# Integral $P\int_0^\infty \frac{x^{\lambda-1}}{1-x} dx$

I am trying to calculate the following principle value integral $$P\int_0^\infty \frac{x^{\lambda-1}}{1-x} dx$$ 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 $$P\int_{C_1}+\ P\int_{C_2}+\ P\int_{C_3}=\lim_{R\to\infty} \lim_{\epsilon \to 0}\int_{Re^{i(2\pi-\epsilon)}}^{(1+\epsilon)2\pi i} \frac{z^{\lambda-1}}{1-z}dz-\frac{1}{2}2\pi i\cdot Res_{z=e^{2\pi i} }[f(z)] +e^{2\pi i(\lambda-1)} \int_{1}^{0} \frac{x^{\lambda-1}}{1-x}dx.$$ 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

• Indeed, $\large\lambda \in \color{#c00000}{\left(\color{#000}{0, 1}\right)}$. – Felix Marin May 21 '14 at 2:15

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

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

• Random Variable once again! The best! – Jeff Faraci May 20 '14 at 23:20

Let the contour $\gamma$ be

$\hspace{3.6cm}$

Since $\gamma$ contains no singularities of $\dfrac{z^{\lambda-1}}{1-z}$, we get $$\int_\gamma\frac{z^{\lambda-1}}{1-z}\,\mathrm{d}z=0\tag{1}$$ The integral along the large circle vanishes as the radius $\to\infty$ since the integrand $\sim-z^{\lambda-2}$.

The integral along the small upper semi-circle, as the radius $\to0$, is $$-\pi i\times\operatorname*{Res}\limits_{z=1}\dfrac1{1-z}=\pi i\tag{2}$$ The integral along the small lower semi-circle, as the radius $\to0$, is $$-\pi i\times\operatorname*{Res}\limits_{z=1}\dfrac{e^{2\pi i(\lambda-1)}}{1-z}=\pi ie^{2\pi i(\lambda-1)}\tag{3}$$ The integral along the upper line segments is $$\mathrm{PV}\int_0^\infty\frac{x^{\lambda-1}}{1-x}\,\mathrm{d}x\tag{4}$$ The integral along the lower line segments is $$-e^{2\pi i(\lambda-1)}\mathrm{PV}\int_0^\infty\frac{x^{\lambda-1}}{1-x}\,\mathrm{d}x\tag{5}$$ Summing $(2)-(4)$ and applying $(1)$ yields \begin{align} \mathrm{PV}\int_0^\infty\frac{x^{\lambda-1}}{1-x}\,\mathrm{d}x &=-\pi i\frac{1+e^{2\pi i(\lambda-1)}}{1-e^{2\pi i(\lambda-1)}}\\ &=\pi i\frac{e^{\pi i(\lambda-1)}+e^{-\pi i(\lambda-1)}}{e^{\pi i(\lambda-1)}-e^{-\pi i(\lambda-1)}}\\[9pt] &=\pi\cot(\pi(\lambda-1))\\[15pt] &=\pi\cot(\pi\lambda)\tag{6} \end{align}

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\pp\int_{0}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x:\ {\large ?}\,,\qquad \lambda \in \pars{0,1}}$.

\begin{align} &\color{#c00000}{\pp\int_{0}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x}= \lim_{\epsilon \to 0^{+}}\pars{% \int_{0}^{1 - \epsilon}{x^{\lambda - 1} \over 1 - x}\,\dd x \int_{1 + \epsilon}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x} \\[3mm]&= \lim_{\epsilon \to 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{x^{\lambda - 1} \over 1 - x}\,\dd x +\int_{1/\pars{1 + \epsilon}}^{0}{\pars{1/x}^{\lambda - 1} \over 1 - 1/x}\,\pars{-\,{\dd x \over x^{2}}}} \\[3mm]&=\lim_{\epsilon \to 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{x^{\lambda - 1} \over 1 - x}\,\dd x -\int^{1/\pars{1 + \epsilon}}_{0}{x^{-\lambda} \over 1 - x}\,\dd x} \\[3mm]&=\lim_{\epsilon \to 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{x^{\lambda - 1} - x^{-\lambda} \over 1 - x}\,\dd x -\int^{1/\pars{1 + \epsilon}}_{1 - \epsilon}{\dd x \over x^{\lambda}\pars{1 - x}}} \\[3mm]&=\lim_{\epsilon \to 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{1 - x^{-\lambda} \over 1 - x}\,\dd x -\int_{0}^{1 - \epsilon}{1 - x^{\lambda - 1} \over 1 - x}\,\dd x -\int^{1/\pars{1 + \epsilon}}_{1 - \epsilon}{\dd x \over x^{\lambda}\pars{1 - x}}} \end{align}

