pr.probability - Reference request for a "well-known identity" in a paper of Shepp and Lloyd

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 $Gamma(x+1) = xGamma(x)$.
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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