111.
Since xxx, one finds:
1,
and
21.These course notes were prepared while consulting standard references
in the subject, which included those that follow1courant R.Courant, Differential and Integral
Calculus (2 volumes), English translation by E.J. McShane,
Interscience Publishers, New York, 19612whittakerWatson E.T. Whittaker G.N. Watson,
ACourse of Modern Analysis, 4th edition, Cambridge University
Press, 19693widder David Widder, Advanced Calculus, 2nd edition,
Prentice Hall, 1961