change of integrand in integral representation of euler constant In Julian Havil's book "Gamma" he gives an integral representation of gamma as:
$$-\gamma = \int_0^\infty \!\! e^{-u}\ln u\,\mathrm du = \int_0^1 e^{-u}\ln u\,\mathrm du + \int_1^\infty \!\! e^{-u}\ln u\,\mathrm du\\$$
He then transforms the first integral so that the equation reads
$$-\gamma =  -\int_0^1 \!\! (-e^{-u}+1)\ln u\,\mathrm du + \int_1^\infty \!\! e^{-u}\ln u\,\mathrm du$$ 
and keeps the second integrand the same. I am stumped on how he makes that change of integrands, and would appreciate any help.
 A: As Robert Israel points out, the equation given would imply that $\int_0^1 \log(u)\,\mathrm{d}u=0$.
A common method for evaluating that integral (without referencing the Gamma function) is
$$
\begin{align}
\int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t
&=\lim_{n\to\infty}\int_0^n\log(t)\,\left(1-\tfrac tn\right)^n\,\mathrm{d}t\tag{1}\\
&=\lim_{n\to\infty}n\int_0^1\log(nt)\,(1-t)^n\,\mathrm{d}t\tag{2}\\
&=\lim_{n\to\infty}\left[\frac{n}{n+1}\log(n)+n\int_0^1\log(1-t)\,t^n\,\mathrm{d}t\right]\tag{3}\\
&=\lim_{n\to\infty}\left[\frac{n}{n+1}\log(n)-\frac{n}{n+1}\int_0^1\frac{1-t^{n+1}}{1-t}\,\mathrm{d}t\right]\tag{4}\\
&=\lim_{n\to\infty}\frac{n}{n+1}\left[\log(n)-\sum_{k=1}^{n+1}\frac1k\right]\tag{5}\\[6pt]
&=-\gamma\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\lim\limits_{n\to\infty}\left(1-\tfrac tn\right)^n[t\le n]=e^{-t}$ monotonically; apply Lebesgue Monotone Convergence
$(2)$: substitute $t\mapsto nt$
$(3)$: $\log(nt)=\log(n)+\log(t)$ then substitute $t\mapsto1-t$
$(4)$: integrate by parts
$(5)$: divide the polynomials and integrate
$(6)$: $\gamma=\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^n\frac1k-\log(n)\right)$
A: I think this appears on page 107 of the paperback version of Gamma.  I think that you may have misread what it says, unless there is a typo in a version I do not have.  The first integral you give is not such, but a square bracket of evaluation after integration by parts has been carried out.
Best wishes,
Julian Havil
