From: cdavenport@higher-me.com (Craig Davenport) Subject: Re: Analytic Function? Date: Tue, 15 Aug 2000 05:49:06 GMT Newsgroups: sci.math Summary: [missing] On 14 Aug 2000 daryl@cogentex.com (Daryl McCullough) wrote: >Consider the following function of x: > f(z) = integral from 0 to infinity of exp(-t)/(1 + zt) dt >I believe that f(z) is well-defined for any complex value of >z except when z is real and negative. Does anyone know if f(x) >can be analytically continued to the entire complex plane >(except for isolated singularities)? The interesting thing >about this function is that if you try to expand it in a >Taylor series, you get something wildly divergent: > f(z) = sum from j=0 to infinity of (-z)^j j! This is one of the standard examples of divergent series, treated in detail by Laguerre in 1879, and discussed in just about every survey of divergent series. The function exp(-t)/(1+zt) is indeed analytic for all complex z except real and negative values. I don't believe it can be analytically extended any further. Of course, you are taking liberties by assigning the label "f(z)" both to this function and to the divergent series. We should really put the word "sum" in quotation marks when referring to a valuation of a divergent series. ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Analytic Function? Date: Tue, 15 Aug 2000 14:44:06 GMT Newsgroups: sci.math On 14 Aug 2000 09:19:05 -0700, daryl@cogentex.com (Daryl McCullough) wrote: >Consider the following function of x: > > f(z) = integral from 0 to infinity of exp(-t)/(1 + zt) dt > >I believe that f(z) is well-defined for any complex value of >z except when z is real and negative. Does anyone know if f(x) >can be analytically continued to the entire complex plane >(except for isolated singularities)? Nope. The function is not even continuous near any point of the negative real axis. Take g(z) = z * f(1/z) and you get a formula that looks a little more familiar. If you fix an x < 0 you can show that Im(g(x + iy)) has two different limits as y tends to 0 from above or from below. (Looks to me like Im(g(x+iy)) = - integral_0^infinity (exp(-t) * y/((x+t)^2+y^2) dt) You may or may not recognize the y/((x+t)^2+y^2) as the Poisson kernel - in any case it's easily seen to be an approximate identity (except for a factor of Pi) so for x < 0 the above tends to -Pi * exp(x) as y decreases to 0. But y is negative for y < 0, so the limit as y increases to 0 is Pi * exp(x).) >The interesting thing about this function is that >if you try to expand it in a Taylor series, you >get something wildly divergent: > > f(z) = sum from j=0 to infinity of (-z)^j j! > >(To tell the truth, I discovered f(z) by starting >with this divergent power series, and working >backwards to the integral definition.) > >I suspect that means that f(z) has an essential >singularity at z=0, but I don't know how to prove it. > >Daryl McCullough >CoGenTex, Inc. >Ithaca, NY ============================================================================== From: Ayatollah Potassium Subject: Re: Analytic Function? Date: Tue, 15 Aug 2000 14:10:19 -0400 Newsgroups: sci.math Daryl McCullough wrote: > f(z) = integral from 0 to infinity of exp(-t)/(1 + zt) dt > > I believe that f(z) is well-defined for any complex value of > z except when z is real and negative. Does anyone know if f(x) > can be analytically continued to the entire complex plane > (except for isolated singularities)? It doesn't extend continuously to any negative real z. Pulling out a factor of exp(1/z) from the numerator and z from the denominator, we get (with u = t+1/z): f(z)=1/z * exp(1/z) * g(z) where g(z) = integral exp(-u)/u du, from 1/z to 1/z + infinity. This is an easier integral to analyze because only the path of integration depends on z. For nonzero z, the integration path can be changed to 1/z --> 1 --> +infinity, so the dependence of the integral on z is reduced to one endpoint (the integral from 1 to infinity is some positive constant C). Thus, g(z) = C + integral from 1/z to 1 of (exp(-u)/u du). Write the integrand as (exp(-u) -1)/u + 1/u. The first term has as integral an entire function H(u), and g(z) = C + H(1) - H(1/z) + log(z). Conclusion: f(z) = 1/z * exp(1/z) * [A - H(1/z) + log(z)] where H is entire and A is constant. Due to the log(z) if you approach a point z on the negative real axis from above and below the two values will differ by 2*pi*i * exp(1/z)/z, so no extension exists. > The interesting thing about this function is that > if you try to expand it in a Taylor series, you > get something wildly divergent: > > f(z) = sum from j=0 to infinity of (-z)^j j! > > (To tell the truth, I discovered f(z) by starting > with this divergent power series, and working > backwards to the integral definition.) Yes, it follows from the integral for the gamma function. n! = integral exp(-t)*t^n dt, from t=0 to infinity. > I suspect that means that f(z) has an essential > singularity at z=0, but I don't know how to prove it. The calculation of f(z) above doesn't give the asymptotic behavior near 0, maybe the original integral is better for this. Does f(z) --> f(0) as z-->0 in the half-plane Re(z) >= 0? ============================================================================== From: baez@galaxy.ucr.edu Subject: Re: Analytic Function? Date: 17 Aug 2000 20:45:04 GMT Newsgroups: sci.math In article <8n965p$hir@edrn.newsguy.com>, Daryl McCullough wrote: > f(z) = sum from j=0 to infinity of (-z)^j j! This is closely akin to the exponential integral function Ei(z) = -integral_{-z}^infinity exp(-t)dt / t (1) where we take the principal value of this integral to deal with the singularity at t = 0. Ei(x) has the asymptotic formula Ei(-z) = exp(-z) sum from j to infinity of (-z)^j (j-1)! So if you want the answer to your question, you can take results about about Ei(x) and mess around with them a bit. There's a bunch in Abramowitz and Stegun, the bible of special functions. Here's what they say: Analytic continuation yields a multivalued function with branch points at z = 0 and z = infinity. They are single-valued functions in the z-plane cut along the negative real axis. I too once fell in love with the idea of a function whose asymptotic expansion looks like a goofy version of the exponential: 1 + 1/z + 2!/z^2 + 3!/z^3 + .... It led me straight to Ei. There's a lot of fun to be had with Ei, Li, Si and the like.