I am trying to find the inverse Laplace transform of $\ln\left(s^{2}+1\right)$ using contour integration, Which could be found by:
$$ \frac{1}{2i\pi}\int_{a-i\infty}^{a+i\infty}\ln\left(s^{2}+1\right)e^{st}ds $$
As the function has two branch points, at $s = \pm i$, I considered the contour:
Whereat the end I get that the integral of the arc vanishes, and the integrals over the negative real part of the contour cancel each other out. I'm just left with the integrals over the branch points, which I'm going to call $ \alpha_{1} $ and $ \alpha_{2} $.
For $ \alpha_{1} $ I get $ \int_{0}^{i}\left[\ln\left(s^{2}+1\right)+\frac{5i\pi}{2}\right]e^{st}ds + \int_{i}^{0}\left[\ln\left(s^{2}+1\right)+\frac{i\pi}{2}\right]e^{st}ds $ which is equal to $ 2i\pi\int_{0}^{i}e^{st}ds $
For $ \alpha_{2} $ I get $ \int_{0}^{-i}\left[\ln\left(s^{2}+1\right)-\frac{i\pi}{2}\right]e^{st}ds + \int_{-i}^{0}\left[\ln\left(s^{2}+1\right)+\frac{3i\pi}{2}\right]e^{st}ds $ which is equal to $ -2i\pi\int_{0}^{-i}e^{st}ds $
In my calculations, the semi-circle integrals over +i and -i are 0, so at the end my final answer would be $ \int_{\alpha_{1}}^{ }+\int_{\alpha_{2}}^{ } $, and I get $ -\frac{2i\pi}{t}\left(e^{it}-e^{-it}\right) $. As the final answer for the inverse Laplace I get $ \frac{-2i}{t}\sin\left(t\right) $, which is not the correct answer. My question is what am I missing in the countour? Maybe I set up the arguments of $ \ln\left(s^{2}+1\right) $ wrong, or maybe the integrals over the real part does not cancel out, any help or hint would be welcome, thank you!
The correct answer is $ \frac{2\cos\left(t\right)}{t} $
Best Answer
Something's definitely wrong in the question. Assume that $\mathcal{L}^{-1}\log(1+s^2) = f(x)$. Then
$$ (\mathcal{L}f)(s)=\int_{0}^{+\infty}f(x) e^{-sx} dx = \log(s^2+1) $$
so by considering the limit as $s\to 0^+$ we have $\int_{0}^{+\infty}f(x) dx = 0$, no particular concern. The actual concern comes from the fact that $\log(s^2+1)$ is increasing and diverges as $s\to +\infty$, while usually $\int_{0}^{+\infty}f(x) e^{-sx} dx$ converges to $0$. By differentiating with respect to $s$ we have
$$ \int_{0}^{+\infty}\left( x f(x) \right) e^{-sx} = -\frac{2s}{s^2+1} $$
so the only chance is $x f(x) = -2\cos(x)$, or $f(x)=-2\frac{\cos(x)}{x}$. However this function has a simple pole at the origin, hence $\int_{0}^{+\infty}f(x) e^{-sx}\,dx $ does not converge for any $s\in\mathbb{R}$ and $\log(s^2+1)$ is not a Laplace transform.
There are several criteria for establishing that a function is not a Laplace transform, also depending on the chosen domain for $\mathcal{L}$. Even without considering the derivative as I did above, $\log(s^2+1)$ does not meet criterion (3) here (Paley-Wiener-type) - this is essentially the objection in the first comment.
Another simple criterion is that any measurable $f$ with a sub-exponential growth is such that $(\mathcal{L}f)(s)$ converges to zero as $s\to +\infty$. Additionally, if $f\in L^2(\mathbb{R}^+)$ is non-negative, its Laplace transform is log-convex. In particular the Laplace transform of a function in $L^2(\mathbb{R}^+)$ can be written as $e^{g_1(s)}-e^{g_2(s)}$ with $g_1,g_2$ being convex functions with the same limit as $s\to +\infty$.