Edit 1:
With the hint of Ron, we can simplify the question to :
$$\bar{f}(s)=\frac{1}{(s^2+1)\arctan s }$$
So what about this function's inverse Laplace Transform? Or can anyone tell me that the inverse transform is nonexist?
So the function is $$\bar{f}(s)=\frac{4\Omega}{(\pi-2\arctan\frac{2\Omega}{s})(s^2+4\Omega^2)}$$ with $\Omega>0$. I am expecting an ocillating function of angular frequency $2\Omega$ in the asymptotic form.
Using the Bromwich contour, I can find that we have a pole of first order at $s=0$. Two branch points $s=\pm 2i\Omega$, two poles also at $\pm 2i\Omega$.
I've got that $\mathrm{Res}f(0)=1$, and I've defined that the branch cut to be $(2i\Omega,i\infty)$ and $(-2i\Omega,-i\infty)$. However during the calculation I found there is divergence I don't know how to handle.
Is there any tool that can do this numerically, for example, set $\Omega=1$ and obtain a numerical result.
I would like to give a partial answer to my question. The inverse fourier transformation can be obtained using the Bromwich integral:
$$f(t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{e^{st}}{(s^2+1)\arctan s } \mathrm{d}s$$
Define $g(s)$ for writing convenience:
$$g(s)=\frac{e^{st}}{(s^2+1)\arctan s } $$
To integrate this, we choose a contour as below:
Note that we have a pole at the origin and two branch points at $\pm i$.
Using this contour, we can apply cauchy theorem: $$\mathrm{Res}g(0)=f(t)+\frac{1}{2\pi i}(\oint_{c_1}+\cdots\oint_{c_4}+\oint_\Gamma+\oint_{\gamma_1}+\oint_{\gamma_2})g(s)\mathrm{d}s$$
$$\oint_{\gamma_1}g(s)\mathrm{d}s=\int_{\frac{\pi}{2}}^{-\frac{3\pi}{2}}\frac{e^{(i+\varepsilon e^{i\phi})t}\mathrm{d}\phi}{2\arctan(i+\varepsilon e^{i\phi})}\to 0$$ because $\arctan i \to \infty$. Similarly $\oint_{\gamma_2}\to 0$.
Integration along $c_1\cdots c_4$:
Because we have the branch cuts, we need only to consider the principal value of $\arctan$. The principal value is denoted as the first letter to be capital.
\begin{align} \mathrm{Arctan}(z)=&\frac{1}{2i}\left[\mathrm{Ln}(1+iz)-\mathrm{Ln}(1+iz)\right] \\ =& \begin{cases} \frac{1}{2i}\mathrm{Ln}\left|\frac{1-y}{1+y}\right| +\frac{\pi}{2},& \text{if $z$ on $c_1$and $c_4$}, \text{ i.e.,when $x\to 0^+$ and $|y|>1$} \\ \frac{1}{2i}\mathrm{Ln}\left|\frac{1-y}{1+y}\right| -\frac{\pi}{2},& \text{if $z$ on $c_2$and $c_3$}, \text{ i.e.,when $x\to 0^-$ and $|y|>1$} \end{cases} \end{align}
So:
$\oint_{c_1}$ $$\oint_{c_1}g(s)\mathrm{d}s=\int_{+\infty}^{1}\frac{ie^{iyt}\mathrm{d}y}{(1-y^2)(\left(\frac{1}{2i}\mathrm{Ln}\left|\frac{1-y}{1+y}\right|+\frac{\pi}{2}\right)}$$
$\oint_{c_2}$ $$\oint_{c_2}g(s)\mathrm{d}s=\int_{1}^{+\infty}\frac{ie^{iyt}\mathrm{d}y}{(1-y^2)(\left(\frac{1}{2i}\mathrm{Ln}\left|\frac{1-y}{1+y}\right|-\frac{\pi}{2}\right)}$$
$\oint_{c_3}$ $$\oint_{c_3}g(s)\mathrm{d}s=\int_{-\infty}^{-1}\frac{ie^{iyt}\mathrm{d}y}{(1-y^2)(\left(\frac{1}{2i}\mathrm{Ln}\left|\frac{1-y}{1+y}\right|-\frac{\pi}{2}\right)}$$
$\oint_{c_4}$ $$\oint_{c_4}g(s)\mathrm{d}s=\int_{-1}^{-\infty}\frac{ie^{iyt}\mathrm{d}y}{(1-y^2)(\left(\frac{1}{2i}\mathrm{Ln}\left|\frac{1-y}{1+y}\right|+\frac{\pi}{2}\right)}$$
sum up: $$\oint_{c_1}+\cdots+\oint_{c_4}=2\pi i \int_1^\infty \frac{4\cos xt}{(x^2-1)\left[\left(\ln\left|\frac{1-x}{1+x}\right|\right)^2+\pi^2\right]}\mathrm{d}x$$ Where I've change the principal value $\mathrm{Ln}$ to the normal $\ln$, because the argument is positive real, we have no confusion about which branch we are using. i.e. we are back to the real argument integral.
The asymptotic form of the above integration?
Now the whole problem is how can we get the asymptotic form of the above integral. Does it even exist? If exist, what form it takes when $t$ is very large? I've post this question to a seperate post: How to get the asymptotic form of this oscilatting integral?