By Hall E. H.

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_+ ~f:L 2p. e-3pt/2{s' + Pei('l/3/2)1+ s2+ Pe-t('l/3/2)1} 2p. p. 3 sr s2 . ;3 2 v3cos (ptv3)} -- . 2 (9) The moments of the number of renewals 51 For the a-stage Erlangian distribution we have to consider the a- 1 non-zero roots of (s + p)a = pa and these are p{exp(27Tki/a)-1} (k = l, ... ,a-1). 3 (iii). The line of integration is converted into a closed contour by a large semicircle to the left of the imaginary axis, and the various terms in Ho(t) arise from the poles of H~(s) at s = 0, s" s2, •••• This argument extends with little difficulty if H~(s) has an infinite number of poles in the negative half-plane.

3 (iii). The line of integration is converted into a closed contour by a large semicircle to the left of the imaginary axis, and the various terms in Ho(t) arise from the poles of H~(s) at s = 0, s" s2, •••• This argument extends with little difficulty if H~(s) has an infinite number of poles in the negative half-plane. The partial fraction expansion in a finite series is replaced by an infinite series, the so-called Mittag-Leffler series, and provided some convergence conditions are satisfied, the only change in Ho(t) is that the finite series of exponential terms becomes an infinite series.

Where A. = 1IE(T). f. ofT is a linear combination of exponential terms. 44 Renewal theory Another simple special case occurs when the underlying renewal process is a Poisson process,/*(s) = p/(p + s). k (7) a negative binomial distribution. More generally, if the distribution of failure-time is of the special Erlangian type with a stages, fairly simple results can be obtained from (5) provided that k and a are both rather small. 1. The renewal function In the present chapter we consider the moments, and especially the mean value, of N 1, the number of renewals in the time interval (0, t).