Difficult Laplace inverse transform Let $$F(s)=-\frac{1}{s}+\arctan(s+2),\ \ \mathrm{with\ } \ \Re  (s)>0$$
I want to compute the inverse Laplace transform.
attempt: I need to find $f$ such that $L(f)(s)=F(s)$, i.e.,
$$\int_0^\infty e^{-ts}f(t)\ dt = -\frac{1}{s}+\arctan(s+2) $$
but how can I clear $f(t)$ from here? I know for example that if $F(s)=\frac{1}{s}$ for $s\geq 0$ then we should have $f(t)=1$, for $t\geq 0$.
 A: Let's consider the inverse Fourier transformation. By definition $L^{-1}(F)(x)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}F(s)\exp(sx)ds$, where the integral is evaluating along the line parallel the imaginary axis $Y$ and crossing the $X$ axis at some $a$ ($a$ - real positive number). $\arctan(s+2)$ has two branch points: $s=-2+i$ and $-2-i$. We can connect these brunch points by the cut; as soon as we do not cross the cut $\arctan(s+2)$ is a single-valued function.
We have to evaluate
$L^{-1}(F)(x)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\left(-\frac{1}{s}+\arctan(s+2)\right)\exp(sx)ds$
After integrating the second term by part we get
$L^{-1}(F)(x)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\left(-\frac{1}{s}-\frac{1}{x(1+(s+2)^2)}\right)\exp(sx)ds+\frac{1}{2\pi{i}x}\exp(sx)\arctan(s+2)|_{a-i\infty}^{a+i\infty}=$$=I_1+I_2+I_3$
To get $I_3$ we just have to take the limit:
$I_3=$$\frac{1}{2\pi{i}x}\lim_{N\to\infty}\left(\exp(ax+iNx)(-\frac{i}{2})\log\frac{1+i(a+2+iN)}{1-i(a+2+iN)}-\exp(ax-iNx)(-\frac{i}{2})\log\frac{1+i(a+2-iN)}{1-i(a+2-iN)}\right)$$=\frac{\exp(ax)}{2\pi{i}}(-\frac{i}{2})(\pi{i})\lim_{N\to\infty}\frac{\exp(iNx)-exp(-iNx)}{x}=\frac{\exp(ax)}{2}\lim_{N\to\infty}\frac{\exp(iNx)-exp(-iNx)}{2xi}=$$=\frac{\exp(ax)}{2}\lim_{N\to\infty}\frac{\sin(Nx)}{x}=\frac{\exp(ax)}{2}\pi\delta(x)=\frac{\pi}{2}\delta(x)$
To evaluate $I_1$ and $I_2$ we integrate along the line parallel the imaginary axis $Y$ (and crossing the $X$ axis at real and positive $a$), but we can close the contour along the part of big circle of radius R counter clockwise on the left side (and, partially, on the right side) of the complex plane. It can be easily evaluated that integrals along the parts of the circle (both in the left and right sides) $\to0$ as soon $R\to\infty$ (Jordan's lemma is applicable here, because integrand $\to0$ at $R\to\infty$). When evaluating $I_2$ we don't have to care about any cut - because we got rid of $\arctan$ and the integrand now is a single-valued function inside the contour.
$I_1=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\left(-\frac{1}{s}\right)\exp(sx)ds=Res_{s=0}(-\frac{1}{s}\exp(sx))=-1$
$I_2=-\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\frac{1}{x}\frac{\exp(sx)}{1+(s+2)^2}ds=$$-$$\frac{1}{x}(Res_{s=-2+i}\frac{\exp(sx)}{1+(s+2)^2}+Res_{s=-2-i}\frac{\exp(sx)}{1+(s+2)^2})=$$=-\frac{\exp(-2x)}{x}\frac{\exp(ix)-\exp(-ix)}{2i}=-\frac{\exp(-2x)\sin(x)}{x}$
$$L^{-1}(F)(x)=-1+\frac{\pi}{2}\delta(x)-\frac{\exp(-2x)\sin(x)}{x}$$
A: I have not studied inverse Laplace transforms much, so my knowledge on the matter is very limited. But giving Mathematica the code
InverseLaplaceTransform[-1/s + ArcTan[s + 2], s, t]

It outputs (in regular typesetting)
$$-1+e^{-2t}\left(\frac{\pi}{2}\delta(t)-\operatorname{sinc}(t)\right)$$
