How does one show that $\lim_{n\to\infty}\ln{n}-\int_{0}^{n}{e^x-x-1\over x(e^x+1)}=\ln{\pi}-\gamma$
Write
\begin{align*} \int_{0}^{n} \frac{e^x - x - 1}{x(e^x + 1)} \, dx &= \int_{0}^{n} \frac{e^x - 1}{e^x + 1} \frac{dx}{x} - \int_{0}^{n} \frac{dx}{e^x + 1} \\ &= \left[ \frac{e^x - 1}{e^x + 1} \log x \right]_{0}^{n} - \int_{0}^{n} \frac{2e^x}{(e^x + 1)^2} \log x \, dx + \left[ \log(1 + e^{-x}) \right]_{0}^{n} \\ &= \frac{e^n - 1}{e^n + 1} \log n + \log\left( \frac{1 + e^{-n}}{2} \right) - \int_{0}^{n} \frac{2e^x}{(e^x + 1)^2} \log x \, dx. \end{align*}
From this, we have
$$ \lim_{n\to\infty}\left( \log n - \int_{0}^{n} \frac{e^x - x - 1}{x(e^x + 1)} \, dx \right) = \int_{0}^{\infty} \frac{2e^x}{(e^x + 1)^2} \log x \, dx + \log 2. $$
In order to evaluate the last integral, we adopt the Feynman's trick. Specifically, we introduce the following function
$$ I(s) = \int_{0}^{\infty} \frac{2e^x}{(e^x + 1)^2} x^s \, dx. $$
The value we are interested in is $I'(0)$. In order to compute this, assume first that $\Re(s) > 1$ and we perform the following computation:
\begin{align*} I(s) &= 2 \sum_{n=1}^{\infty} (-1)^{n-1} n \int_{0}^{\infty} x^s e^{-nx} \, dx \\ &= 2\Gamma(s+1) \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} \\ &= 2\Gamma(s+1)(1 - 2^{1-s})\zeta(s). \end{align*}
(The assumption $\Re(s) > 1$ is essential when applying the Fubini's theorem.) Since both $I(s)$ and $2\Gamma(s+1)(1 - 2^{1-s})\zeta(s)$ are analytic for $\Re(s) > -1$, by the principle of analytic continuation the identity above extends to $\Re(s) > -1$. Thus by the log-differentiation together with known values
$$\zeta(0) = -\frac{1}{2}, \qquad \zeta'(0) = -\frac{1}{2}\log(2\pi), \qquad \psi(1) = -\gamma$$
(where $\psi$ is the digamma function), we obtain
$$ I'(0) = -\gamma - 2\log 2 + \log(2\pi). $$
Plugging this back, we obtain the desired equality.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
\begin{align} &\lim_{n\to\infty}\bracks{\ln\pars{n} - \int_{0}^{n}{\expo{x} - x - 1 \over x\pars{\expo{x} + 1}}\,\dd x} \\[5mm] = &\ \lim_{n\to\infty}\bracks{\ln\pars{n} - \int_{0}^{n}{1 - \expo{-x} \over 1 + \expo{-x}}\,{\dd x \over x}\ +\ \overbrace{\int_{0}^{n}{\dd x \over \expo{x} + 1}} ^{\ds{\ln\pars{2} - \ln\pars{1 + \expo{-n}}}}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{% \ln\pars{2n \over 1 + \expo{-n}} - \pars{% \int_{0}^{n}{1 - \expo{-x} \over 1 + \expo{-x}}\,{\dd x \over x} - \int_{0}^{n}{1 - \expo{-x} \over x}\,\dd x} - \int_{0}^{n}{1 - \expo{-x} \over x}\,\dd x} \\[5mm] & = \lim_{n \to \infty}\braces{% \ln\pars{2n \over 1 + \expo{-n}} + \int_{0}^{n}\!\!{\expo{-x} - \expo{-2x} \over 1 + \expo{-x}}\,{\dd x \over x} - \bracks{\ln\pars{n}\pars{1 - \expo{-n}} - \int_{0}^{n}\!\!\!\!\ln\pars{x}\expo{-x}\,\dd x}} \\[5mm] = &\ \ln\pars{2}\ +\ \underbrace{\int_{0}^{\infty}{\expo{-x} - \expo{-2x} \over 1 + \expo{-x}} \,{\dd x \over x}}_{\ds{\ln\pars{\pi \over 2}}}\ +\ \underbrace{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}_{\ds{-\gamma}} = \bbx{\ds{\ln\pars{\pi} - \gamma}} \end{align}
Note that $\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \left.\totald{}{\mu}\int_{0}^{\infty}x^{\mu}\expo{-x}\,\dd x\, \right\vert_{\ \mu\ =\ 0} = \left.\totald{\Gamma\pars{\mu + 1}}{\mu}\right\vert_{\ \mu\ =\ 0} = \Psi\pars{1} = -\gamma}$.
The remaining integral is evaluated as follows: \begin{align} &\bbx{\ds{% \int_{0}^{\infty}{\expo{-x} - \expo{-2x} \over 1 + \expo{-x}}\,{\dd x \over x}}} = \int_{0}^{\infty}{\expo{-3x} - 2\expo{-2x} + \expo{-x} \over 1 - \expo{-2x}}\,{\dd x \over x} \\[5mm] \stackrel{\expo{-2x}\ \mapsto\ x}{=}\,\,\,& -\int_{0}^{1}{x^{1/2} - 2 + x^{-1/2} \over 1 - x}\,{\dd x \over \ln\pars{x}} = -\int_{0}^{1}{x^{1/2} - 2 + x^{-1/2} \over 1 - x} \pars{-\int_{0}^{\infty}x^{t}\,\dd t}\,\dd x \\[5mm] = &\ \int_{0}^{\infty}\int_{0}^{1} {x^{t + 1/2} - 2x^{t} + x^{t - 1/2} \over 1 - x}\,\dd x\,\dd t \\[5mm] = &\ \int_{0}^{\infty}\bracks{% -\int_{0}^{1}{1 - x^{t + 1/2} \over 1 - x}\,\dd x + 2\int_{0}^{1}{1 - x^{t} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{t - 1/2} \over 1 - x}\,\dd x}\,\dd t \\[5mm] = &\ \int_{0}^{\infty}\bracks{-\Psi\pars{t + {3 \over 2}} + 2\Psi\pars{t + 1} - \Psi\pars{t + {1 \over 2}}}\dd t \\[5mm] = &\ \left.-\ln\pars{\Gamma\pars{t + 3/2}\Gamma\pars{t + 1/2} \over \Gamma^{2}\pars{t + 1}}\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} = \ln\pars{\Gamma\pars{3/2}\Gamma\pars{1/2} \over \Gamma^{2}\pars{1}} = \ln\pars{{1 \over 2}\,\Gamma^{2}\pars{1 \over 2}} \\[5mm] = &\ \bbx{\ds{\ln\pars{\pi \over 2}}} \end{align}