Modified Bessel function near zero Let $K_0$ be the modified Bessel function of the second kind, that is for instance
$$ K_0(x)=\int_0^{+\infty}e^{-x\cosh(t)}dt $$
How can we prove that $K_0(x)\underset{x\rightarrow 0^+}{\sim}-\log x$ ? I tried to modify the integral by doing the change of variable $u=x\cosh(t)$ but it didn't help me, and Laplace method only gives the equivalent when $x\rightarrow +\infty$.
 A: We can write (take $s=\cosh t$ and then $s=t+1$)
$$
K_0 (x) = \int_1^{ + \infty } {e^{ - xs} \frac{{dt}}{{\sqrt {s^2  - 1} }}}  = e^{ - x} \int_0^{ + \infty } {e^{ - xt} t^{ - 1/2} \frac{{dt}}{{\sqrt {t + 2} }}} $$
for $x>0$. Now
\begin{align*}
\frac{d}{{dx}}\int_0^{ + \infty } {e^{ - xt} t^{ - 1/2} \frac{{dt}}{{\sqrt {t + 2} }}}  &=  - \int_0^{ + \infty } {e^{ - xt} t^{1/2} \frac{{dt}}{{\sqrt {t + 2} }}}  \\ & =  - \frac{1}{x}\int_0^{ + \infty } {e^{ - s} s^{1/2} \frac{{ds}}{{\sqrt {s + 2x} }}} = - \frac{1}{x} +\mathcal{O}(1)
\end{align*}
as $x\to 0+$. Thus, integrating between $x$ and $1$,
$$
\int_0^{ + \infty } {e^{ - xt} t^{ - 1/2} \frac{{dt}}{{\sqrt {t + 2} }}}  =  - \log x +\mathcal{O}(1),
$$
as $x\to 0+$. Finally,
$$
K_0 (x) = e^{ - x} ( - \log x +\mathcal{O}(1)) =  - \log x + \mathcal{O}(1)
$$
as $x\to 0+$.
A: It was suggested in a comment to an unanswered question to use the first term of the series representation of the modified Bessel function of the first kind in the definition $$K_{0}(x) = \frac{\pi}{2}\lim_{\nu \to 0} \frac{I_{-\nu}(x) -I_{\nu}(x)}{\sin(\nu \pi)}. $$
I'm going to present a slightly different approach.

An integral representation of the modified Bessel function of the first kind is  $$I_{\nu}(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} \cos(\nu \theta) \, d \theta - \frac{\sin(\nu \pi)}{\pi} \int_{0}^{\infty} e^{-x \cosh t - \nu t} \, \mathrm dt, \quad x>0. $$
For $\nu \ge 0$, differentiation under the integral sign with respect to $\nu$ is justified by the dominated convergence theorem since $$\left|\frac{\partial}{\partial \nu} \, e^{x \cos \theta} \cos(\nu \theta) \right| \le \theta e^{x \cos \theta},$$ which is integrable on $(0, \pi)$, and $$\left| \frac{\partial}{\partial \nu} \, e^{-x \cosh t-\nu t}\right| = te^{-x\cosh t-\nu t} \le te^{-x\cosh t},$$ which is integrable on $(0, \infty)$.
Differentiating both sides of the integral representation with respect to $\nu$ and then evaluating the result at $\nu=0$, we get $$\frac{\partial }{\partial \nu} I_{\nu}(x)\Bigg|_{\nu=0} = - \int_{0}^{\infty} e^{-x \cosh t} \, \mathrm dt = - K_{0}(x).$$
As $x \to 0^{+}$, $I_{\nu}(x)$ has the asymptotic form $\frac{1}{\Gamma(\nu +1)} \left(\frac{x}{2} \right)^{\nu} $, which is the first term of the series representation.
Since term-by-term differentiation of the series representation with respect to $\nu$ is permissible (see section 3$\cdot$13 in A Treatise on the Theory of Bessel Functions and this question), we get  $$ \begin{align} K_{0}(x) &\underset{\phantom{x \to 0^{+}}}= - \frac{\partial }{\partial \nu} I_{\nu}(x)\Bigg|_{\nu=0} \\ &\underset{x \to 0^{+}}\sim - \frac{\partial }{\partial \nu}\frac{1}{\Gamma(\nu +1)} \left(\frac{x}{2} \right)^{\nu} \Bigg|_{\nu =0} \\ &\underset{\phantom{x \to 0^{+}}}= \frac{\Gamma'(\nu+1)}{\Gamma^{2}(\nu+1)}\left(\frac{x}{2} \right)^{\nu} - \frac{1}{\Gamma(\nu +1)}\left(\frac{x}{2} \right)^{\nu} \ln \left(\frac{x}{2} \right) \Bigg|_{\nu =0} \\ &\underset{\phantom{x \to 0^{+}}}= \Gamma'(1) - \ln \left(\frac{x}{2} \right) \\ &\underset{\phantom{x \to 0^{+}}}= - \ln x - \gamma + \ln 2. \end{align}$$

The discussion in section 3$\cdot$13 is about $J_{\nu}(z)$, but the argument for $I_{\nu}(z)$ is virtually identical.
