Inverse Laplace Transform of $1/(s+1)$ without table The pole is on the left half plane, so $\gamma =0$
$$\frac{1}{2i\pi}\int ^{i\infty}_{-i\infty}\frac {e^{st}}{s+1}ds$$
substituting $iu=s$
$$\frac{1}{2i\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}idu$$
(EDIT: I managed to get the answer $e^{-t}$ using feynman technique, but I dropped some conditional somewhere I guess.)I am unable to integrate this, however mathematica gives
$$\frac12  e^{-\left| t\right| } (\text{sgn}(t)+1)$$
which is correct for $t>0$ How do I get this result?
I also have a more general request:
I was unable to find a site which shows how to perform these inverse laplace transforms of simple functions, like $\frac{1}{s-a}$. When I search for inverse laplace transform, I either see the formula for it (which isn't all that clear to me right now) or a table. I would like to learn to how to do these transforms.
 A: Let continue your calculus :
$$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}du$$
$$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {(\cos(ut)+i \sin(ut))(-iu+1)}{u^2+1}du$$
$$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {\cos(ut)+u \sin(ut)}{u^2+1}du + \frac{i}{2\pi}\int ^{\infty}_{-\infty}\frac {-u\cos(ut)+ \sin(ut)}{u^2+1}du$$
$\int ^{\infty}_{\infty}\frac {-u\cos(ut)}{u^2+1}du =0$ (integral of odd function)
$\int ^{\infty}_{-\infty}\frac {\sin(ut)}{u^2+1}du =0$ (integral of odd function)
Then, with the known integrals :
$\int ^{\infty}_{-\infty}\frac {\cos(ut)}{u^2+1}du=2\int ^{\infty}_0\frac {\cos(ut)}{u^2+1}du=2\left(\frac{\pi}{2}e^{-t}\right)$  in $t>0$
$\int ^{\infty}_{-\infty}\frac {u \sin(ut)}{u^2+1}du=2\int ^{\infty}_0\frac {u \sin(ut)}{u^2+1}du=2\left(\frac{\pi}{2}e^{-t}\right)$  in $t>0$
Finally, in $t>0$ :
$$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}du=\frac{1}{2\pi}\left( 2\left(\frac{\pi}{2}e^{-t}\right) + 2\left(\frac{\pi}{2}e^{-t}\right) \right) = e^{-t}$$
.
Note : in my first edition, there was a mismatch of notations : In fact $y=t$.
