TITLE: Asymptotic form of Bessel $Y_0(x)$ for small $x$
QUESTION [6 upvotes]: The relevant part of the integral definition of $Y_0$ is
$$-\frac{2}{\pi}\int_0^{\infty } e^ {-x \sinh t} \, dt$$
which should be asymptotic to
$$\frac{2}{\pi}\big( \ln\frac x 2 + \gamma \big)$$
Where $\gamma$ is the Euler-Mascheroni Constant. How can I evaluate that integral for small x?
It would be sufficient to find the logarithmic term, but I would also be interested in where the Euler-Mascheroni Constant comes from.

REPLY [3 votes]: The constant might be fixed the following way:
Denote the integral in question is $D=\frac{\pi C}{2}=\color{blue}{J_1}+\color{red}{J_2}$ with
$$
\color{blue}{J_1}=\color{blue}{\int_0^1\frac{1-e^{-z}}{z}\quad},\quad\color{red}{J_2}=\color{red}{-\int_1^{\infty}\frac{e^{-z}}{z}\quad}
$$
Let us start with $\color{blue}{J_1}$ and integrate by parts:
$$
\color{blue}{J_1}=\color{blue}{\lim_{\epsilon\rightarrow0}\left(\log{\epsilon}-\int_{\epsilon}^1\frac{e^{-z}}{z}\right)}=\color{blue}{\lim_{\epsilon\rightarrow0}\left(\log{\epsilon}-\log{\epsilon}-\int_{0}^1\log(z)e^{-z}\right)}=\color{blue}{-\int_{0}^1\log(z)e^{-z}}
$$
Integrating now $\color{red}{J_2}$ by parts yields
$$
\color{red}{J_2}=\color{red}{-\int_{1}^{\infty}\log(z)e^{-z}}
$$
therefore 
$$
-D=-(\color{blue}{J_1}+\color{red}{J_2})=\color{blue}{\int_{0}^1\log(z)e^{-z}}+\color{red}{\int_{1}^{\infty}\log(z)e^{-z}}={\int_{0}^{\infty}\log(z)e^{-z}}
$$
or
$$
D=\gamma
$$
which implies

$$
C=\frac{2 \gamma}{\pi}
$$