How can we prove this matrix exponential formula I am working on the reduction of second ordre differential equation to a first order system of equations, and I encounter this formula which we are trying to generalise.
It is known that for $\alpha \in \mathbb{C}$ then,
$$\exp \left(t
   \begin{bmatrix}
     0 & 1 \\
     \alpha & 0 \\
   \end{bmatrix}\right) = 
   \begin{bmatrix}
     cosh(t\alpha^\frac{1}{2}) & \int_0^t{cosh(s\alpha^\frac{1}{2})ds} \\
     \frac{d}{dt}cosh(t\alpha^\frac{1}{2}) & cosh(t\alpha^\frac{1}{2}) \\
   \end{bmatrix} \ \ (t \in \mathbb{R})\ \ \ (1.1)$$
where
$$cosh(t\alpha^\frac{1}{2}) = \sum_{n=0}^\infty\frac{t^{2n}}{(2n)!}\alpha^n$$
I proved it when $\alpha=1$ and $\alpha=-1$, but how to do it in the general case?
You can find this formula in this paper.
Edit:
I found this result in Klaus Nagel book, but is there any other nice method to prove the formula?
2.7 More Examples. (iii) Take an arbitrary 2 x 2 matrix $A = \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}$, define $\delta := ad - bc$, $\tau := a + d$, and take $\gamma \in \mathbb{C}$ such that $\gamma^2 = \frac{1}{4}(\gamma^2 - 4\delta)$.  Then the (semi) group generated by A is given by the matrices
(2.5)
$$e^{tA} = \begin{cases}
   e^{\frac{t\tau}{2}}(\frac{1}{\tau}sinh(t\gamma)A + (cosh(t\gamma) - \frac{2\tau}{\gamma}sinh(t\gamma))I) & if\ \gamma \ne 0, \\
   e^{\frac{t\tau}{2}}(tA + (1 - \frac{t\tau}{2})I) & if \gamma = 0
  \end{cases}$$
 A: Generally, the ODE
$$
\frac{\partial}{\partial t} f(t) = kf(t)
$$
has a solution which looks like $f(t) = e^{tk}f(0) = Ce^{tk}$ for arbitrary constant $C$. This solution corresponds to the initial condition $f(0)=C$ and $\dot f(0) = Ck$.
In our case, the function $f(t) = C e^{t k}$ for $k=\begin{bmatrix}0 & 1 \\ \alpha & 0\end{bmatrix}$ and $C=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ is the solution to
$$
\frac{\partial}{\partial t} f(t) = \begin{bmatrix}0 & 1 \\ \alpha & 0\end{bmatrix} f(t)
$$
with the initial condition $f(0) = C$ and $\dot f(0) = Ck = k$.
The solution to the equation is $f(t)=\begin{bmatrix}p(t) & q(t) \\ r(t) & s(t)\end{bmatrix}$, where
$$
\begin{cases}
\dot p(t) = r(t), & p(0) = 1, & \dot p(0) = 0, \\
\dot q(t) = s(t), & q(0) = 0, & \dot q(0) = 1, \\
\dot r(t) = \alpha p(t), & r(0) = 0, & \dot r(0) = \alpha, \\
\dot s(t) = \alpha q(t), & s(0) = 1, & \dot s(0) = 0.
\end{cases}
$$
From which you quickly get $\ddot p(t) = \alpha p(t)$ and $\ddot s(t) = \alpha s(t)$, hence $p(t) = s(t) = \cosh (t\sqrt\alpha)$, and everything else follows from it, so we obtain the result
$$
f(t) = \begin{bmatrix}\cosh(t \sqrt \alpha) & \frac{1}{\sqrt\alpha}\sinh(t\sqrt \alpha) \\ \sqrt\alpha\sinh(t \sqrt \alpha) & \cosh(t \sqrt \alpha) \end{bmatrix}.
$$
Taking into consideration $\frac{\partial}{\partial x} \cosh x = \sinh x$ and $\frac{\partial}{\partial x} \sinh x = \cosh x$ it leads to the same answer, as in your initial question.
A: $$
A = \left(\begin{array}{cc} 0 & 1 \\  a & 0\end{array}\right)
\;\;\;
A^2 = \left(\begin{array}{cc} a & 0 \\  0 & a\end{array}\right)
\;\;\;
A^3 = \left(\begin{array}{cc} 0 & a \\  a^2 & 0\end{array}\right)
\;\;\;
A^4 = \left(\begin{array}{cc} a^2 & 0 \\  0 & a^2\end{array}\right)
  $$
$$
  A^{2k} = \left(\begin{array}{cc} a^k & 0 \\  0 & a^k\end{array}\right)
\;\;\;
  A^{2k+1} = \left(\begin{array}{cc} 0 & a^k \\  a^{k+1} & 0\end{array}\right)
    $$
$$
e^{tA} = \sum_{n=0}^\infty \frac{t^n A^n}{n!}  =
\sum_{k=0}^\infty \frac{t^{2k} A^{2k}}{(2k)!}
 + \sum_{k=0}^\infty \frac{t^{2k+1} A^{2k+1}}{(2k+1)!}
 $$
$$
 \left(\begin{array}{cc}
   \sum_{k=0}^\infty \frac{t^{2k}a^k}{(2k)!} &
   \sum_{k=0}^\infty \frac{t^{2k+1}a^k}{(2k+1)!}  \\
   \sum_{k=0}^\infty \frac{t^{2k+1}a^{k+1}}{(2k+1)!}  &
   \sum_{k=0}^\infty \frac{t^{2k}a^k}{(2k)!} 
 \end{array}\right)
 $$
Assuming that integration and differentiation commute with these inifinite sums
we can write:
$$
 \frac{d}{dt}
 \sum_{k=0}^\infty \frac{t^{2k}a^k}{(2k)!}  =
 \sum_{k=1}^\infty \frac{2k t^{2k-1}a^k}{(2k)!}  =
 \sum_{k=1}^\infty \frac{ t^{2k-1}a^k}{(2k-1)!}  =
 \sum_{k=0}^\infty \frac{ t^{2k+1}a^k}{(2k+1)!}  =
 $$
$$
 \int_0^t
 \sum_{k=0}^\infty \frac{s^{2k}a^k}{(2k)!}  =
 \sum_{k=0}^\infty \frac{s^{2k+1}a^k}{(2k+1)(2k)!}\arrowvert_0^t  =
 \sum_{k=0}^\infty \frac{t^{2k+1}a^k}{(2k+1)!}  =
 $$
