Show that $\displaystyle {\int_{0}^{\infty} \frac {x^{-c}} {1+x}\ dx = \frac {\pi} {\sin \pi c}}$ if $0 \lt c \lt 1.$
I have found this example in Conway's book on complex analysis in page no. $119.$ The author evaluates this integral by using of residue theorem. Here's how it is solved in the book.
To evaluate the the integral we must consider a branch of the function $z^{-c}.$ The point $z = 0$ is called a branch point of $z^{-c},$ and the process used to evaluate this integral is sometimes called integration around a branch point.
Let $G = \{z\ :\ z \neq 0\ \text {and}\ 0 \lt \arg z \lt 2 \pi\};$ define a branch of the logarithm on $G$ by putting $\ell (re^{i\theta}) = \log r + i \theta$ where $0 \lt \theta \lt 2\pi.$ For $z$ in $G$ put $f(z) = \exp\ [-c\ell(z)];$ so $f$ is a branch of $z^{-c}.$ We now select an appropriate curve $\gamma$ in $G.$ Let $0 \lt r \lt 1 \lt R$ and let $\delta \gt 0.$ Let $L_1$ be the line segment $[r + \delta i, R + \delta i];$ $\gamma_{R}$ the part of the circle $|z| = R$ from $R + \delta i$ counterclockwise to $R - \delta i;$ $L_2$ the line segment $[R - \delta i, r - \delta i];$ and $\gamma_{r}$ the part of the circle $|z| = r$ from $r - \delta i$ clockwise to $r + \delta i.$ Put $\gamma = L_1 + \gamma_{R} + L_2 + \gamma_{r}.$
Since $\gamma \sim 0$ in $G$ and $\text {Res}\ (f(z) (1+z)^{-1};-1) = f(-1) = e^{-i \pi c},$ the Residue Theorem gives $$\int_{\gamma} \frac {f(z)} {1 + z}\ dz = 2 \pi i e^{-i \pi c}.$$
Using the definition of a line integral $$\int_{L_1} \frac {f(z)} {1+z}\ dz = \int_{r}^{R} \frac {f(t+i\delta)} {1 + t + i\delta}\ dt.$$
Let $g(t,\delta)$ be defined on the compact set $[r,R] \times \left [0, \frac {1} {2} \pi \right ]$ by $$g(t, \delta) = \left |\frac {f(t + i \delta)} {1 + t + i \delta} - \frac {t^{-c}} {1+t} \right |$$ when $\delta \gt 0$ and $g(t,0) \equiv 0.$ Then $g$ is continuous and hence uniformly continuous. If $\varepsilon \gt 0$ there is a $\delta_0 \gt 0$ such that if $(t - t')^2 + (\delta - \delta')^2 \lt \delta_0^2$ then $|g(t,\delta) - g(t', \delta')| \lt \varepsilon /R.$ In particular, $g(t,\delta) \lt \varepsilon / R$ when $r \leq t \leq R$ and $\delta \lt \delta_0.$ Then $$\int_{r}^{R} g(t,\delta)\ dt \leq \varepsilon$$ for $\delta \lt \delta_0.$ This implies that $$\int_{r}^{R} \frac {t^{-c}} {1 + t}\ dt = \lim\limits_{\delta \to 0^+} \int_{L_1} \frac {f(z)} {1 + z}\ dz.$$ Similarly, using the fact that $\ell (\overline {z}) = \overline {\ell (z)} + 2 \pi i$ $$\color {red} {-e^{-2 \pi i c} \int_{r}^{R} \frac {t^{-c}} {1 + t}\ dt = \lim\limits_{\delta \to 0^{+}} \int_{L_2} \frac {f(z)} {1 + z}\ dz.}$$
This is the stage where I got stuck. I observed that $L_1 = \overline {-L_2},$ where $-L_2$ is the path which traces out $L_2$ in the direction opposite to the given direction. This shows that if $z \in -L_2$ then $\overline {z} \in L_1.$ Therefore if $z \in -L_2$ then $f(\overline z) = \exp\ [-c\ell (\overline z)] = e^{-2 \pi i c} \exp\ [-c\overline {\ell (z)}] = e^{-2 \pi i c} \overline {f(z)}.$ From here can we somehow deduce the above equality marked in red by converting the line integral over $L_2$ to the line integral over $L_1\ $?
Any help in this regard would be warmly appreciated. Thanks for investing your valuable time.
We know that $$\begin{equation} \int_{r}^{R} \frac {t^{-c}} {1 + t}\ dt = \lim\limits_{\delta \to 0^{+}} \int_{L_1} \frac {f(z)} {1 + z}\ dz \tag{1} \end{equation}$$ Now $$\begin {align*} \int_{L_2} \frac {f(z)} {1 + z}\ dz & = - \int_{-L_2} \frac {f(z)} {1 + z}\ dz \\ & = - \int_{r}^{R} \frac {f(t - i \delta)} { 1 + t - i \delta}\ dt \tag{2} \end{align*}$$
But $$\begin{align*}f(t - i \delta) & = \exp\ [-c \ell (t - i \delta)] \\ & = \exp\ [-c \ell (\overline {t + i \delta})] \\ & = \exp\ [-c (\overline {\ell(t + i \delta)} + 2 \pi i)] \\ & = e^{-2 \pi i c} \exp\ [-c \overline {\ell(t + i \delta)}] \\ & = e^{-2 \pi i c} \overline {\exp\ [-c \ell (t + i \delta)]} \\ & = e^{-2 \pi i c} \overline {f(t + i \delta)} \tag {3} \end {align*}$$ Combining $(2)$ and $(3)$ we have $$\begin{align*} \int_{L_2} \frac {f(z)} {1 + z}\ dz & = - e^{-2 \pi i c} \int_{r}^{R} \frac {\overline {f (t + i \delta)}} {1 + t - i \delta}\ dt \\ & = - e^{-2 \pi i c} \overline {\int_{r}^{R} \frac {f (t + i \delta)} {1 + t + i \delta}\ dt} \\ & = - e^{-2 \pi i c} \overline {\int_{L_1} \frac {f(z)} {1 + z}\ dz} \tag {4} \end{align*}$$ Therefore from $(1)$ and $(4)$ we have $$\begin{align*} \lim\limits_{\delta \to 0^{+}} \int_{L_2} \frac {f(z)} {1 + z}\ dz & = - e^{-2 \pi i c} \lim\limits_{\delta \to 0^{+}} \overline {\int_{L_1} \frac {f(z)} {1 + z}\ dz} \\ & = - e^{-2 \pi i c} \overline {\lim\limits_{\delta \to 0^{+}} \int_{L_1} \frac {f(z)} {1+z}\ dz}\ \ (\because z \mapsto \overline {z}\ \text {is a continuous function}) \\ & = - e^{-2 \pi i c} \overline {\int_{r}^{R} \frac {t^{-c}} {1+t}\ dt}\ \ (\text {by}\ (1)) \\ & \color {blue} {= - e^{-2 \pi i c} \int_{r}^{R} \frac {t^{-c}} {1+t}\ dt}\ \ \left (\because \int_{r}^{R} \frac {t^{-c}} {1+t}\ dt \in \mathbb R \right ) \end{align*}$$
The blue equality is exactly the last equality in the question marked in red.