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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
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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
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3 Answers

up vote 6 down vote accepted

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
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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
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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
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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
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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
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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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