[proofplan] Write $y^s=e^{s\log y}$ and split the proof according to the sign of $\log y$. If $y>1$, close the vertical segment to the left; the exponential factor decays on the far left side, and the pole at $s=0$ contributes residue $1$. If $0<y<1$, close to the right; there is no pole in the rectangle, and the exponential factor decays on the far right side. The boundary case $y=1$ is computed directly from the symmetric principal value of $1/s$. [/proofplan]

[step:Introduce the meromorphic integrand and its residue at the origin] Let \begin{align*} L := \log y. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. For each $T>0$, define the symmetric truncation \begin{align*} I_T(y,c) := \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{y^s}{s}\,ds. \end{align*} Define the [meromorphic function](/page/Meromorphic%20Function) \begin{align*} F_y: \mathbb{C}\setminus\{0\} &\to \mathbb{C} \\ s &\mapsto \frac{e^{Ls}}{s}. \end{align*} The only singularity of $F_y$ is a simple pole at $s=0$. Since \begin{align*} e^{Ls} = 1 + Ls + O(s^2) \end{align*} as $s \to 0$, the Laurent expansion of $F_y$ at $0$ is \begin{align*} F_y(s) = \frac{1}{s} + L + O(s), \end{align*} so \begin{align*} \operatorname{Res}(F_y,0)=1, \end{align*} where $\operatorname{Res}(F_y,0)$ denotes the residue of $F_y$ at the isolated singularity $0$. We shall use the [Cauchy Residue Theorem](/theorems/??? ) for rectangles. [/step]

[step:Close the contour to the left when $y>1$] Assume $y>1$, so $L>0$. Fix $T>0$ and let $R>0$. Consider the positively oriented rectangle with vertices \begin{align*} c-iT,\quad c+iT,\quad -R+iT,\quad -R-iT. \end{align*} For $R>0$, the pole $0$ lies inside this rectangle. By the Cauchy residue theorem, \begin{align*} \int_{c-iT}^{c+iT}F_y(s)\,ds + \int_{c+iT}^{-R+iT}F_y(s)\,ds + \int_{-R+iT}^{-R-iT}F_y(s)\,ds + \int_{-R-iT}^{c-iT}F_y(s)\,ds = 2\pi i. \end{align*} We estimate the left vertical side. Parametrize it by \begin{align*} \gamma_R: [-T,T] &\to \mathbb{C} \\ t &\mapsto -R+it. \end{align*} Then \begin{align*} \left|\int_{-R+iT}^{-R-iT}F_y(s)\,ds\right| &\leq \int_{-T}^{\,T}\frac{e^{-LR}}{|-R+it|}\,d\mathcal{L}^1(t) \\ &\leq \frac{2T e^{-LR}}{R}. \end{align*} For fixed $T$, this tends to $0$ as $R\to\infty$. For the two horizontal sides, using $|\,\sigma \pm iT\,|\geq T$, we obtain \begin{align*} \left|\int_{c+iT}^{-R+iT}F_y(s)\,ds\right| &\leq \int_{-R}^{c}\frac{e^{L\sigma}}{|\sigma+iT|}\,d\mathcal{L}^1(\sigma) \\ &\leq \frac{1}{T}\int_{-R}^{c}e^{L\sigma}\,d\mathcal{L}^1(\sigma) \\ &\leq \frac{e^{Lc}}{LT}, \end{align*} and the same estimate holds for the lower horizontal side: \begin{align*} \left|\int_{-R-iT}^{c-iT}F_y(s)\,ds\right| \leq \frac{e^{Lc}}{LT}. \end{align*} For fixed $T>0$, the improper horizontal-side integrals as $R\to\infty$ exist absolutely, because \begin{align*} \int_{-\infty}^{c}\frac{e^{L\sigma}}{|\sigma+iT|}\,d\mathcal{L}^1(\sigma) &\leq \frac{1}{T}\int_{-\infty}^{c}e^{L\sigma}\,d\mathcal{L}^1(\sigma) = \frac{e^{Lc}}{LT}, \end{align*} and the same bound holds with $\sigma-iT$ in place of $\sigma+iT$. Letting $R\to\infty$ in the rectangular identity gives \begin{align*} \left|\int_{c-iT}^{c+iT}F_y(s)\,ds - 2\pi i\right| \leq \frac{2e^{Lc}}{LT}. \end{align*} Dividing by $2\pi i$ and then letting $T\to\infty$ yields \begin{align*} \lim_{T\to\infty} I_T(y,c)=1. \end{align*} [/step]

