Inverse Laplace Transform of $\frac{1}{s+1}$ using Mellin's Inverse Formula I am trying to compute the inverse Laplace Transform of $F(s)=\frac{1}{s+1}.$ I know that it is supposed to be $f(t)=e^{-t}$, but I am specifically trying to get that result using Mellin's Inverse Formula, according to which:
$$f(t)=\mathcal{L}^{-1}[F(s)](t)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{e^{st}}{s+1}\ ds$$
According to the formula, I need to choose $\sigma>-1$ (since $s_0=-1$ is the singularity of $F(s)$ and $\Re{\{s_0\}=-1}$) and to integrate on the line $\Re{\{s\}}=\sigma$. I chose $\sigma=0$, thus a proper parametrization would be $s=ix$ when $x\in\mathbb{R}$:
$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{ixt}}{1+ix}\ dx$$
(I also checked that $F(s)$ is bounded on the line, and it is indeed).
I have no idea how to compute this integral. If I were to use the Residue Theorem, I would have ended up with a complex integral again.
Another problem that I have: It doesn't even seem like the integral is convergent. Plugging $t=0$, the result of the integral should be $1$, but the integral I got doesn't seem convergent at all:
$$f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{1+ix}\ dx=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1-ix}{1+x^2}\ dx=\frac{1}{2\pi}\left[\left.\arctan(x)\right|_{-\infty}^{\infty}-\frac{i}{2}\left.\ln(1+x^2)\right|_{-\infty}^{\infty}\right]$$
The integral does not converge because of the imaginary part (the logarithm). But, even if for some reason I should take only the real part, it wouldn't correlate with the desired answer, since I would get $\frac 12$ instead of $1$.
I must be missing something here.
 A: Partial answer.
The proper inverse Laplace transform is:
$$f(t) = \mathcal{L}^{-1} \{F(s)\}(t) =  \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\sigma-iT}^{\sigma+iT}e^{st}F(s)\,ds$$
And the desired inverse transform for $\frac{1}{s+1}$ is:
$$f(t) = \mathcal{L}^{-1} \left\{\frac 1{s+1}\right\}(t)=e^{-t}u(t)$$
where $u(t)$ is the unit step or Heaviside step function. Note that $f(t)$ must be $0$ for negative $t$.
For the case $t=0$ with $\sigma=0$ and $s=ix$ we get:
$$f(0) = \frac{1}{2\pi i}\lim_{T\to\infty}\int_{-iT}^{iT}\frac{1}{s+1}\,ds
=\frac{1}{2\pi}\lim_{T\to\infty}\left[\arctan(x)-\frac{i}{2}\ln(1+x^2)\right]_{-T}^{T} = \frac 12$$
As you can see, the imaginary part cancels as you already surmised.
Since $f$ must be $0$ for negative $t$, it has a discontinuity at $0$ where it jumps from $0$ to $1$. So any value for $f(0)$ that is between $0$ and $1$ is acceptable as solution.
One of the possible conventions for the unit step function is the half-maximum convention:
$$u(t)=\begin{cases}0 & \text{if } t<0 \\ \frac 12 &\text{if }t=0\\ 1&\text{if }t>0\end{cases}$$
It ensures that the inverse Laplace transform "pans out" in our case at $t=0$.
