How to solve differential equation involving an integral The integro-differential equation takes the form 
$$c′(t)+iac(t)+\int_{0}^{\infty} f(t−\tau)c(\tau)d\tau=0$$
in which
$$f(t−\tau)=\frac{\pi}{4}∫_{0}^{\infty} J(\omega)e^{−i\omega(t−\tau)} d\omega,$$
$$J(ω)=2\pi \alpha \omega_{c}^{1−b}\omega^{b}\Theta(\omega_{c} −\omega),$$ 
and $a,\alpha,\omega_{c},b>0$ are all real constants. $\Theta(x)$ is the usual step function. To solve this equation, it seems the Laplace transform does not work.
 A: It seems to me that Laplace transform leads to a dead-end indeed. So I am posting an community-wiki answer, anyone is welcome to edit it.
Let's replace your constant $s$ by $b$ here. Taking Laplace transform of:
$$c'(t)+iac(t)+\int_{0}^{\infty} f(t−\tau)c(\tau)d\tau=0$$
gives:
$$
\newcommand{\L}[1]{\mathcal{L}\{#1\}}s\L{c}-c(0) + ia \L{c} + \L{f}\cdot \L{c} = 0.\tag{1}
$$
Now onto computing $\L{f}$, first rewritting the $f(t)$  as:
$$
f(t)=\frac{\alpha\pi^2}{2}\int_{0}^{\infty} \omega_{c}^{1−b}\omega^{b}\Theta(\omega_{c} −\omega)e^{−i\omega t} d\omega.
$$
The step function is:
$$
\Theta(\omega_{c} −\omega)=\left\{\begin{aligned}
1\quad \text{when } \omega_{c} −\omega\ge 0,
\\
0\quad \text{when } \omega_{c} −\omega <0.
\end{aligned}\right.
$$
Hence, $f$ is:
$$
f(t)=\frac{\alpha\pi^2}{2}\int_{0}^{\omega_c} 
\omega_{c}^{1−b}\omega^{b}e^{−i\omega t}\, d\omega.
$$
The Laplace transform is with respect to $t$, and we can interchange it with the integral with respect to $\omega$ for everything is smooth here:
$$
\L{f}(s) = \frac{\alpha\pi^2}{2}\omega_{c}^{1−b}\int_{0}^{\omega_c} \omega^{b} 
\L{e^{−i\omega t} }(s)\,d\omega = \frac{\alpha\pi^2}{2}\omega_{c}^{1−b}\int_{0}^{\omega_c} \omega^{b} 
\frac{1}{s-i\omega}\,d\omega.
$$
For general $b>0$ (the $s$ in the question), the integral
$$
\int_{0}^{\omega_c} \omega^{b} 
\frac{1}{s-i\omega}\,d\omega
$$
involves special function. (WolframAlpha gives hypergeometric function) 
If $b=1,2,\ldots$, we can have a closed form of above integral. Let's try letting $b=1$: 
$$
\L{f}(s) = \frac{\alpha\pi^2}{2} \left(\omega_c + is\ln\Big|\frac{s}{s-i\omega_c} \Big|\right).
$$
Plugging back to (1):
$$
\L{c} = \frac{c(0)}{s + ia + \L{f}},
$$
and it doesn't look like we can get a reasonable inverse Laplace transform out of the expression, I am suggesting do certain qualitative analysis of this equation and then numerically solve it if you would like to know what the solution is like for different parameters.  
