(z + n) 2. n=0. 2 (z + 2) (z + n). 2 We d like to compute this as an integral. To this end, we d like a function with residues

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1 8. Stirling Formula Stirling s Formula is a classical formula to compute n! accurately when n is large. We will derive a version of Stirling s formula using complex analysis and residues. Recall the formula for the second logarithmic derivative of the gamma function: d d ( ) Γ () = Γ() Let s start with the partial sum n= ( + n). + ( + ) + ( + ) + + ( + n). We d like to compute this as an integral. To this end, we d like a function with residues ( + ν) at the integral points ν. We pick π cot πw Φ(w) = ( + w). w = u + iv is the variable of integration. For the time being we keep = x + iy fixed and we assume that x >. We integrate around the rectangle whose vertical sides are u = and u = n + / and whose horiontal sides are v = ±Y. Call this contour K. Note that this contour contains a pole of Φ(w), at w =, so that we need to take the principal value: pr. v. πi K Φ(w) = + n m= ( + m). We will first let Y go to infinity and then n. Note that cot πw tends uniformly to ±i as v goes to. Since /( + w) goes to ero the integral over the horiontal sides goes to ero. Over the infinite vertical line u = n + /, cot πw is bounded, and by periodicity this bound is independent of n. The integral over the line u = n + / is therefore at most a constant multiple of w +. u=n+/ Note that on the line u = n + / we have w = n + w

2 and so i u=n+/ w + = i u=n+/ (w + )(n + w + ). We can calculate the last integral using residues. The poles are at w = and w = n + +. Only the first = w is to the left of the line u = n + /. The residue at w = is n + + x. The integral is therefore π n + + x. This goes to ero as n goes to infinity. It remains to deal with the principal value of the integral over the imaginary axis. [ ] cot πiv (iv + ) v dv = coth πv (iv ) (v + ) Putting all of this together we get ( ) d Γ () = d Γ() + We use the expression coth πv = + e πv. Now the integral from the first term is v (v + ) dv = v coth πv (v + ) [ (v + ) ] =. Thus ( ) d Γ () = d Γ() + + 4v dv (v + ) e πv, where the integral is strongly very convergent. If we restrict to the right half plane we can integrate this formula. Γ () Γ() = C + log + v (v + ) dv e πv, where log is the principal branch and C is a constant. Note that if we restrict to a compact subset of the right half plane then the integral converges uniformly and so we are allowed to differentiate under the integral sign.

3 We want to integrate once more. If we did this directly we would get tan(/v) which is multi-valued. Instead we integrate by parts first, v (v + ) dv e πv = π v (v + ) log( e πv ) Now we integrate with respect to again to get log Γ() = C + C + ( ) log + π (v + ) log e πv Here C is a new constant of integration and C has been replaced by C. This formula means that log Γ() is single-valued on the right half plane and given by the expression on the right. We choose C so that the LHS is real on the real axis. To determine the constants C and C we need to consider the integral π (v + ) log e πv We first check that J() tends to ero as goes to infinity but stays away from the imaginary axis. Suppose that we restrict to the half plane x c >. We break the integral into two parts In the first integral J() = + = J + J. v + / = 3 4 and so J 4 log 3π e πv In the second integral and so v + = iv + iv > c J < πc / log e πv Since the integral of log is convergent J e πv and J tend to ero as goes to infinity. To determine C we use the functional equation Γ( + ) = Γ() or log Γ( + ) = log + log Γ(), which is valid provided we stay in the right half plane. Hence C +C+C +(+ ) log(+)+j(+) = C +C+(+ ) log +J() 3

4 and so C = ( + ) ( log + ) + J() J( + ). Letting go to infinity we get C =. Using the other functional equation Γ()Γ( ) = (with some work!) that C = log π. Thus π sin π we can deduce log Γ() = log π + ( /) log + J(). Equivalently Γ() = π / e e J(). This is Stirling s formula. As J() approaches as approaches infinity we can use Stirling s formula to estimate n! for large values of n. Theorem 8.. for any x >. Proof. Let Γ() = F () = Note that, integrating by parts, F ( + ) = e t t dt = e t t dt e t t dt. e t t dt = F (). It follows that F () is in fact an entire function and that F ( + ) Γ( + ) = F () Γ(), so that the ratio is periodic with period one. The goal is to apply Liouville s theorem to the ratio. The key point is to bound the absolute value of the ratio F/Γ on some strip, say on x. For a start F () e t t x dt = F (x), so that F () is bounded in the strip. Now we apply Stirling s formula to find a lower bound for Γ() : log Γ() = log π x + (x /) log y arg + Re J(). 4

5 The only term that bothers us is y arg which can go to negative infinity, comparable to π y /. Thus F/Γ grows no faster than e π y /. Note that F/Γ is a function of the variable q = e πi. It has singularities at q = and q = but F/Γ grows at most like q / as q approaches ero and q / as q approaches infinity. Thus the singularities there are removable and F/Γ is constant by Liouville. Since F () = Γ() = it follows that F () = Γ(). 5