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.
[Math] Inverse Laplace transform of $\frac{\log(s)}{1 + s}$
complex-analysiscontour-integrationlaplace transform
Related Solutions
There is no pole at $z=-1$; it is merely a branch point. The Bromwich contour from which the ILT may be found must be deformed so as to avoid this branch point, like this:
You may show that the integrals over $C_2$, $C_4$, and $C_6$ all vanish. The result is, letting $z=-1+e^{i \pi} u$ on $C_3$ and $z=-1+e^{-i \pi} u$ on $C_5$,
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} + e^{-i \pi/2} \int_{\infty}^0 du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} \\ + e^{i \pi/2} \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} = i 2 \pi$$
as the residue at the pole $z=0$ is $1$.
From this, you may rearrange to get that the ILT is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} = 1-\frac1{\pi} \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} $$
The integral may be evaluated by differentiating with respect to $t$ and subbing $u=v^2$. The result, which I leave to the reader, is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{s \sqrt{1+s}} = \operatorname{erf}{\sqrt{t}} $$
ADDENDUM
A little more detail on the evaluation of the integral on the RHS above. Let
$$I(t) = \int_0^{\infty} du \frac{e^{-(1+u) t}}{(1+u) \sqrt{u}} = \int_{-\infty}^{\infty} dv \frac{e^{-t (1+v^2)}}{1+v^2}$$
Then
$$I'(t) = -\int_{-\infty}^{\infty} dv\, e^{-t (1+v^2)} = \sqrt{\pi} t^{-1/2} e^{-t} $$
$$\implies I(t) = I(0) - \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{-t'} = \pi - 2 \sqrt{\pi} \int_0^{\sqrt{t}} du \, e^{-u^2} = \pi - \pi \, \operatorname{erf}{\sqrt{t}}$$
The result follows.
It would be the situation in B: you would deform around the pole. It works as follows.
The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole $z=-1$ and the branch point $z=0$ is $\epsilon$:
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + e^{i \pi} \int_{\infty}^{1+\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}\\+ e^{i \pi} \int_{1-\epsilon}^{\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t \epsilon e^{i \phi}}}{\sqrt{\epsilon e^{i \phi}} (1+\epsilon e^{i \phi})} +e^{-i \pi} \int_{\epsilon}^{1-\epsilon} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{-i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}+ e^{-i \pi} \int_{1+\epsilon}^{\infty} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)} = 0$$
Note that the integrals about the semicircular detours above and below the axis (the 3rd and the 7th integrals) cancel. In the limit as $\epsilon \to 0$, the integral about the branch point (the 5th integral) also vanishes. We are then left with, as $\epsilon \to 0$,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + \frac1{2 \pi} PV \int_{\infty}^0 dx \frac{e^{-t x}}{\sqrt{x} (1-x)} - \frac1{2 \pi} PV \int_0^{\infty} dx \frac{e^{-t x}}{\sqrt{x} (1-x)} = 0$$
where $PV$ denotes the Cauchy principal value of the integral. Thus, the ILT is given by (subbing $x=u^2$)
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = \frac1{\pi} PV \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{1-u^2} $$
To evaluate the integral, we rewrite as
$$e^{-t} PV \int_{-\infty}^{\infty} du \, \frac{e^{t (1- u^2)}}{1-u^2} = e^{-t} I(t)$$
where
$$I'(t) = e^{t} PV \int_{-\infty}^{\infty} du \, e^{-t u^2} = \sqrt{\pi} t^{-1/2} e^{t} $$
and $I(0) = 0$. Thus,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t}\frac1{\pi} \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{t'} = e^{-t} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} dv \, e^{v^2} $$
or, finally,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t} \operatorname{erfi}{\left (\sqrt{t} \right )} $$
Best Answer
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.