%ComputedEntitiesFile; ]>
About the Gamma FunctionNotes for Honors Calculus II, Originally Prepared in Spring 1995
&SecRef-1;Basic Facts about the Gamma FunctionThe Gamma function is defined by the improper integral (x) 0tx et dtt . The integral is absolutely convergent for x 1 since tx1 et et2 , t 1 and 0et2 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 limr 0 r1tx1 et dt exists for x 0 since tx1 et tx1 for t 0, and, therefore, the limiting value of the preceding integral is no larger than that of limr 0 r1tx1 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) 0tx et dt , writing 0udv u()v() u(0)v(0) 0vdu , with dv etdt and u tx, one obtains the functional equation (x1) x (x) , x 0 . Obviously, (1) 0et 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) ex2 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 () et2 dt (12) by observing that et2 dt 2 0et2 dt , and that upon making the substitution tu12 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 tx tx(log t) , one finds ddx (x) 0tx (log t) et dtt , and, differentiating again, d2dx2 (x) 0tx (log t)2 et 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) k0n1k . 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 et 1e for 0 t 1, one has (x)01(1e)tx1dt (1e)txxt0t11ex. 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
&SecRef-2;Product FormulasIt will be recalled, as one may show using lHopitals Rule, that et n lim 1tnn . 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)0ntx11tnn dt,n 1. The substitution in which t is replaced by nt leads to the formula (x,n) nx 01tx1 (1 t)n dt . This integral for (x,n) is amenable to integration by parts One finds thereby: (x,n)1xnn1x1(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) nx nx(x1)(x2)(xn) , n 1 . Thus, one arrives at the formula (x) n lim nx nx(x1)(x2)(xn) . This last formula is not exactly in the form of an infinite product k1pk n limk1npk . But a simple trick enables one to maneuver it into such an infinite product One writes n as a collapsing product: n1n1nnn1 3221 or n1 k1n1 1k and, taking the xth power, one has (n1)x k1n1 1kx Since limn nx(n1)x 1 one may replace the factor nx by (n1)x in the last expression above for (x) to obtain (x) 1xlimn k1n1 1kx1 xk , or (x)1xk11 1kx1 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)r1 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 k1uk(x) , where uk(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 uk(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 k11k2 . Now as k approaches , t 1k approaches 0, and so limk uk(x)1k2 limt 0xlog(1t)log(1xt)t2 limt 0x1tx1xt2t limt 0x[(1xt)(1t)]2t(1t)(1xt) x(x1)2 Hence, the limit of uk(x)k2 is x(x1)2, and the series uk(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 k1log1 1k 1k1 xk n limk1nlogk1k1xk n limlog(n1)k1n1xk n limlog(n1)k1n1k k1n1k1xk n limlog(n1)k1n1k xk1n1k(xk) xk11k(xk) , where denotes Eulers constant n limk1n1k log n . When x 1 one has (1) 1 k11k(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
&BibRef-courant;courant R.Courant, Differential and Integral Calculus (2 volumes), English translation by E.J. McShane, Interscience Publishers, New York, 1961 &BibRef-whittakerWatson;whittakerWatson E.T. Whittaker G.N. Watson, ACourse of Modern Analysis, 4th edition, Cambridge University Press, 1969 &BibRef-widder;widder David Widder, Advanced Calculus, 2nd edition, Prentice Hall, 1961