About the Gamma FunctionNotes for Honors Calculus II, Originally Prepared in Spring 1995
1Basic 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 equation x1xx,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 rapidly
2Product 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 nt leads 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 k approaches , t1k approaches 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 follow
1courant R.Courant, Differential and Integral Calculus (2 volumes), English translation by E.J. McShane, Interscience Publishers, New York, 1961 2whittakerWatson E.T. Whittaker G.N. Watson, ACourse of Modern Analysis, 4th edition, Cambridge University Press, 1969 3widder David Widder, Advanced Calculus, 2nd edition, Prentice Hall, 1961