I ran into a "wellknown 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.

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] 


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. 


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 reexposed 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. 

