Expansion of the modified Bessel function $K_\nu(z)$ around $z\to 0$ Here, the definition (Fradkin, Quantum field theory an integrated approach) of the modified Bessel function is
$$K_{\nu}(z)=\frac{1}{2}\int^{\infty}_{0}dt\,t^{\nu-1}e^{-z(t+1/t)/2}.$$ It is said that when $z\to 0$
$$K_{\nu}(z)=\frac{\Gamma(\nu)}{2(\frac{z}{2})^{\nu}}+O\!\left(\frac{1}{z^{\nu-2}}\right).\tag{1}$$
But, in another book (Jean Zinn-Justin, phase transitions and renormalization group), the author gives
$$K_{0}(z)=\frac{1}{2}\int^{\infty}_{0}\frac{dt}{t}e^{-z(t+1/t)/2}=_{z\to0}-\ln(z/2)-\gamma+O(z), \tag{2}$$ where $\gamma$ is the Euler's constant $\gamma=-\int^{\infty}_{0}dt\ln(t)\, e^{-t}$.
It seems that eq. $(1)$ is not consistent with eq. $(2)$ (since $\Gamma(\nu)=1/\nu-\gamma+O(\nu)$, and $(z/2)^{\nu}=1-\nu \ln(\frac{z}{2})$, then eq. $(1)$ gives $\frac{1}{\nu}-\ln(\frac{z}{2})-\gamma+O(\nu)$), may be the limit $z\to 0$ and $\nu\to 0$ can not be exchanged.
How to derive eqs. $(1)$ and $(2)$? Eq. $(2)$ seems relate to Modified Bessel function near zero, but can we just derive the result from the integral above?
 A: You are right: the point is that the limits $\nu\to 0$ and $z\to0; \nu\neq0$ cannot be exchanged. Roughly speaking, we have different values:
$\displaystyle \Big(\frac{z}{2}\Big)^\nu\to 0\,\,\text{at}\, \,\nu\neq0; z\to 0$
$\displaystyle\Big(\frac{z}{2}\Big)^\nu\to1+\nu\ln\frac{z}{2}\to 1\,\,\text{at}\, \,z\neq0; \nu\to 0$
We can show this explicitly. Splitting the interval of integration and making in the second integral the substitution $t=\frac{1}{x}$
$$2K_{\nu}(2z)=\int_0^1t^{\nu-1}e^{-z\big(t+\frac{1}{t}\big)}dt+\int_1^\infty t^{\nu-1}e^{-z\big(t+\frac{1}{t}\big)}dt=\int_0^1\big(t^{\nu-1}+t^{-\nu-1}\big)e^{-z\big(t+\frac{1}{t}\big)}dt$$
$$=\int_0^1\big(t^{\nu-1}+t^{-\nu-1}\big)e^{-\frac{z}{t}}\Big(1-zt+\frac{(zt)^2}{2!}+...\Big)dt$$
$$=\int_0^1t^{-\nu-1}e^{-\frac{z}{t}}dt+\int_0^1t^{\nu-1}e^{-\frac{z}{t}}dt-z\int_0^1t^{-\nu}e^{-\frac{z}{t}}dt-z\int_0^1t^\nu e^{-\frac{z}{t}}dt+...\tag{1}$$
All other terms are regular at both $z=0$ and $\nu=0$. Evaluating the third and the fourth terms, we have
$$z\int_0^1t^{-\nu}e^{-\frac{z}{t}}dt<z\int_0^1t^{-\nu}dt=\frac{z}{1-\nu}$$
$$z\int_0^1t^\nu e^{-\frac{z}{t}}dt<z\int_0^1t^\nu dt=\frac{z}{1+\nu}$$
so they are also regular at $z=0;\,\nu=0$.
Evaluating the first two terms and making the substitution $x=\frac{1}{t}$, we get (for the first term)
$$\int_0^1t^{-\nu-1}e^{-\frac{z}{t}}dt=\int_1^\infty x^{\nu-1}e^{-zx}dx=\frac{\Gamma(\nu)}{z^\nu}-\frac{1}{z^\nu}\int_0^zx^{\nu-1}e^{-x}dx$$
Decomposing the exponent into the series,
$$=\frac{\Gamma(\nu)}{z^\nu}-\frac{1}{\nu}+\frac{z}{1+\nu}+...\tag{2}$$
Only first two terms are the main asymptotic terms; all others are regular at both $z=0;\,\nu=0$
Now, we evaluate the second term in (1) in the similar way. Integrating by part,
$$\int_0^1t^{\nu-1}e^{-\frac{z}{t}}dt=z^\nu\int_z^\infty e^{-x}x^{-\nu-1}dx=-\frac{z^\nu x^{-\nu}e^{-x}}{\nu}\,\bigg|_{x=z}^\infty-\frac{z^\nu}{\nu}\int_z^\infty x^{-\nu}e^{-x}dx$$
$$=\frac{e^{-z}}{\nu}-z^\nu\frac{\Gamma(1-\nu)}{\nu}+\frac{z}{\nu}\Big(\frac{1}{1-\nu}-\frac{z}{2-\nu}+\frac{z^2}{2!(3-\nu)}-+...\Big)\tag{3}$$
Taking together (2) and (3)
$$\boxed{\,\,2K_{\nu}(2z)=\frac{\Gamma(\nu)}{z^\nu}-\frac{1}{\nu}-z^\nu\frac{\Gamma(1-\nu)}{\nu}+\frac{e^{-z}}{\nu}+\frac{z}{\nu}\Big(\frac{1}{1-\nu}-\frac{z}{2-\nu}+...\Big)+\text{regular part}\,\,}$$
Now we can investigate different cases:
$\displaystyle 1.\qquad \nu\neq0;z\to0$
$$2K_{\nu}(2z)\to\frac{\Gamma(\nu)}{z^\nu}$$
$\displaystyle 2.\qquad \nu\to0;z\neq0$
$$2K_{\nu}(2z)\to\frac{1-\gamma\nu}{\nu(1+\nu\ln z)}-\frac{1}{\nu}-\frac{(1+\gamma\nu)(1+\nu\ln z)}{\nu}+\frac{e^{-z}}{\nu}+\frac{z}{\nu}\Big(1-\frac{z}{2!}+\frac{z^2}{3!}-...\Big)$$
$$\to\,-2\ln z-2\gamma$$
