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I need some help to solve the following integral by contour integration.

$$\int_{0}^{1} x^a (1-x)^{1-a}\,\mathrm{d}x$$

I attached my ideas and a picture of the paths to fix the labels.

Kind regards, Jonas

sketch of actual contours + labels

Use contour integration to show that:

$\int_0^1 x^a (1-x)^{1-a}\,\mathrm{d}x = \dfrac{a\pi(1-a)}{2\sin(a\pi)}$ for $-1<a<2$ First we make a transition to a complex valued function $f: \mathbb{C} \rightarrow \mathbb{C}$, $z \longmapsto f(z)$.

Recognize that the integrand is a multi-valued function since $a \not\in \mathbb{Z}$. It has branch points at $z=0$ because of $z^a$ and at $z=1$ because of $(1-z)^{1-a}$. We exclude the straight line between the two branch points ($\mathbb{C} \setminus [0,1]$) to make the function single valued. To prove that it is single-valued now, we take a closer look at the phase when we circle around both of the branch points.

$$\begin{align*} z&=\rho e^{i\theta}\\[1ex] z^a&=\rho^a e^{ia\theta} \end{align*}$$ $$\begin{align*} z&=\epsilon e^{i\phi}+1\\[1ex] (1-z)^{1-a}&=\epsilon^{1-a}e^{i(1-a)\phi} \end{align*}$$

$\Rightarrow f(z) = \rho^a e^{ia\theta} \epsilon^{1-a} e^{i(1-a)\phi} \propto e^{i(a(\theta - \phi) + \phi)}$

$$\begin{array}{c|c|c|c} \text{points} & \theta & \phi & e^{i(a(\theta - \phi) + \phi)}\\ \hline A&0&0&0\\ B&0&\pi&-a\pi + \pi \\ C& 0&\pi&-a\pi + \pi \\ D& \pi &\pi& \pi \\ E& 2\pi & \pi & a\pi + \pi \\ F & 2\pi & \pi & a\pi + \pi\\ A' & 2\pi & 2\pi & 2\pi\\ \end{array}$$

Since the phase changes from $0$ to $2\pi$ as we go from $A$ to $A'$, we conclude that the branch cut makes the function single valued.

$$\Gamma = \Gamma_R \cup \Gamma_2 \cup \Gamma_1 \cup \Gamma_3 \cup \Gamma_0$$

  • For $\Gamma_0$ and $\Gamma_R$ we use the following parameterization. $$z = r e^{i\alpha}\quad\text{and}\quad\mathrm{d}z = r i e^{i\alpha} \mathrm{d}\alpha$$ $$\int f(z)\,\mathrm{d}z=\int r^a e^{ia\alpha} \; (1-r e^{i\alpha}) ^{1-a} \; r i \: e^{i\alpha} \mathrm{d}\alpha$$highest order is $r$: $r^ar^{1-a}r=r^2$.
  • For $\Gamma_0$ we need the $\lim_{r \to 0}$:

$$\int_{\Gamma_0} f(z) \,\mathrm{d}z = 0$$

  • For $\Gamma_R$ we need the limit $\lim_{r \to \infty}$:

$$\int_{\Gamma_R} f(z) \,\mathrm{d}z = \infty$$

  • For $\Gamma_1$ we use this parameterization. $$z = \epsilon e^{i\phi} +1\quad\text{and}\quad\mathrm{d}z = \epsilon i \: e^{i\phi} \mathrm{d}\phi$$ $$\int f(z)\,\mathrm{d}z=\int (\epsilon e^{i\phi}+1)^a(\epsilon e^{i\phi})^{1-a}\epsilon ie^{i\phi}\,\mathrm{d}\phi$$highest orders $\epsilon^a \epsilon^{1-a}\epsilon$
  • For $\Gamma_1$ we use the $\lim_{\epsilon \to 0}$:

$$\int_{\Gamma_1} f(z) \,\mathrm{d}z = 0$$ $$\int_\Gamma f(z) \mathrm{d}z = \text{Res}(f(z),\infty) = -\text{Res}\left(f\left(\frac{1}{z}\right)\frac{1}{z^2},0\right)$$ $$\int_{\Gamma_0} f(z) \,\mathrm{d}z = 0$$ $$\int_{\Gamma_1}f(z)\,\mathrm{d}z=0$$ $$\lim_{R\to\infty}\int_{\Gamma_R}f(z)\,\mathrm{d}z=\infty$$

