# Reference request for a “well-known identity” in a paper of Shepp and Lloyd

I ran into a "well-known identity" on page 345 of Shepp and Lloyd's On ordered cycle lengths in a random permutation: $$\int_x^{\infty} \frac{\exp(-y)}y dy = \int_0^x \frac{1-\exp(-y)}y dy - \log x - \gamma,$$ where $\gamma$ is the Euler constant. I am clueless as to how it is derived. Any reference to the derivation of such formulae would suffice, but an explicit solution will also be appreciated.

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This refers to an earlier question: mathoverflow.net/questions/19392/… – Michael Lugo Mar 27 '10 at 18:23
A link to an online copy of the paper (preferably not behind a paywall) would be highly appropriate here. – Scott Morrison Mar 27 '10 at 18:46
I added a JSTOR link. The article is also at ifile.it/tfja0wc, but the link is probably temporary. – Anton Geraschenko Mar 27 '10 at 19:09
Meta discussion: tea.mathoverflow.net/discussion/313/… – Harry Gindi Mar 27 '10 at 19:15

You can apply WZ theory to such identities. In particular, both sides satisfy $$x*z''(x) + (x+1)z'(x)$$ Picking $x=1$ as the initial condition (since the DE is regular there, that helps), we see that both sides evaluate to $Ei(1,1)$ and their derivatives both evaluate to $-1/e$, so they are equal.

I got that differential equation using Maple's PDEtools[dpolyform] function, which uses Groebner bases over differential polynomials to 'solve' this problem. All the rest is classical analysis (as in A course of modern analysis by Whittaker and Watson, 1926 - which is unfortunately not material that is taught very much anymore, I certainly had to learn a lot of that 'on my own').

[Edit: fixed an error in the evaluation of the derivative, I pasted in the wrong line]

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Yes that's a tremendously rich text much of which I am not familiar with. Thanks for the hard work! – John Jiang Mar 27 '10 at 21:08
What is "WZ theory"? – Mariano Suárez-Alvarez Mar 27 '10 at 21:59
Wilf-Zeilberger, probably, see en.wikipedia.org/wiki/Wilf-Zeilberger_pair – Reid Barton Mar 27 '10 at 22:02
@Reid: correct. – Jacques Carette Mar 27 '10 at 22:04

This identity appears on the Wikipedia page for the "exponential integral": http://en.wikipedia.org/wiki/Exponential_integral#Definition_by_Ein

I imagine you can get it by integrating the Taylor series and playing around. Wikipedia, and several other places on the web, point to the book by Abramovitz and Stegun.

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Thanks for the nice reference on wiki! I was looking at the wrong place. – John Jiang Mar 27 '10 at 19:46
You can prove the identity up to a constant factor by differentiating with respect to x, so it only remains to prove it for x = 1. This should be a little easier. – Qiaochu Yuan Mar 27 '10 at 19:55
Indeed, it makes it a lot easier. Using integration by part, one can show that $\int_1^{\infty} \exp(-y)/y dy - \int_0^1 \frac{1-\exp(-y)}{y}dy = \int_0^{\infty} \exp(-y) \log y dy which is listed as equal to$-\gamma$in the following wiki page: en.wikipedia.org/wiki/… I am yet to figure out why that formula in the wiki page is true. – John Jiang Mar 27 '10 at 20:51 The integral$\int_0^\infty e^{-t}\log t\,dt$equals$\Gamma'(1)$. This can be evauated as$-\gamma$using the infinite product for the gamma function. – Robin Chapman Mar 28 '10 at 7:58 Thanks Robin. I will write a short summary of the proof combining all the ingredients given so far. – John Jiang Mar 28 '10 at 18:02 So one of the approaches to proving the equality in the question is via the following three steps: First differentiate both sides of the equation to see that they agree up to a constant. This reduces to showing the case of$x = 1$, for which$\log x = 0$. Next we apply integration by parts to get $$\int_1^{\infty} \exp(-y)/y dy - \int_0^1 \frac{1-\exp(-y)}{y}dy = \int_0^{\infty} \exp(-y) \log y dy$$ Finally observe that$\Gamma'(1)$equals the RHS, by differentiating under the integral sign, valid because things are decaying fast enough at infinity. So it remains to show$\Gamma'(1) =\gamma$. I saw a soft argument (i.e., without using infinite product) in the link scipp.ucsc.edu/~haber/ph116A/psifun_10.pdf This is re-exposed below: first we establish that for$\Psi(x) = \log \Gamma(x)$, $$\Psi'(x+1) = \Psi'(x) + 1/x$$ This is easy enough since we have we have the functional equation$\Gamma(x+1) = x\Gamma(x)$. Next using stirling approximation we get $$\Psi(x+1) = (x+1/2)\log x -x + 1/2 \log 2 \pi + O(1/x)$$ and then they differentiate this and claim that$O(1/x)' = O(1/x^2)$, which is clearly false (take$f(x) = 1/x cos(e^x)$). But I found in Wikipedia another formula that gives the precise error term in terms of an integral of the monotone function$arctan(1/x)$. So this is enough to establish$O(1/x^2)$for the error term in the derivative of$\Psi$. So we get the asymptotics$\lim_{x \to \infty} \Psi'(x+1) = \log(x)$, from which we get$\Psi'(1) = \gamma$. Now notice$\Psi'(x) = \Gamma'(x)/ \Gamma(x)$, and$\Gamma(1) = 1$, so$\Gamma'(1) = \gamma\$ also.

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