inverse Laplace transform by integral

Question

I've seen this formula for the inverse Laplace transform in several books: $$f(t)=\mathcal{L}^{-1}\{F\}(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{\alpha -iT}^{\alpha +iT}e^{st}F(s)ds$$ where $f$ is the original function, $F$ is its Laplace transform, $\alpha\geq 0$ is bigger than the real part of every singularity of $F$, and $t>0$ is fixed. I'm trying to write this contour integral out as an ordinary integral but it's not working. Here is a simple example of my problem with $f(t)=e^{-t}$ and $F(s)=\int_0^{\infty}e^{-st}e^{-t}dt=\frac{1}{1+s}$. Since the singularities of $F$ are in the left half-plane we can take $\alpha=0$. Now the formula for the inverse Laplace transform becomes $$f(t)=\mathcal{L}^{-1}\{F\}(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{-iT}^{iT}e^{st}F(s)ds=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{-T}^{T}e^{ist}F(is)ds.$$ Substituting in $f$ and $F$ and computing the integral in Mathematica gives $$e^{-t}=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{-T}^{T}e^{ist}\frac{1}{1+is}ds=-ie^{-t}.$$ What went wrong?

Answer 1

When you changed $\frac{1}{2\pi i}\lim_{T\to\infty}\int_{-iT}^{iT}e^{st}F(s)ds$ to $\frac{1}{2\pi i}\lim_{T\to\infty}\int_{-T}^{T}e^{iut}F(iu)du$ (I've replaced $s$ with $u$ is the second part to make this clear) you did the substitution $s= iu$, but you did not do $ds=i\cdot du$, which is why you get $-ie^{-t}$. Multiplying the limit by $i$ obtains the correct answer of $e^{-t}$.