# Asymptotic Expansion of Digamma Function

I noticed that it said the asymptotic expansion for the digamma function ($$\psi(z)$$) can be obtained from using
$$\begin{equation} \psi(z+1) = -\gamma + \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+z}) \end{equation}$$ (where $$\gamma$$ is the Euler–Mascheroni constant) combined with Euler–Maclaurin formula to conclude
$$\begin{equation} \psi(z) \approx \log(z) - \frac{1}{2z} \end{equation}$$ My main confusion about this is that when I tried to use the Euler–Maclaurin formula to approximate the sum as an integral, I could not figure out how to obtain a $$\gamma$$ to cancel out the $$\gamma$$ term in the series expansion. Any help on how to obtain the above relationship using the formula or just where the $$\gamma$$ comes from when I apply the Euler-Maclaurian formula would be greatly appreciated!
I shall assume that $$|\arg (z+1)|<\pi$$. By the Euler–Maclaurin formula, \begin{align*} \psi (z + 1) = & - \gamma + \int_1^{ + \infty } {\left( {\frac{1}{t} - \frac{1}{{t + z}}} \right)dt} + \frac{1}{2}\left( {1 - \frac{1}{{1 + z}}} \right) \\ & + \int_1^{ + \infty } {\left( {\left\{ {1 - t} \right\} - \frac{1}{2}} \right)\left( {\frac{1}{{t^2 }} - \frac{1}{{(t + z)^2 }}} \right)dt} \\ = & \; \log (z + 1) - \gamma + \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} - \frac{1}{{2(z + 1)}} - \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt}. \end{align*} Thus, it remains to show that $$\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} = \gamma$$ and $$\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt} = \mathcal{O}\!\left( {\frac{1}{{(z + 1)^2 }}} \right).$$ The first one can be shown as follows: \begin{align*} \int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} & =\sum\limits_{k = 1}^\infty {\int_k^{k + 1} {\frac{{\left\{ {1 - t} \right\}}}{{t^2 }}dt} } = \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{1 - t}}{{(k + t)^2 }}dt} } \\ & = \sum\limits_{k = 1}^\infty {\left[ {\frac{1}{k} - \log \left( {1 + \frac{1}{k}} \right)} \right]} = \mathop {\lim }\limits_{n \to + \infty } \sum\limits_{k = 1}^n {\left[ {\frac{1}{k} - \log \left( {1 + \frac{1}{k}} \right)} \right]} \\ & = \mathop {\lim }\limits_{n \to + \infty } \left(\sum\limits_{k = 1}^n {\frac{1}{k}} - \log (n + 1) \right)= \gamma . \end{align*} For the second claim, we use integration by parts and some estimations to obtain \begin{align*} & \left| {\int_1^{ + \infty } {\frac{{\left\{ {1 - t} \right\} - \frac{1}{2}}}{{(t + z)^2 }}dt} } \right| = \left| {\int_1^{ + \infty } {\frac{{\left\{ t \right\} - \left\{ t \right\}^2 }}{{(t + z)^3 }}dt} } \right| \\ & \le \frac{1}{4}\int_1^{ + \infty } {\frac{1}{{\left| {t + z} \right|^3 }}dt} = \frac{1}{4}\int_0^{ + \infty } {\frac{1}{{\left| {t + (z + 1)} \right|^3 }}dt} \\ & \le \frac{1}{4}\int_0^{ + \infty } {\frac{1}{{(t + \left| {z + 1} \right|)^3 }}dt} \sec ^3 \left( {\frac{{\arg (z + 1)}}{2}} \right) = \frac{1}{{8\left| {z + 1} \right|^2 }}\sec ^3 \left( {\frac{{\arg (z + 1)}}{2}} \right). \end{align*} In summary, replacing $$z+1$$ by $$z$$, $$\psi (z) = \log z - \frac{1}{{2z}} + R(z)$$ where $$\left| {R(z)} \right| \le \frac{1}{{8\left| z \right|^2 }}\sec ^3 \left( {\frac{{\arg z}}{2}} \right)$$ and $$|\arg z|<\pi$$. With more careful considerations, the constant $$\frac{1}{8}$$ can be replaced by $$\frac{1}{12}$$.