Time integral of a time-dependent Displacement operator The diplacement operator on a bosonic mode with creation and annihilation operators, $\hat{a}^\dagger,\hat{a}$, is usually defined as $$ \hat{D}(\alpha)=\exp(\alpha \hat{a}^\dagger - \alpha^*\hat{a})$$.
Suppose that $\alpha$ is time-dependent, then my question is how can one find the time integral of the displacement operator?
$$
\int_0^t d\tau  \hat{D}(\alpha(\tau))
=
\int_0^t d\tau\exp(\alpha(\tau) \hat{a}^\dagger - \alpha^*(\tau)\hat{a})
$$
My first thought is to write out the explicit defintion of the exponenital and stick to normal ordering, then use the binomial expansion to brute force compute it. So in the end integrating terms like $(\alpha(\tau) \hat{a}^\dagger - \alpha^*(\tau)\hat{a})^k $ using the binomial expansion.
I'm looking for tips if this is a correct way to approach this or possibly existing results before I spend time on this and it turns out to be an exercise in futility.

Edit: This question was closed because it needed more details or clarity - which I'm not sure how would affect the technical part of the question. The background for this question is that I am interested in calculating a first order Dyson series term of a time evolution operator (in the interaction picture) which has the form $ \hat{U}_I^{(1)}(t,0)=-\frac{i}{\hbar}\int_0^t d\tau \hat{H}^{(I)}(\tau)$ where $\hat{H}^{(I)}(t)$ is the effective interaction picture Hamiltonian. In this first order term, there is a term involving exactly this integral $\int_0^t d\tau  \hat{D}(\alpha(\tau))$. This is the background.
As it seems the general case cannot be explicitly tackled, I'm interested in the case where $\alpha(t)$ is linear in time, $\alpha(t)=\alpha_0 t$, and the case where it's as sinusoidal function in time, $\alpha(t)=\alpha_0\cos(\omega t +\phi)$. Any help for these two cases will be plenty.
 A: (Too long for a comment)
Hard to say what you intend to do with the result, but if I were you, I would recommend to expand the time-dependent displacement operator in the (overcomplete) basis of coherent states, in order to work and integrate the operator components only.
Concretely, you can write thanks to closure relations :
$$
\hat{D}(\alpha(t)) = \int_\mathbb{C}\frac{\mathrm{d}^2\mu}{2\pi i}\int_\mathbb{C}\frac{\mathrm{d}^2\nu}{2\pi i}\,D_{\mu\nu}(\alpha(t))|\mu\rangle\langle\nu|
$$
where $\mathrm{d}^2\beta := \mathrm{d}\beta^*\wedge\mathrm{d}\beta = 2i\,\mathrm{d}\Re(\beta)\wedge\mathrm{d}\Im(\beta)$ and
$$
\begin{array}{rcl}
D_{\mu\nu}(\alpha) &=& \langle\mu|\hat{D}(\alpha)|\nu\rangle \\
   &=&\displaystyle
   e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{\sqrt{n!}}\langle\mu|\hat{D}(\alpha)|n\rangle \\
   &=&\displaystyle
   e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}\langle\mu|\hat{D}(\alpha)(\hat{a}^\dagger)^n|0\rangle \\
   &=&\displaystyle
   e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}\langle\mu|\hat{D}(\alpha)(\hat{a}^\dagger)^n\hat{D}(\alpha)^\dagger\hat{D}(\alpha)|0\rangle \\
   &=&\displaystyle
   e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}\langle\mu|(\hat{a}^\dagger+\alpha^*)^n|\alpha\rangle \\
   &=&\displaystyle
   e^{-\frac{1}{2}|\nu|^2}\sum_{n\ge0}\frac{\nu^n}{n!}(\mu^*+\alpha^*)^n\langle\mu|\alpha\rangle \\
   &=&\displaystyle
   \exp\left(-\frac{1}{2}|\mu|^2-\frac{1}{2}|\nu|^2-\frac{1}{2}|\alpha|^2+\mu^*\alpha+\nu(\mu+\alpha)^*\right)
\end{array}
$$
where we used the facts that $\hat{D}(\alpha)^\dagger\hat{a}\hat{D}(\alpha) = \hat{a}+\alpha$ and $\langle\mu|\alpha\rangle = e^{-\frac{1}{2}|\mu|^2-\frac{1}{2}|\alpha|^2+\mu^*\alpha}$. Then you can integrate the components $D_{\mu\nu}(\alpha)$ with respect to time; in the case where $\alpha(t) = at+b$, you will be facing a gaussian integral.