Questions:

  • Are the 4 equations in the Summary correct?
  • Is this a good way to calculate the integral?
  • How do I find the phase relation between $\Gamma_2$ and $\Gamma_3$?
  • How do I evaluate the residual at $\infty$ ($\int_\Gamma$)
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    $\begingroup$ Hi and welcome to Math.SE. Please reformat your question. Just pasting a LaTeX document does not make it very clear or readable to other users. $\endgroup$
    – mickep
    Dec 4, 2015 at 20:49
  • $\begingroup$ Thanks, I hope it works now.. $\endgroup$
    – user45494
    Dec 4, 2015 at 20:58
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    $\begingroup$ @user45494, no it doesn't work. The community needs to see the outcomes of $\LaTeX$, not the commands, so the try to post the outcomes. See this page. $\endgroup$
    – user249332
    Dec 4, 2015 at 21:04
  • $\begingroup$ Integrate around the dogbone contour and determine the residue at infinity. $\endgroup$
    – Mark Viola
    Dec 4, 2015 at 21:12
  • $\begingroup$ @user45494 please review my edit in case I've made any grievous mistakes. $\endgroup$
    – user170231
    Dec 4, 2015 at 21:45

1 Answer 1

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The integral $I$ around the dog bone contour is given by

$$\begin{align} I(a)&=\int_0^1 x^a\,(1-x)^{1-a}\,dx+\int_1^0 x^a\,e^{i2\pi a}\,(1-x)^{1-a}\,dx\\\\ &=\left(1-e^{i2\pi a}\right)\int_0^1 x^{a}\,(1-x)^{1-a}\,dx\tag 1 \end{align}$$

The integral around the dog bone contour is also equal to $2\pi i$ times the residue at infinity. The residue at infinity is

$$-\lim_{z\to 0}\frac{1}{2!}\frac{d^2}{dz^2}\left(z^3\frac{(z-1)^{1-a}}{z^3}\right)=\frac12 a(a-1)e^{i\pi a}\tag 2$$

Setting $(1)$ equal to $2\pi i$ time $(2)$ reveals

$$\int_0^1 x^{a}\,(1-x)^{1-a}\,dx=\frac{\pi a(1-a)}{2\sin(\pi a)}$$

as expected!


Another approach relies only on established properties of the Beta and Gamma functions. The Beta function is defined as

$$B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx$$

Therefore, the integral of interest can be expressed in terms of the Beta function as

$$\int_0^1 x^{a}\,(1-x)^{1-a}\,dx=B(a+1,2-a)$$

Now, we use the Relationship Between the Beta and Gamma Functions, $B(x,y)=\frac{\Gamma (x)\Gamma(y)}{\Gamma(x+y)}$, to write

$$\begin{align} \int_0^1 x^{a}\,(1-x)^{1-a}\,dx&=\frac{\Gamma(a+1)\Gamma(2-a)}{\Gamma(3)}\\\\ &=\frac12 \Gamma(a+1)\Gamma(2-a) \end{align}$$

Now, we use the functional relationship $\Gamma(z+1)=z\Gamma(z)$ along with Euler's Reflection Formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ to obtain

$$\begin{align} \int_0^1 x^{a}\,(1-x)^{1-a}\,dx&=\frac12 a\Gamma(a)(1-a)\Gamma(1-a)\\\\ &=\frac{\pi a(1-a)}{2\sin(\pi z)} \end{align}$$

and we are done!

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  • $\begingroup$ thank you for your answer! But I still have some questions concerning the first approach: 1. why can you simply multiply the phase to the integrand, instead of multipliying it to every z? 2. and why can you treat the residue at 0 (after inversion) as a pole of order 2 (I thought it is a pole of order a+2) $\endgroup$
    – user45494
    Dec 6, 2015 at 13:35
  • $\begingroup$ You're welcome. My pleasure. I don't understand the first question, so help me better understand please. For the second question, the residue at infinity maps to the residue at $0$ when transforming $z \to 1/z$. So, we don't have a residue at $1$ $\endgroup$
    – Mark Viola
    Dec 6, 2015 at 17:29

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