How to evaluate $\int_{c-i\infty}^{c+i\infty}\frac{\log(z)}{z}e^{zt}\,dz$ I am trying to compute
$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\log s}{s}e^{st}\,ds$$
in relation to
$$\mathcal{L}^{-1}\left\{\frac{-\gamma-\log s}{s}\right\}$$
which certainly evaluates to $\log t$. But the integrand "has no poles" according to Wolfram Alpha. So how could I compute this integral without using the residue theorem? Do I have to expand $\log s$ into it's power series involving the harmonic numbers? Or am I missing something? Thank you.
 A: First, note that the function $F(s)=\frac{\log(s)}{s}$ has a branch point at $s=0$.  Therefore, we choose the branch cut that extends from $s=0$ to $-\infty$.
Then, we deform the Bromwich contour with the classical keyhole contour along the negative real axis.  Applying Cauchy's Integral Theorem, we find that for $t>0$
$$\begin{align}
2\pi i\mathscr{L}^{-1}\{F\}(t)&=\lim_{\varepsilon\to 0^+}\left(\int_{-\infty}^{-\varepsilon}\frac{\log(s-i0^+)}{s}e^{st}\,ds\right.\\\\
&+\int_{-\pi}^\pi \frac{\log(\varepsilon e^{i\phi})}{\varepsilon e^{i\phi}}e^{\varepsilon e^{i\phi}t}\,i\varepsilon e^{i\phi}\,d\phi\\\\
&\left.-\int_{-\infty}^{-\varepsilon}\frac{\log(s+0^+)}{s}e^{st}\,ds\right)\\\\
&=i2\pi \lim_{\varepsilon\to 0^+}\left(\log(\varepsilon)+\int_\varepsilon^\infty \frac{e^{-st}}{s}\,ds+O\left(\varepsilon\log(\varepsilon)\right)\right)\\\\
&=i2\pi \lim_{\varepsilon\to 0^+}\left(\log(\varepsilon)(1-e^{-\varepsilon t})+\int_\varepsilon^\infty e^{-st}\log(s)\,ds\right)\\\\
&=i2\pi \int_0^\infty te^{-st}\log(s)\,ds\\\\
&=i2\pi \int_0^\infty e^{-s}(\log(s)-\log(t))\,ds\\\\
&=i2\pi(-\gamma-\log(t))
\end{align}$$
Dividing by $2\pi i$, we find that
$$\mathscr{L}^{-1}\{F\}(t)=-\gamma-\log(t)$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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The integration is performed by "closing" the contour with a key-hole one which takes care of the $\ds{\ln}$-branch cut along $\ds{\left(-\infty,0\right]}$. Namely,
\begin{align}
&\bbox[5px,#ffd]{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\ln\pars{s} \over s}\expo{ts}\,{\dd s \over 2\pi\ic}}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&
-\int_{-\infty}^{-\epsilon}
{\ln\pars{-s} + \ic\pi \over s}\expo{ts}\,{\dd s \over 2\pi\ic} -
\int_{-\pi}^{-\pi}{\ln\pars{\epsilon} + \ic\theta \over
\epsilon\expo{\ic\theta}}
\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 2\pi\ic}
\\[2mm] &
-\int_{-\epsilon}^{-\infty}
{\ln\pars{-s} - \ic\pi \over s}\expo{ts}\,{\dd s \over 2\pi\ic}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,&
\int_{\epsilon}^{\infty}
{\ln\pars{s} + \ic\pi \over s}\expo{-ts}\,{\dd s \over 2\pi\ic} +
\ln\pars{\epsilon}
\\[2mm] &\ -\int_{\epsilon}^{\infty}
{\ln\pars{s} - \ic\pi \over s}\expo{-ts}\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
\int_{\epsilon}^{\infty}
{\expo{-ts} \over s}\,\dd s + \ln\pars{\epsilon}
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
\bracks{-\ln\pars{\epsilon} -\int_{\epsilon}^{\infty}\ln\pars{s}\bracks{\expo{-ts}\pars{-t}}
\dd s} + \ln\pars{\epsilon}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\Large\to}\,\,\,&
t\int_{0}^{\infty}\ln\pars{s}\expo{-ts}\,\dd s =
t\bracks{\nu^{1}}\int_{0}^{\infty}s^{\nu}\expo{-ts}\,\dd s
\\[5mm] = &\
\bracks{\nu^{1}}t^{-\nu}\int_{0}^{\infty}s^{\nu}\expo{-s}\,\dd s
=
\bracks{\nu^{1}}t^{-\nu}\,\Gamma\pars{\nu + 1}
\\[5mm] = &\
-\ln\pars{t} + \Psi\pars{1} = \bbx{-\ln\pars{t} - \gamma} \\ &\
\end{align}
