I have a doubt about Laplace antitransform and Jordan lemma.
Let's say I have a certain Laplace transform $U(s)$ and want to compute the antitransform by means of the formula $$u(t)\theta(t)=\frac{1}{2\pi i}\int_{r-i\infty}^{r+i\infty} U(s)e^{st}ds, \space\space\space r>\alpha_{0}$$ where $\alpha_{0}$ is the convergence abscissa.
What I usually do is to use the Jordan lemma to transform the integral along the parallel to the imaginary axis $s=r$ in a contour integral and use the residue theorem to calculate $u(t)$ explicitely, completing the path once for $t<0$ on the right of $s=r$ (obtaining zero) and once on the left (obtaining $u(t)\theta(t)$).
My problem arises with what is written in my course teaching material. There it is stated the following about the Jordan lemma:
a) If $U(s)\in\omicron(1),s\rightarrow\infty$ uniformly with respect to the argument of $s$ in $\frac{\pi}{2}\leq\ Arg(z)\leq\frac{3\pi}{2}$, and $t>0$, then $\lim_{R\to\infty}\int_{\gamma_R}U(s)e^{st}=0$, where $\gamma_R$ is a semicirconference of radius $R$ on the left of $s=r$.
b) If $U(s)\in\omicron(1),s\rightarrow\infty$ uniformly with respect to the argument of $s$ in $\frac{3\pi}{2}\leq\ Arg(z)\leq2\pi\cup 0\leq\ Arg(z)\leq\frac{\pi}{2}$, and $t<0$, then $\lim_{R\to\infty}\int_{\gamma_R}U(s)e^{st}=0$, where $\gamma_R$ is a semicirconference of radius $R$ on the right of $s=r$.
I only know a theorem that guarantees that $U(s)$ is $o(1),s\rightarrow\infty$ for all directions on the semiplane on the right of the convergence abscissa, and that's all right, that's the case of b), I complete the path with a semicircunference on the right for $t<0$ and since there are no singularities the integral is $0$.
What I don't understand is why for $t>0$ I can apply the lemma completing the path on the left since in a) is required for $U(s)$ to be $o(1),s\rightarrow\infty$ uniformly on the left semiplane, while I only know that $U(s)$ is such in the right one.