Integrating elements of a Lie group with respect to parameters of the corresponding Lie algebra I am working with an operator $\textbf{M}$ that is represented by the Lie group SO(1,3), thus it can be written as,
$$
 \textbf{M} = \exp{\textbf{L}}
$$
where,
$$
\textbf{L} = \begin{bmatrix}
 0&a&b&c\\
 a&0&d&e\\
 b&-d&0&f\\
 c&-e&-f&0
\end{bmatrix}.
$$
I need to integrate $\textbf{M}$ with respect to one or more of the parameters of $\textbf{L}$, for example,
$$
 \textbf{B}=\int_{0}^{T}\textbf{M}\,da
$$
is the integral of $\textbf{M}$ with respect to the differential parameter $a$. This integration can be performed component-wise as,
$$
B_{i,j}=\int_{0}^{T}M_{i,j}\,da
$$
But, the analytic form of $\textbf{M}$ is very cumbersome to work with when all the parameters of $\textbf{L}$ are non-zero. I am having a very difficult time obtaining analytic solutions for the elements of $\textbf{B}$ in terms of the elements of $\textbf{L}$. It is possible to perform the integration on the Lie algebra before the exponential map? Or are there any simplifications that arise from knowledge of the group symmetry? I searched for a long time but cannot find a way to write the integration in terms of $\textbf{L}$ directly.
 A: Apologies for not producing a most general answer for arbitrary Lie groups,
(which you might tease with great effort out of WP ), but only a trail-map for your particular (charmed!) problem. 
I call it charmed because it should remind you of the Lorentz group, with a,b,c parameterizing Kx,Ky,Kz boosts and d,e,f the three J rotation angles. Decent treatments of the representations of the Lorentz group for starters would remind you that you may take linear combinations of the Ks and Js that commute with each other, and so then your exponential is really the direct product of two exponentials, $\exp({\bf \theta \cdot A}) \otimes \exp ({\bf \phi \cdot B})$ each in a 2x2 complexified rep, with θ and φ awful complex angles you are meant to map your 6 angles to. Once you have done that, though, since exponentiated Pauli matrices have a standard 
resolution linear in Pauli matrices, your integral is tractable---and, more interestingly, expressible as the direct product of exponentials, whence, reversing your steps, as an exponential of a 4x4 matrix, if desired. 
It may be a lot of work, but it is straightforward. (Try with all parameters vanishing except a and f first: you have ${\bf M}=\exp(a \sigma_1,~ if\sigma_2) $ in the two separable 2x2 blocks; you can then see the second block factors out and is unaffected by the integration, while the first one is not, and the integral of M is thus ${\bf B}=\int {\bf M}=\exp(\frac{T}{2} \sigma_1 +1\!\!1 \log(2 \sinh(T/2)),~ if\sigma_2) $.) There might be a slick physical, Wigner rotation-style argument too, but it might take as much time. 
