Integral along $\Gamma_c := \{c + i t \mid c>0 , -\infty < t < \infty\}$ I have a Complex Analysis homework problem which I've been working on for some time, and have become stuck.  I am asked to compute
$$
I \equiv
{1 \over 2\pi{\rm i}}\int _{\Gamma_c}{a^{s} \over s\left(s - 1\right)}\,{\rm d}s
\quad\mbox{where}\quad
\Gamma_c \equiv \left\{c + {\rm i}t \mid c > 0\,, -\infty < t < \infty\right\}
$$
and show that $I = 0$ for $0 < a <1$, but $\displaystyle{I = 1 - {1 \over a}}$ for
$a \geq 1$.
I have set up the integral as follows:
I create a closed curve $\Gamma$ consisting of a portion of $\Gamma_c$, which I'll call $\Gamma_R '$ stretching between $-R,R \in \mathbb{R}$ for $|R|>1$, i.e., $\Gamma _R ' := \{c + i t \mid c>0 , -R \le t \le R\}$, and also consisting of the semi-circle $\Gamma_R := \{ R e ^{i\theta} + c \mid \frac{\pi}{2} \le \theta \le \frac{3 \pi}{2} \}$.  I call this closed curve $\Gamma$ and take it to be the union of $\Gamma_R$, and $\Gamma_R '$.  I would have liked to have included a picture for simplicity, but not really sure how so I hope you'll bare with me...
Then in terms of this set up, 
$$\int _{\Gamma_c} \frac{a^s}{s (s-1) } ds = \lim _{R\rightarrow \infty} \left( \int _{\Gamma} \frac{a^s}{s (s-1) } ds - \int _{\Gamma_R } \frac{a^s}{s (s-1) } ds \right) $$
Now, $\int _{\Gamma} \frac{a^s}{s (s-1) } ds$ can be computed nicely with residues- I get that 
$$ \int_ {\Gamma} \frac{a^s}{s (s-1) } ds = 2 \pi i \left( 1 - \frac{1}{a} \right)$$
as I believe I should.  My difficulty comes from dealing with the piece 
$$\lim_{R \rightarrow \infty} \left( \int _{\Gamma_R } \frac{a^s}{s (s-1) } ds \right) $$
I have the parameterization for the semi-circle $s(\theta) = Re ^{i \theta} +c$ for $\frac{\pi}{2} \le \theta \le \frac{3 \pi}{2}$, so it seems that what I need to compute is 
$$\lim_{R\rightarrow \infty} \left( \int_{\frac{\pi}{2}} ^{\frac{3 \pi}{2}} \frac{a^{R e^{i \theta} +c} R \theta e^{i \theta} }{(R e^{i \theta} + c + 1) (R e^{i \theta} + c)} d\theta \right)$$ which to me looks like no fun to work with; I hope you'll agree!  Any help that you might offer would be greatly appreciated!  
 A: Where's the factor of $\theta$ in the numerator coming from?  Anyway, for a little simplicity, you can join the endpoints of the vertical line with the imaginary axis to form two small segments, along which the integral vanishes as $R \to \infty$.  Thus,
$$\int_{\Gamma_R} ds \frac{a^s}{s(s-1)} = \int_c^0 dx \frac{e^{(\log{a}) (x+i R)}}{(x+i R)(x+i R-1)} + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{(\log{a}) R e{i \theta}}}{R e^{i \theta} (R e^{i \theta}-1)}\\ + \int_0^c dx \frac{e^{(\log{a}) (x-i R)}}{(x-i R)(x-i R-1)}$$
The magnitude of each of the first and third integrals (along those little segments) is bounded by $c a^c/R^2$, which clearly vanishes as $R \to \infty$.  For the second integral, the magnitude is bounded by
$$\frac1{R} \int_{\pi/2}^{3 \pi/2} d\theta \, e^{(\log{a}) R \cos{\theta}}$$
Note that $\log{a} \gt 0$ when $a \gt 1$, and $\cos{\theta} \lt 0$ when $\theta \in (\pi/2, 3 \pi/2)$.  In fact, shifting the interval by $\pi/2$ and using symmetry and the inequality $\sin{\theta} \ge 2 \theta/\pi$, we get the following bound:
$$\left | \int_{\Gamma_R} ds \frac{a^s}{s(s-1)} \right | \le \frac{2}{R} \int_0^{\pi/2} d\theta \, e^{-(\log{a}) R \sin{\theta}} \le \frac{2}{R} \int_0^{\pi/2} d\theta \, e^{-2 (\log{a}) R \theta/\pi} \le \frac{\pi}{R^2 \log{a}}$$
When $a \in (0,1)$, $\log{a} \lt 0$, and you close to the right where $\cos{\theta} \gt 0$, so that the above arguments apply all the same.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{I \equiv
    {1 \over 2\pi\ic}\int_{\Gamma_c}{a^{s} \over s\pars{s - 1}}\,\dd s
    \quad\mbox{where}\quad
    \Gamma_c \equiv
    \braces{c + {\rm i}t \mid c > 0\,,\ t \in {\mathbb R}}:\ {\large ?}.\quad
    a > 0.}$

