integration - Integral $Pint_0^infty frac{x^{lambda-1}}{1-x} dx$

I am trying to calculate the following principle value integral
\begin{equation}
P\int_0^\infty \frac{x^{\lambda-1}}{1-x} dx

\end{equation}
for $\lambda \in [0,1].$ I tried to turn this into a contour integral so our complex function is given by
$$
f(z)=\frac{z^{\lambda-1}}{1-z}
$$
which has a simple pole at $z=1$ and branch points at $z=0$ and $z=\infty$. We integrate over a contour with two indented paths thus we pick up half residues at these two contours), thus we can write the contour C as
$$
C=\sum_{i=1}^6 C_i + C_{\epsilon}+C_{R}=C_1+C_2+C_3+C_4+C_5+C_6+C_{\epsilon}+C_R.
$$
Since the contour encloses no poles and $f(z)$ is holomorphic, by the Cauchy-Goursat theorem we know that

$$
\oint_C f(z) dz=0.
$$
I can show that the integrals of the contours $C_R$ and $C_\epsilon$ vanish since
$$
\bigg|\int_{C_R}\bigg| \leq \bigg| \int_{0}^{2\pi} d\theta \frac{R^{\lambda-1} R}{R} \bigg|=\bigg| \frac{2\pi}{R^{1-\lambda}}\bigg| \to 0 \ \text{as} \ R\to\infty \ \text{for} \ \lambda < 1
$$
and
$$
\bigg|\int_{C_\epsilon}\bigg| \leq \bigg| \int_{0}^{2\pi} d\theta \epsilon^{\lambda-1}\cdot \epsilon = \big|2\pi\epsilon^\lambda\big| \to 0 \ \text{as} \ \epsilon \to 0 \ \text{for} \ \lambda > 0.

$$
Now write the contour integral as
$$
0=\oint_C f(z)dz=P\int_{C_1} + \ P\int_{C_2} +\ P\int_{C_3}+\ P\int_{C_4}+\ P\int_{C_5}+\ P\int_{C_6}.
$$
Explicitly we can now calculate three contour integrals over $C_1, C_2, C_3$ by using $z=xe^{2\pi i}$, $dz=dxe^{i2\pi}=dx$ and we obtain
\begin{equation}
P\int_{C_1}+\ P\int_{C_2}+\ P\int_{C_3}=\lim_{R\to\infty} \lim_{\epsilon \to 0}\int_{Re^{i(2\pi-\epsilon)}}^{(1+\epsilon)2\pi i} \frac{z^{\lambda-1}}{1-z}dz-\frac{1}{2}2\pi i\cdot Res_{z=e^{2\pi i} }[f(z)] +e^{2\pi i(\lambda-1)} \int_{1}^{0} \frac{x^{\lambda-1}}{1-x}dx.
\end{equation}
Note the first integral in terms of $z$ can just be written as

$$
\lim_{R\to\infty} \lim_{\epsilon \to 0}\int_{Re^{i(2\pi-\epsilon)}}^{(1+\epsilon)2\pi i} \frac{z^{\lambda-1}}{1-z}dz=\int_{\infty}^{1} \frac{x^{\lambda-1}}{1-x}dx
$$
however I am stuck as to how go from here. Thanks

Answer

I'll evaluate the more general case $$\text{PV} \int_{0}^{\infty} \frac{x^{\lambda-1}}{x^{b}-1} \ dx \ \ (b >\lambda > 0, \ b\ge 1) .$$

Let $ \displaystyle f(z) = \frac{z^{\lambda-1}}{z^{b}-1}$, where the branch cut is along the positive real axis.

Now integrate around a wedge of radius $R$ that makes an angle of $ \displaystyle \frac{2 \pi }{b}$ with the positive real axis and is indented around the simple poles at $z=1$ and $z=e^{2 \pi i /b}$, and the branch point at $z=0$.

The integral obviously vanishes along the arc of the wedge as $R \to \infty$.

And there is no contribution from the indentation around the branch point at $z=0$ since $$\Big| \int_{0}^{\frac{2 \pi}{b}} f(re^{it}) ire^{it} \ dt \Big| \le \frac{2 \pi}{b} \frac{r^{\lambda}}{1-r^{b}} \to 0 \ \text{as} \ r \to 0.$$

Then going around the contour counterclockwise,
$$ \text{PV} \int_{0}^{\infty} f(x) \ dx - \pi i \ \text{Res} [f, \pi i] + \ \text{PV}\int_{\infty}^{0} f(te^{\frac{2 \pi i }{b}}) e^{\frac{2 \pi i}{b}} \ dt - \pi i \ \text{Res}[f,e^{\frac{2 \pi i}{b}}] = 0 .$$

Looking at each part separately,

$$ \text{Res}[f,1] = \lim_{z \to 1} \frac{z^{\lambda-1}}{bz^{b-1}} = \frac{1}{b}$$

$ $

$$\text{PV} \int_{\infty}^{0} f(te^{\frac{2 \pi i }{b}}) e^{\frac{2 \pi i}{b}} \ dt = - e^{\frac{2 \pi i}{b}} \text{PV} \int_{0}^{\infty} \frac{t^{\lambda-1} e^{\frac{2 \pi i(\lambda-1)}{b}}}{t^{b} e^{2 \pi i} - 1} \ dt = - e^{\frac{2 \pi i \lambda}{b}} \text{PV} \int_{0}^{\infty} \frac{t^{\lambda-1}}{t^{b}-1} \ dt $$

$ $

$$ \text{Res}[f, e^{\frac{2 \pi i}{b}}] = \lim_{z \to e^{\frac{2 \pi i}{b}}} \frac{z^{\lambda-1}}{bz^{b-1}} = \frac{e^{\frac{2 \pi i (\lambda-1)}{b}}}{b e^{\frac{2 \pi i(b-1)}{b}}} = \frac{1}{b} e^{\frac{2 \pi i \lambda }{b}} $$

Plugging back in and rearranging,

$$\text{PV} \int_{0}^{\infty} \frac{x^{\lambda-1}}{x^{b}-1} \ dx = \frac{\pi i}{b} \frac{1 + e^{\frac{2 \pi i \lambda}{b}}}{1-e^{\frac{2 \pi i \lambda}{b}}} = - \frac{\pi}{b} \cot \left(\frac{\pi \lambda}{b} \right) $$

or

$$ \text{PV} \int_{0}^{\infty} \frac{x^{\lambda-1}}{1-x^{b}} \ dx = \frac{\pi}{b} \cot \left(\frac{\pi \lambda}{b} \right) $$