Prove that

$$\int^{1}_{0} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{8}$$


Consider the function

$$f(z) = \sqrt{z-z^2} = e^{\frac{1}{2}\log(z-z^2)}$$

Consider the branch cut on the x-axis $$x(1-x)\geq 0\,\, \implies \, 0\leq x \leq 1 $$

Consider $ w= z-z^2 $ then

$$\log(w) = \log|w|+i\theta,\,\, \theta\in[0,2\pi)$$

Consider the contour

enter image description here Consider the integral

$$\int_{c_0}f(z)\,dz+\int_{c_1}f(z)\,dz+\int^{1-\epsilon}_{\epsilon} e^{\frac{1}{2}\log|x-x^2|}\,dx-\int^{1-\epsilon}_{\epsilon} e^{\frac{1}{2}\log|x-x^2| +\pi i}\,dx = 2\pi i \mathrm{Res}(f,\infty)$$

Consider the Laurent expansion of

$$\sqrt{z-z^2} =i \sqrt{z^2} \sqrt{1-\frac{1}{z}}= iz\sum_{k=0}^\infty{\frac{1}{2} \choose k} \left(-\frac{1}{z} \right)^k$$

Hence we deuce that

$$ \mathrm{Res}(f,\infty) = -\frac{i}{8}$$

That implies

$$\int_{c_0}f(z)\,dz+\int_{c_1}f(z)\,dz+2\int^{1-\epsilon}_{\epsilon} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{4}$$

Considering integrals the contours around $c_0$ and $c_1$ go to zero . Finally we get

$$\int^{1}_{0} \sqrt{x}\sqrt{1-x}\,dx = \frac{\pi}{8}$$


I have my concerns about the expansion at infinity

$$\sqrt{z-z^2} =i \sqrt{z^2} \sqrt{1-\frac{1}{z}}= iz\sum_{k=0}^\infty{\frac{1}{2} \choose k} \left(-\frac{1}{z} \right)^k$$

First I am assuming that $\sqrt{z^2} = z$ which seems to be wrong on the chosen branch cut. Also it is wrong to assume that $\sqrt{zw} = \sqrt{z}\sqrt{w}$.

  • $\begingroup$ You can define the branch cuts of the two square roots $\sqrt{z}$ and $\sqrt{1-z}$ separately. In this case you should choose the negative real axis for $\sqrt{z}$ and $(-\infty,1]$ for $\sqrt{1-z}$. In the product you end up with a branch cut between $0$ and $1$ while on the negative real axis the two branch cuts effectively cancel each other out. This means that on $\mathbb{C}\[0,1]$ the function is analytic, and the residue at infinity can be computed without problems. $\endgroup$ – Count Iblis Feb 14 '17 at 0:33
  • 1
    $\begingroup$ $\sqrt{z^2}=|z|$. $\endgroup$ – Takahiro Waki Feb 14 '17 at 0:38
  • $\begingroup$ @CountIblis, I appreciate your help but my interest is not solving using another approach. I want to understand what i did wrong and how to resolve it? $\endgroup$ – Zaid Alyafeai Feb 14 '17 at 0:53
  • $\begingroup$ @TakahiroWaki What are you thinking? $z\in \mathbb{C}$ here. $\endgroup$ – Mark Viola Feb 14 '17 at 3:24

There is a conceptual error in the OP.

Note that if $w=z(1-z)$, then the condition $\arg(w)\in[0,2\pi)$ restricts the domain of the complex $z$ plane to a half space.

To see this, we write $z=|z|e^{i\arg(z)}$ and $1-z=|1-z|e^{i\arg(1-z)}$ so that $w=|z||1-z|e^{i\left(\arg(z)+\arg(1-z)\right)}$.

Inasmuch as $\arg(z)+\arg(1-z)$ spans a range of $4\pi$ in the complex $z$-plane, then $\arg(w)$ does likewise.

To address the concerns in the OP in detail, we begin with a short primer.