\begin{align}
I&={1 \over 2\pi\ic}\int_{-\infty}^{\infty}
{a^{c + \ic t} \over \pars{c + \ic t}\bracks{\pars{c + \ic t} - 1}}\,
\pars{\ic\,\dd t}
=-\,{1 \over 2\pi}\int_{-\infty}^{\infty}\
{a^{c + \ic t}
 \over \pars{t - \ic c}\bracks{t - \ic\pars{c - 1}}}\,\dd t
\end{align}

$\ds{\LARGE 0 < a < 1}$
In this case, we must close the integration in a lower complex half plane contour:
\begin{align}
&\color{#c00000}{\left.\vphantom{\large A}I\,\right\vert_{0\ <\ a\ <\ 1}}
=-\,{a^{c} \over 2\pi}\pars{-2\pi\ic}\times
\\[3mm]&\braces{\Theta\pars{-c}
\bracks{%
{a^{c + \ic\pars{\ic c}} \over \pars{\ic c} - \ic\pars{c - 1}}
+
{a^{c + \ic\bracks{\ic\pars{c - 1}}} \over \ic\pars{c - 1} - \ic c}}
+ \Theta\pars{c}\Theta\pars{1 - c}\,
{a^{c + \ic\bracks{\ic\pars{c - 1}}} \over \ic\pars{c - 1} - \ic c}}
\\[3mm]&=a^{c}\ic\bracks{%
\Theta\pars{-c}\,{1 - a \over \ic} + \Theta\pars{c}\Theta\pars{1 - c}\,
{a \over -\ic}}
=\color{#c00000}{%
\bracks{\Theta\pars{-c}\pars{1 - a} - \Theta\pars{c}\Theta\pars{1 - c}a}a^{c}}
\end{align}

$\ds{\LARGE a\ \geq\ 1}$
In this case, we must close the integration in a upper complex half plane contour:
\begin{align}
\color{#c00000}{\left.\vphantom{\large A}I\,\right\vert_{a\ >\ 1}}&
=-\,{a^{c} \over 2\pi}\pars{-2\pi\ic}\Theta\pars{c}\times
\\[3mm]&\braces{\Theta\pars{1 - c}\,
{a^{c + \ic\pars{\ic c}} \over \ic c - \ic\pars{c - 1}}
+
\Theta\pars{c - 1}\bracks{%
{a^{c + \ic\pars{\ic c}} \over \pars{\ic c} - \ic\pars{c - 1}}
+
{a^{c + \ic\bracks{\ic\pars{c - 1}}} \over \ic\pars{c - 1} - \ic c}}}
\\[3mm]&=\Theta\pars{c}a^{c}\ic\bracks{%
{\Theta\pars{1 - c} \over \ic} + {\Theta\pars{c - 1} \over \ic}\,\pars{1 - a}}
\\[3mm]&=\color{#c00000}{%
\Theta\pars{c}\bracks{\Theta\pars{1 - c} + \Theta\pars{c - 1}\pars{1 - a}}a^{c}}
\end{align}

$$
I=\left\lbrace%
\begin{array}{rcl}
\vphantom{\large A}&&
\\
\color{#00f}{\left.\begin{array}{lcl}
\pars{1 - a}a^{c} & \mbox{if} & c < 0
\\
-a^{c} & \mbox{if} & 0 < c < 1
\\ 
0 & \mbox{if} & c > 1
\\
\mbox{diverges} && \mbox{otherwise}
\end{array}\right\rbrace}
& \mbox{if} & 0 < a < 1
\\[5mm]
\color{#c00000}{\left.\begin{array}{lcl}
0     & \mbox{if} & c < 0
\\
a^{c} & \mbox{if} & 0 < c < 1
\\
\pars{1 - a}a^{c} & \mbox{if} & c > 1
\\
\mbox{diverges} && \mbox{otherwise}
\end{array}\right\rbrace} & \mbox{if} & a \geq 1
\\
\vphantom{\large A}&&
\end{array}\right.
$$
