Let $\eta=$ diag$\,(-1,1,1,1) $. Consider the proper orthochronous Lorentz subgroup $SO^+(1,3)$ which contains all matrix $L=(a_{ij})_{0\leq i,j\leq3}$ such that $L^\top\eta\,L=\eta $, $\det L=1$ and $a_{00}\geq1 $. This group $SO^+(1,3) $ is a connected matrix Lie group, and its Lie algebra is \begin{align} \mathfrak{so}(1,3)=\Big\{\hspace{.03cm} X\in M_4(\mathbb R)\,\big|\, e^{tX}\in SO^+(1,3),\ \forall\hspace{.02cm}t\in\mathbb R \hspace{.03cm}\Big\}. \end{align} Moreover, the exponential map \begin{align*} \exp:\ \mathfrak{so}(1,3)\,&\longrightarrow\, SO^+(1,3) \\ X\,&\longmapsto\, e^X \end{align*} is a surjective, hence $\forall L\in SO^+(1,3),\ L=e^X$ for some $X\in \mathfrak{so}(1,3)$. We have \begin{align} \begin{array}{rclcrcl} \eta & \hspace{-.15cm} = & \hspace{-.15cm} e^{X^\top}\eta\,e^X & \hspace{-.15cm} \Leftrightarrow & \hspace{-.15cm} I_4 & \hspace{-.15cm} = & \hspace{-.15cm} e^{X^\top}\eta\,e^X\eta^{-1}\,=\,e^{X^\top}e^{\eta X\eta^{-1} } \\ & & & \hspace{-.15cm} \Leftrightarrow & \hspace{-.15cm} X^\top & \hspace{-.15cm} = & \hspace{-.15cm} -\eta X\eta .\tag1 \end{array} \end{align} This implies \begin{align*} X\,&=\,\begin{bmatrix} \,0 & v_1 & v_2 & v_3\, \\ \,v_1 & 0 & -\theta_3 & \theta_2\, \\ \,v_2 & \theta_3 & 0 & -\theta_1\, \\ \,v_3 & -\theta_2 & \theta_1 & 0\, \end{bmatrix} \\ &=\,\theta_1 J_1+\theta_2 J_2+\theta_3 J_3+v_1 K_1+v_2K_2+v_3 K_3\,=\,\theta\cdot J+v\cdot K, \end{align*} with \begin{align} J_1=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&-1 \\ 0&0&1&0 \end{pmatrix}\ \ \ J_2=\begin{pmatrix} 0&0&0&0 \\ 0&0&0&1 \\ 0&0&0&0 \\ 0&-1&0&0 \end{pmatrix}\ \ \ J_3=\begin{pmatrix} 0&0&0&0 \\ 0&0&-1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{pmatrix} \end{align} and \begin{align} K_1=\begin{pmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}\ \ \ K_2=\begin{pmatrix} 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}\ \ \ K_3=\begin{pmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{pmatrix}. \end{align} Consider $\theta=0$, one obtains \begin{align} L\,&=\,e^{v\cdot K} \\ &=\,I_4+|v|\,\widehat v\cdot K+ \frac{|v|^2}{2}( v\cdot K)^2+\frac{|v|^3}{6}( v\cdot K)^3+\cdots \\ &=\,I_4+\left(|v|+\frac{|v|^3}{3!}+\frac{|v|^5}{5!} +\cdots \right)(v\cdot K) + \left(\frac{|v|^2}{2!}+\frac{|v|^4}{4!} +\cdots \right)(v\cdot K)^2 \\ &=\,I_4+\big(\sinh|v|\big)(v\cdot K) +\big(\cosh|v|-1\big)(v\cdot K)^2.\tag2 \end{align} But I feel like I have done wrong somewhere in my calculation, because in the materials that I consult, the expression of $L$ should be $e^{\psi(\widehat v\cdot K)} $ where $\tanh\psi=|v| $ and thus $|v|<1$. However I have no clue in where of my solution this issue comes from.
I hope someone will help me to clarify my isse or enlight me with some helpful idea. Thanks.
UPDATE
I found a solution for my issue above. Let $\,u=\displaystyle\frac{\tanh|v|}{|v|}\,u$, we have $|u|=tanh|v| $ and $\widehat{\,v\,}=\widehat u$. Then, \begin{align*} \cosh|v|\,&=\,\frac{1}{\sqrt{1-|u|^2}}\,=:\,\gamma, \\ \sinh|v|\,&=\,\frac{|u|}{\sqrt{1-|u|^2}}\,=\,\gamma|u| \end{align*} and \begin{align*} L\,&=\,I_4+\gamma|u|\big(\widehat u\cdot K\big)+(\gamma-1)\big(\widehat u\cdot K\big)^2 \\ &=\,I_4+\gamma\big(u\cdot K\big)+\frac{\gamma-1}{|u|^2}\big( u\cdot K\big)^2 \\ &=\, \begin{bmatrix} \gamma & \gamma u_1 & \gamma u_2 & \gamma u_3 \\ \gamma u_1 & 1+(\gamma-1)\displaystyle\frac{u_1^2}{u^2} & (\gamma-1)\displaystyle\frac{u_1u_2}{u^2} & (\gamma-1)\displaystyle\frac{u_1u_3}{u^2} \\ \gamma u_2 & (\gamma-1)\displaystyle\frac{u_1u_2}{u^2} & 1+(\gamma-1)\displaystyle\frac{u_2^2}{u^2} & (\gamma-1)\displaystyle\frac{u_2u_3}{u^2} \\ \,\gamma u_3 & (\gamma-1)\displaystyle\frac{u_1u_3}{u^2} & (\gamma-1)\displaystyle\frac{u_2u_3}{u^2} & 1+(\gamma-1)\displaystyle\frac{u_3^2}{u^2} \, \end{bmatrix}. \end{align*}