[step:Close the contour to the right when $0<y<1$] Assume $0<y<1$, so $L<0$. Define \begin{align*} A := -L > 0. \end{align*} Fix $T>0$ and let $R>c$. Consider the rectangle with vertices \begin{align*} c-iT,\quad c+iT,\quad R+iT,\quad R-iT. \end{align*} The function $F_y$ has no pole inside this rectangle, because its only pole is at $0$ and $0$ lies to the left of the line $\operatorname{Re}s=c$. Hence the Cauchy residue theorem gives \begin{align*} \int_{c-iT}^{c+iT}F_y(s)\,ds + \int_{c+iT}^{R+iT}F_y(s)\,ds + \int_{R+iT}^{R-iT}F_y(s)\,ds + \int_{R-iT}^{c-iT}F_y(s)\,ds = 0. \end{align*} On the right vertical side, \begin{align*} \left|\int_{R+iT}^{R-iT}F_y(s)\,ds\right| &\leq \int_{-T}^{\,T}\frac{e^{-AR}}{|R+it|}\,d\mathcal{L}^1(t) \\ &\leq \frac{2T e^{-AR}}{R}, \end{align*} which tends to $0$ as $R\to\infty$ for fixed $T$. For the upper horizontal side, \begin{align*} \left|\int_{c+iT}^{R+iT}F_y(s)\,ds\right| &\leq \int_c^R \frac{e^{-A\sigma}}{|\sigma+iT|}\,d\mathcal{L}^1(\sigma) \\ &\leq \frac{1}{T}\int_c^R e^{-A\sigma}\,d\mathcal{L}^1(\sigma) \\ &\leq \frac{e^{-Ac}}{AT}. \end{align*} The lower horizontal side satisfies the same estimate: \begin{align*} \left|\int_{R-iT}^{c-iT}F_y(s)\,ds\right| \leq \frac{e^{-Ac}}{AT}. \end{align*} For fixed $T>0$, the improper horizontal-side integrals as $R\to\infty$ exist absolutely, because \begin{align*} \int_c^{\infty}\frac{e^{-A\sigma}}{|\sigma+iT|}\,d\mathcal{L}^1(\sigma) &\leq \frac{1}{T}\int_c^{\infty}e^{-A\sigma}\,d\mathcal{L}^1(\sigma) = \frac{e^{-Ac}}{AT}, \end{align*} and the same bound holds with $\sigma-iT$ in place of $\sigma+iT$. Letting $R\to\infty$ in the rectangular identity gives \begin{align*} \left|\int_{c-iT}^{c+iT}F_y(s)\,ds\right| \leq \frac{2e^{-Ac}}{AT}. \end{align*} After division by $2\pi i$ and passage to the limit $T\to\infty$, we obtain \begin{align*} \lim_{T\to\infty} I_T(y,c)=0. \end{align*} [/step]

[step:Compute the symmetric principal value when $y=1$] Assume $y=1$. Then $L=\log 1=0$, so $y^s=e^{0}=1$ and \begin{align*} I_T(1,c) &= \frac{1}{2\pi i}\int_{c-iT}^{c+iT}\frac{1}{s}\,ds. \end{align*} Parametrize the vertical segment by \begin{align*} \alpha_T: [-T,T] &\to \mathbb{C} \\ t &\mapsto c+it. \end{align*} Since $d\alpha_T(t)=i\,d\mathcal{L}^1(t)$, we get \begin{align*} I_T(1,c) &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{1}{c+it}\,d\mathcal{L}^1(t) \\ &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{c-it}{c^2+t^2}\,d\mathcal{L}^1(t). \end{align*} The imaginary part vanishes because $t\mapsto t/(c^2+t^2)$ is odd on $[-T,T]$. Therefore \begin{align*} I_T(1,c) &= \frac{1}{2\pi}\int_{-T}^{\,T}\frac{c}{c^2+t^2}\,d\mathcal{L}^1(t) \\ &= \frac{1}{\pi}\arctan\left(\frac{T}{c}\right). \end{align*} Letting $T\to\infty$ gives \begin{align*} \lim_{T\to\infty}I_T(1,c) = \frac{1}{\pi}\cdot \frac{\pi}{2} = \frac{1}{2}. \end{align*} Combining the three cases proves the stated formula. [/step]