Find the value of the improper integral

(1)

Since $k$ is not an integer, a circuit around the origin changes the argument of $$z^{-k} \;=\; e^{-k\log z} \;=\; e^{-k(\ln|z|+i\arg{z})},$$ by the amount $-2k\pi$ , thus giving to $z^{-k}$ a new value. Consequently the integrand $\displaystyle\frac{z^{-k}}{z+1}$ of (1) has the origin as a branch point. In the annulus between the origin-centered circles $\Gamma$ and $\gamma$ , cut open along the positive real axis, any branch of the integrand is single-valued. Let's take the branch with $0 \leqq \arg{z} \leqq 2\pi$ and suppose $0 < r < 1 < R$ . Then the integrand function is analytic in the cut annulus except at the point $z = -1$ which is a simple pole.

(2)

One has first $$\operatorname{Res}\!\left(\frac{z^{-k}}{x\!+\!1};\,-1\right) \;=\; \lim_{z\to-1}(z+1)\frac{z^{-k}}{x\!+\!1} \;=\; (-1)^{-k} \;=\; e^{-i\pi k}.$$

The integral (2) is split to four parts, two of them are the integrals around $\Gamma$ and $\gamma$ , and the other two the integrals along the cut on the real axis, in two directions. Using the estimating theorem of contour integral, one obtains $$\left|\int_\Gamma\!\frac{z^{-k}}{z\!+\!1}\,dz\right| \;\leqq\; \frac{R^{-k}}{R\!-\!1}\cdot2\pi R \;=\; \frac{2\pi R^{-k}}{1\!-\!\frac{1}{R}} \;\to\; 0 \quad \mbox{as} \quad R \to \infty,$$ $$\left|\int_\gamma\!\frac{z^{-k}}{z\!+\!1}\,dz\right| \;\leqq\; \frac{r^{-k}}{1\!-\!r}\cdot2\pi r \;=\; \frac{2\pi r^{1-k}}{1\!-\!r} \;\to\; 0 \,\quad \mbox{as} \,\quad r \to 0\!+\!.$$ Thus the two first parts of the integral (2) have the limits 0 as $R \to \infty$ and $r \to 0\!+$ . When one integrates on the real axis from $R$ to $r$ along the ``lower edge'' of the cut, denoting $z = x\!+\!iy$ ($x,\,y \in \mathbb{R}$ ), one has $\arg{z} = 2\pi$ and $$z^{-k} \;=\; e^{-k(\ln{x}+i\cdot2\pi)} \;=\; x^{-k}e^{-2i\pi k},$$ whence $$\int_R^r\!\frac{z^{-k}}{z\!+\!1}\,dz \;=\; e^{-2i\pi k}\int_R^r\!\frac{x^{-k}}{x\!+\!1}\,dx \;=\; -e^{-2i\pi k}\int_r^R\!\frac{x^{-k}}{x\!+\!1}\,dx.$$ But the last written integral equals to the integral gotten as one integrates from $r$ to $R$ along the ``upper edge'' of the cut, because on this path one has $\arg{z} = 0$ and thus $z^{-k} = x^{-k}$ . Accordingly, the sum of the integrals taken along the cut has the value $$\left(1-e^{-2i\pi k}\right)\!\int_r^R \frac{x^{-k}}{x\!+\!1}\,dx.$$ Accordingly, (2) gives the limit equation $$\left(1-e^{-2i\pi k}\right)\!\int_0^\infty\!\frac{x^{-k}}{x\!+\!1}\,dx \;=\; 2i\pi e^{-i\pi k},$$ and this implies moreover $$\int_0^\infty\!\frac{x^{-k}}{x\!+\!1}\,dx \;=\; 2i\pi\frac{e^{-i\pi k}}{1-e^{-2i\pi k}} \;=\; \frac{2i\pi}{e^{i\pi k}\!-\!e^{-i\pi k}}.$$ Thus the Euler formula lastly yields the result

(3)