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Is it possible to find the inverse laplace transform $$\mathcal{L}^{-1}\frac{\log(s)}{1 + s}$$ using the Bromwich integral formula $$\mathcal{L}^{-1} \{F(s)\}(t) = f(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{st}F(s)\,ds$$ I'm having trouble coming up with a suitable contour to use. If the denominator were $1 - s$, then the pole would be in the right hand plane and the residue theorem would reduce the integral to the sum of the residues plus the integral around the branch cut. But with this one, the pole is in the left hand plane where the branch cut should be. Instead of setting $\gamma = 1$ as it must be in the case where the denominator is $1 - s$, can I set $\gamma = 0$, run the contour down the real axis and detour around the origin? I'm not sure if this is allowed, or will work. Thanks for any advice.

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You should be able to let the cut be the ray from the origin to $e^{i(\pi + \epsilon)}\infty$, for example, so you end up with an integral around the pole at $s=-1$ and another around the cut. – user26872 Dec 29 '12 at 6:43
up vote 3 down vote accepted

The result below assumes $t>0$ (as usual for Laplace transforms).

In the Bromwich contour $\gamma$ has to be chosen large enough that it is to the right of all singularities (poles and branch points) so $\gamma = 0^+$ is perfectly valid. The singularities of $\log s/(1+s)$ are a pole at $s=-1$ and two branch points at $s=0$ and $s=\infty$ which we connect via a branch cut along the negative real line.

Then we can deform the contour further to a path which starts at $-\infty -i 0^+$. Runs along the negative real line just below the branch cut. Ends at $0-i 0^+$ in a little semi-circle and then runs back from $0+i 0^+$ to $-\infty +i0^+$ just above the branch cut.

The Bromwich integral thus is given by $$f(t)=\frac1{2\pi i} \int_{-\infty}^0\!dx\, \left( \frac{\log (x-i0^+)}{1+x-i0^+ } - \frac{\log (x+i0^+)}{1+x+i0^+ } \right) e^{x t} $$ as the small circle around the branch point at $0$ does not contribute ($|z|\log z \to 0$ for $|z|\to0$).

In the remaining integral, we use $\log(x \pm i 0^+) = \log |x| \pm i \pi$ valid for $x<0$: $$\begin{align} f(t) &= \frac1{2\pi i} \int_{-\infty}^0\!dx\, \left( \frac{\log |x|-i\pi}{1+x-i0^+ } - \frac{\log |x|+i\pi}{1+x+i0^+} \right) e^{x t}\\ &= \frac1{2\pi i} \overbrace{\int_{-\infty}^0\!dx\, \log |x| \underbrace{\left( \frac1{1+x-i0^+ } - \frac1{1+x+i0^+}\right)}_{2\pi i \delta(x+1)} e^{x t}}^{=0}\\ &\quad -\frac12 \int_{-\infty}^0\!dx \underbrace{\left( \frac1{1+x-i0^+ } + \frac1{1+x+i0^+}\right)}_{2\mathcal{P}\,(1+x)^{-1}} e^{x t} \\ &= -\int_{-\infty}^0\!dx \,\mathcal{P} \frac{e^{x t}}{1+x} =- e^{-t} \int_{-\infty}^t\!ds \,\mathcal{P} \frac{e^{s}}{s}\\ &=- e^{-t} \mathop{\rm Ei}(t) \end{align}$$ with $s=(1+x)t$ and Ei the exponential integral.

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Nice work. (+1) – user26872 Dec 29 '12 at 18:50
Thanks very much! – Bitrex Dec 30 '12 at 7:27
@oen: thanks. Of course once one knows the solution there are simpler ways to obtain it (e.g., it is much simpler to check that it is the right result than to obtain it in the first place). – Fabian Dec 30 '12 at 8:15

$$ f(t)=\mathcal{L}^{-1} \left[ \frac{\log(s)}{s+1} \right]$$ $$ f(f) \space \underleftrightarrow{\mathcal{L}} \space F(s) \hspace{10mm} e^{-at}f(t) \space \underleftrightarrow{\mathcal{L}} \space \frac{\log(s)}{s+1} $$ $$ e^{t}f(t)=\mathcal{L}^{-1} \left[ \frac{\log(s-1)}{s} \right] \hspace{10mm} \frac{1}{s} \triangleq \int_0^t $$ $$ e^tf(t)=\int_0^t \mathcal{L^{-1}} \left[ \log(s-1) \right] d \tau \hspace{10mm} -t.f(t) \space \underleftrightarrow{\mathcal{L}} \space \frac{d}{ds}F(s)$$ $$G(s=)\mathcal{L^{-1}} \left[ \log(s-1) \right] \rightarrow -t.g(t)=\mathcal{L} \left[ \frac{1}{s-1} \right] \rightarrow g(t)=\frac{-e^t}{t}$$ $$f(t)=e^{-t} \int_0^t \frac{-e^{\tau}}{\tau} d \tau = \int_0^t \frac{-e^{-(t-\tau)}}{\tau} d \tau$$

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Something is wrong here, the last integral doesn't converge. – mrf Dec 29 '12 at 11:13

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