Given $\frac{e^{st}}{s\log s}$ where we choose the branch cut to lie on the negative real axis I am trying to find the integral $$\int_C \frac{e^{s y}}{s \log s} ds$$ where $C$ denotes the contour that starts at $-\infty$ above the real axis, wraps around the origin (not surpassing $1$),then loops back to $-\infty$ below the real axis all with increasingly small distance to $\mathbb{R}^-$.
Trying to show that $$\int_C \frac{e^{s y}}{s \log s} ds=2\pi i \int_0^\infty \frac{e^{-y x}}{x\{\pi^2+\log^2x\}}dx$$ I started of with the line integral above $\mathbb{R}^-$ like so: \begin{align*} I^r_+ &=\int_{-\infty+ir}^{0+ir} \frac{e^{s y}}{s\log s} ds\\ &=\int_{-\infty}^0 \frac{e^{(s+ir)y}}{(s+ir)\log (s+ir)} ds \end{align*} For the line integral below $\mathbb{R}^-$ this then leads to: \begin{align*} I_-^r &=\int_{0-ir}^{-\infty-ir} \frac{e^{s y}}{s\log s} ds\\ &=\int_0^{-\infty} \frac{e^{(s-ir)y}}{(s-ir)\log (s-ir)} ds\\ &=\left[-\int_{-\infty}^0 \frac{e^{(s+ir)y}}{(s+ir)\log (s+ir)} ds\right]^\times\\ &=\left[-I_+^r\right]^\times \end{align*} Where $[\cdot]^\times$ denotes the complex conjugate. We now have: $$I_+^r+I_-^r=2i\;\Im\left(I_+^r\right)$$ Calculating the imaginary part of $e^{(s+ir)y}/\{(s+ir)\log (s+ir)\}$ yields: $$\frac{e^{(s+ir)y}}{(s+ir)\log (s+ir)}\sim \frac{-e^{sy}\pi}{s(\frac14 \log(s^2)+\pi^2)}$$ as $r\rightarrow 0$. By substituting $s\mapsto-s$ it follows that: \begin{align*} I^0_++I^0_- &=2i\int_{-\infty}^0 \frac{-e^{sy}\pi}{s(\frac14 \log(s^2)+\pi^2)} ds\\ &=2\pi i \int_0^\infty \frac{e^{-sy}}{s(\log^2(s)+\pi^2)} ds \end{align*} Now for the semicircle, starting at $0+ir$ and ending at $0-ir$: \begin{align*} I^r_{arc} &=\int_{\pi/2}^{-\pi/2} \frac{e^{re^{i\theta}y}}{re^{i\theta}\log(re^{i\theta})}ire^{i\theta}d\theta\\ &=\int_{\pi/2}^{-\pi/2} \frac{e^{re^{i\theta}y}}{\log(r)+i\theta} d\theta\\ \end{align*} So $I^r_{arc}\sim 0$ as $r\rightarrow 0$.
This together yields: \begin{align*} \int_C \frac{e^{s y}}{s \log s} ds &=I^0_++I^0_-+I^0_{arc}\\ &=2\pi i \int_0^\infty \frac{e^{-y x}}{x\{\pi^2+\log^2x\}}dx \end{align*} This works but is quite ugly, as the mentioned imaginary part is tideous to calculate and many "interchange limit and integral" arguments have to be made.
Do you know of another, more elegant way to show this?
Regards :)