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What part don't you understand?

\begin{aligned}
\ln \Gamma(z)=& \int_{0}^{\infty}\left[(z-1) e^{-t}-\frac{e^{-t}-e^{-z i}}{1-e^{-t}}\right] \frac{d t}{t} \quad(\mathscr{R} z>0) \\
=&\left(z-\frac{1}{2}\right) \ln z-z+\frac{1}{2} \ln 2 \pi \\
&+2 \int_{0}^{\infty} \frac{\arctan (t / z)}{e^{2 \pi t}-1} d t \quad(\mathscr{R} z>0)
\end{aligned}

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I have never understood Binet's Formula for Logarithm of Gamma Function

Formulation 1

Let $z$ be a complex number with a positive real part.
Then:
$$\operatorname{Ln} \Gamma(z)=\left(z-\frac{1}{2}\right) \operatorname{Ln} z-z+\frac{1}{2} \ln 2 \pi+\int_{0}^{\infty}\left(\frac{1}{2}-\frac{1}{t}+\frac{1}{e^{t}-1}\right) \frac{e^{-t z}}{t} \mathrm{~d} t$$
where:
$\Gamma$ is the Gamma function
$\mathrm{Ln}$ is the principal branch of the complex logarithm.

Formulation 2

Let $z$ be a complex number with a positive real part.
Then:
$$\operatorname{Ln} \Gamma(z)=\left(z-\frac{1}{2}\right) \operatorname{Ln} z-z+\frac{1}{2} \ln 2 \pi+2 \int_{0}^{\infty} \frac{\arctan (t / z)}{e^{2 \pi t}-1} \mathrm{~d} t$$
where:
$\Gamma$ is the Gamma function
$\mathrm{Ln}$ is the principal branch of the complex logarithm.

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Found an elementary proof of Binet's Formula for the Gamma Function s below:

The present note presents an elementary proof of the following important result of J. P. M. Binet [3, p. 249].

Theorem 1. For $x>0$ we have
$$\Gamma(x+1)=\left(\frac{x}{\mathrm{e}}\right)^{x} \sqrt{2 \pi x} \cdot \mathrm{e}^{\theta(x)}$$
where
$$\theta(x)=\int_{0}^{\infty}\left(\frac{1}{\mathrm{e}^{t}-1}-\frac{1}{t}+\frac{1}{2}\right) \mathrm{e}^{-x t} \frac{1}{t} d t$$
Here $\Gamma$ denotes the gamma function defined by
$$\Gamma(x)=\int_{0}^{\infty} t^{x-1} \mathrm{e}^{-t} d t$$
Since $\lim _{x \rightarrow \infty} \theta(x)=0$, from $(1)$ we immediately obtain Stirling's formula
$$n !=\Gamma(n+1) \sim\left(\frac{n}{\mathrm{c}}\right)^{n} \sqrt{2 \pi n} .$$
Binet's formula can also be used to prove a more precise version of Stirling's asymptotic expansion

$$\log \frac{n !}{(n / \mathrm{e})^{n} \sqrt{2 \pi n}}=\sum_{j=1}^{\infty} \frac{B_{2 j}}{2 j(2 j-1) n^{2 j-1}}=\frac{1}{12 n}-\frac{1}{360 n^{3}}+\frac{1}{1260 n^{5}}-\cdots$$

where the $B_{2 j}$ 's denote the Bernoulli numbers defined by

$$\frac{1}{\mathrm{e}^{t}-1}-\frac{1}{t}+\frac{1}{2}=\sum_{j=1}^{\infty} \frac{B_{2 j}}{(2 j) !} t^{2 j-1}$$
For, by problem 154 in Part I, Chapter 4 of [2], the inequalities
$$\sum_{j=1}^{2 N} \frac{B_{2 j}}{(2 j) !} t^{2 j-1}<\frac{1}{\mathrm{e}^{t}-1}-\frac{1}{t}+\frac{1}{2}<\sum_{j=1}^{2 N+1} \frac{B_{2 j}}{(2 j) !} t^{2 j-1}$$

Sasvari, Z. (1999). An Elementary Proof of Binet's Formula for the Gamma Function. The American Mathematical Monthly, 106(2), 156-158. doi:10.2307/2589052

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