Discontinuous integral $\frac1{2\pi i}\int_{c-i \infty}^{c+i\infty}y^s\frac{ds}{s}$ The book tells me to use the following integral,
$$
\frac{1}{2\pi i}\int_{c-i \infty}^{c+i\infty}y^s\frac{ds}{s}=
\begin{cases}
0\quad&\text{if }0<y<1,\\
\frac{1}{2}&\text{if }y=1,\\
1&\text{if }y>1,
\end{cases}
$$
where $c>0$. I don't need to prove this but I wanted to make sense of this integral. Here's my (pseudo)proof for the first case.
Consider a rectangular path consisting of $c+iT$, $c-iT$, $c+S-IT$ and $c+S+IT$ for some $T,S>0$. Call each path, starting from $c+iT$, $C_1,C_2,C_3$ and $C_4$ respectively - so $C_1$ and $C_3$ are two vertical paths and the other two are horizontal paths. If we integrate $\frac{y^s}{s}$ over this rectangle, the result is 0 since there is no pole or zero inside. As $S,T\to\infty$, integrals along $C_2,C_3,C_4$ vanish since $\left|\frac{y^s}{s}\right|\to0$. Therefore, integral along $C_1$ is also 0 as $S,T\to\infty$.
Now, I know I can use pretty much the same proof for $y>1$ case just by taking a rectangle extending to the left this time. However, I'm having a little bit if of difficulty to prove the case $y=1$ Can anyone help me? (Also, it'd be great if someone can tell me my proof is right)
 A: When $y=1$ you just have $\frac{1}{2\pi i} \left. \ln(s) \right |_{c-i\infty}^{c+i\infty}$. In the sense of Cauchy principal value at least, this evaluation is just $\pi i$ giving an overall result of $1/2$. Without some such regularization you could have a real part persisting, which is not a surprise because the integral is not absolutely convergent.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \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)}
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When $\large\ds{\color{red}{y = 1}}$, you can write $\ds{\pars{~\mbox{with}\ s \equiv c + \ic t~}}$:
\begin{align}
\mc{I} &\equiv \left.{1 \over 2\pi\ic}\int_{-a}^{b}{\ic\,\dd t \over c + \ic t}
\,\right\vert_{\ a,b,c\ >\ 0} =
{1 \over 2\pi}\int_{-a}^{b}{c - \ic t \over t^{2} + c^{2}}\,\dd t =
{1 \over 2\pi}\int_{-a/c}^{b/c}{1 - \ic t/c \over t^{2} + 1}\,\dd t
\\[5mm] & =
{\arctan\pars{b/c} + \arctan\pars{a/c}\over 2\pi} - {\ic \over 4\pi c}
\ln\pars{b^{2} + c^{2} \over a^{2} + c^{2}}
\end{align}

Note that $\ds{\Re\pars{\mc{I}} \to {1 \over 2}}$ as $\ds{a,b \to \infty}$. However, the imaginary part diverges in such case.