However, $\ds{\lim_{\epsilon \to 0^{+}}\int^{1/\pars{1 + \epsilon}}_{1 - \epsilon}{\dd x \over x^{\lambda}\pars{1 - x}} = 0}$ because \begin{align} &\verts{% \int^{1/\pars{1 + \epsilon}}_{1 - \epsilon}{\dd x \over x^{\lambda}\pars{1 - x}}} \leq \int^{1/\pars{1 + \epsilon}}_{1 - \epsilon}{\dd x \over \pars{1 - \epsilon}^{\lambda}\bracks{1 - 1/\pars{1 + \epsilon}}} \\[3mm]&={1 + \epsilon \over \pars{1 - \epsilon}^{\lambda}} \bracks{{1 \over 1 + \epsilon} - \pars{1 - \epsilon}} ={\epsilon^{2} \over \pars{1 - \epsilon}^{\lambda}} \end{align}

and, similarly, $\ds{\lim_{\epsilon \to 0^{+}} \int^{1}_{1 - \epsilon}{x^{\lambda - 1} - x^{-\lambda} \over 1 - x}\,\dd x = 0}$ such that \begin{align} &\color{#c00000}{\pp\int_{0}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x} = \int_{0}^{1}{1 - x^{-\lambda} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{\lambda - 1} \over 1 - x}\,\dd x =\Psi\pars{-\lambda + 1} - \Psi\pars{\lambda} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function ${\bf\mbox{6.3.1}}$ and we used the identity ${\bf\mbox{6.3.22}}$.

With Euler Reflection Formula ${\bf\mbox{6.3.7}}$ $\pars{~\Psi\pars{1 - z} = \Psi\pars{z} + \pi\cot\pars{\pi z}~}$ we'll get: $$\color{#00f}{\large\pp\int_{0}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x} =\color{#00f}{\large\pi\cot\pars{\pi\lambda}}\,,\qquad\qquad \lambda \in \pars{0,1}$$

• Very nice thank you Felix – Jeff Faraci May 22 '14 at 3:42

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Besides my previous answer, there is another way $\ds{\pars{~\mbox{in a "physicist fashion"}~}}$ to evaluate the integral :

$\ds{\large\mbox{With}\quad {\tt\lambda \in \pars{0,1}}}$: \begin{align} &\color{#c00000}{\pp\int_{0}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x} =-\Re\int_{0}^{\infty}{x^{\lambda - 1} \over x - 1 + \ic 0^{+}}\,\dd x \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\pars{1} \\[3mm]&=-\Re\left\lbrack -\ \overbrace{\left.\lim_{R \to \infty}\int_{0}^{2\pi}{z^{\lambda - 1} \over z - 1} \,\dd z\,\right\vert_{z\ \equiv\ R\expo{\ic\theta}}}^{\ds{=\ 0}}\ -\ \int_{\infty}^{0} {x^{\lambda - 1}\pars{\expo{2\pi\ic}}^{\lambda - 1} \over x - 1 - \ic 0^{+}}\,\dd x \right. \\[3mm]&\left.\phantom{-\Re\left\lbrack\right.\,\,\,}\mbox{}-\ \overbrace{\left.\lim_{\epsilon \to 0^{+}}\int^{0}_{2\pi} {z^{\lambda - 1} \over z - 1}\,\dd z\, \right\vert_{z\ \equiv\ \epsilon\expo{\ic\theta}}}^{\ds{=\ 0}}\right\rbrack \\[3mm]&=-\Re\pars{\expo{2\pi\lambda\ic}\int^{\infty}_{0} {x^{\lambda - 1} \over x - 1 - \ic 0^{+}}\,\dd x} =-\Re\pars{\expo{2\pi\lambda\ic}\,\pp\int^{\infty}_{0} {x^{\lambda - 1} \over x - 1}\,\dd x + \ic\pi\expo{2\pi\lambda\ic}} \\[3mm]&=\cos\pars{2\pi\lambda}\,\color{#c00000}{\pp\int^{\infty}_{0} {x^{\lambda - 1} \over 1 - x}\,\dd x} + \pi\sin\pars{2\pi\lambda} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\pars{2} \end{align}

With $\pars{1}$ and $\pars{2}$, we get: \begin{align} &\color{#c00000}{\pp\int^{\infty}_{0}{x^{\lambda - 1} \over x - 1}\,\dd x} =\pi\,{\sin\pars{2\pi\lambda} \over 1 - \cos\pars{2\pi\lambda}} =\pi\,{2\sin\pars{\pi\lambda}\cos\pars{\pi\lambda} \over 2\sin^{2}\pars{\pi\lambda}} =\pi\,{\cos\pars{\pi\lambda} \over \sin\pars{\pi\lambda}} \end{align}

$$\color{#00f}{\large\pp\int_{0}^{\infty}{x^{\lambda - 1} \over 1 - x}\,\dd x} =\color{#00f}{\large\pi\cot\pars{\pi\lambda}}\,,\qquad\qquad \lambda \in \pars{0,1}$$

• +1 thank you very clever I see you are a physicist as I have gone through your lecture notes for statistical mechanics on your website. Thanks. – Jeff Faraci May 22 '14 at 5:09
• @Integrals Thanks a lot. Indeed, right now I'm checking those lectures for finer details I learnt over here ( in MSE ). – Felix Marin May 22 '14 at 5:11