Discontinuous integral $\frac1{2\pi i}\int_{c-i \infty}^{c+i\infty}y^s\frac{ds}{s}$

Question

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)

Answer 1

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.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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.