Approximation of the Laplace Transform of $1 / (1+t^2)$

Question

Is there a good approximation of the Laplace transform of \begin{equation} \mathcal{L}\left\{\frac{1}{1+t^2}\right\}=\int_0^\infty \frac{1}{1+t^2} e^{-st}~\mathrm{d}t = \mathrm{Ci}(s) \sin(s) + 1/2 \cos(s) (\pi - 2 \mathrm{Si}(s)) \end{equation} available in analytical form for real valued $s$?

Answer 1

If you are concerned by rather small values of $s$, you can use the series expansion $$ \mathrm{Ci}(s) \sin(s) +\frac 12 \cos(s) (\pi - 2 \mathrm{Si}(s))= \frac \pi 2 \sum_{n=0}^p(-1)^n\frac {s^{2n}}{(2n)!}+\sum_{m=0}^q (-1)^{m+1}\,T_m \,s^{2m+1}$$ where the first $T_m$ are $$\left(\begin{array}{cc} m & T_m \\ 0 & -\log (s)-\gamma +1 \\ 1 & \frac{-6 \log (s)-6 \gamma +11}{36} \\ 2 & \frac{-60 \log (s)-60 \gamma +137}{7200} \\ 3 & \frac{-140 \log (s)-140 \gamma +363}{705600} \\ 4 & \frac{-2520 \log (s)-2520 \gamma +7129}{914457600} \\ 5 & \frac{-27720 \log (s)-27720 \gamma +83711}{1106493696000} \\ 6 & \frac{-360360 \log (s)-360360 \gamma +1145993}{2243969215488000} \\ \end{array}\right)$$ Using the above terms, that is to say the expansion to $O(s^{14})$, the results seem to be decent up to $s=4$. At this point, the error is $0.0035$.