The complex logarithm $\log(z)$ is defined for $z\ne 0$ as

$$\log(z)=\log(|z|)+i\arg(z) \tag 1$$

It is easy to show that the complex logarithm satisfies

$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag 2$$

which means that any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$.

To see that $(2)$ is true, we simply note that $\log(|z_1||z_2|)=\log(|z_1|)+\log(z_2)$ and $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$.

NOTE: The relationship in $(2)$ is not generally satisfied when the logarithm is restricted on, say, the principal branch for which $\arg(z)=\text{Arg}(z)$, where $-\pi<\text{Arg}\le \pi$.

Using $(2)$, we can write for $z\ne0$, $z\ne 1$

$$\begin{align} f(z)&=\sqrt{z(1-z)}\\\\ &=e^{\frac12 \log(z(1-z))}\\\\ &=e^{\frac12 \left(\log(z)+\log(1-z)\right)}\\\\ &=e^{\frac12\log(z)}e^{\frac12\log(1-z)}\\\\ &=\sqrt{z}\sqrt{1-z} \end{align}$$

SELECTING A BRANCH OF $\displaystyle \sqrt{z(1-z)}$

To obtain a specific branch of $\sqrt{z(1-z)}$, we can use a branch of $\sqrt{z}$ and another branch of $\sqrt{1-z}$.

If we select the branches for $\sqrt{z}$ and $\sqrt{1-z}$ to be such that $-\pi<\arg(z)\le \pi$ and $0<\arg(1-z)\le 2\pi$, then the branch of $\sqrt{z(1-z)}$ is such that

$$\sqrt{z(1-z)}=\sqrt{|z||1-z|}e^{i\frac12 (\arg(z)+\arg(1-z))}$$

with $-\pi<\arg(z)+\arg(1-z)\le 3\pi$.

With this choice, it is straightforward to show that $\sqrt{z(1-z)}$ is analytic on $\mathbb{C}\setminus [0,1]$.


Then, we can write

$$\begin{align} \oint_C \sqrt{z(1-z)}\,dz&=\int_0^1 \sqrt{x(1-x)}e^{i(0+2\pi)/2}\,dx+\int_1^0\sqrt{x(1-x)}e^{i(0+0)}\,dx\\\\ &=-2\int_0^1 \sqrt{x(1-x)}\,dx \tag 3 \end{align}$$

NOTE: We bypassed consideration of the contributions to the integral from the circular deformations around the branch points since their contributions vansish in the limit as the radii go to zero.

Using Cauchy's Integral Theorem, the value of the integral $\oint_C \sqrt{z(1-z)}\,dz$ is unaltered by deforming $C$ into a circular contour, centered at the origin, of radius, $R>1$. Hence, exploiting the analyticity of $\sqrt{z(1-z)}$ for $R>1$, we have

$$\begin{align} \oint_C \sqrt{z(1-z)}\,dz&=\oint_{R>1}\sqrt{z(1-z)}\,dz\\\\ &=\int_{-\pi}^{\pi}\sqrt{Re^{i\phi}(1-Re^{i\phi})}\,iRe^{i\phi}\,d\phi\\\\ &=\int_{-\pi}^{\pi}\sqrt{Re^{i\phi}(1-Re^{i\phi})}\,iRe^{i\phi}\,d\phi\\\\ &=-\int_{-\pi}^{\pi} \left(iRe^{i\phi}\right)^2 \left(1-\frac{1}{Re^{i\phi}}\right)^{1/2}\,d\phi\\\\ &=-\int_{-\pi}^{\pi} \left(iRe^{i\phi}\right)^2 \left(1-\frac{1/2}{Re^{i\phi}}-\frac{1/8}{(Re^{i\phi})^2}-\frac{1/16}{(Re^{i\phi})^3}+O\left(\frac1{(Re^{i\phi})^4}\right)\right)\,d\phi\\\\ &\to -\frac{\pi}{4}\,\,\text{as}\,\,R\to \infty \tag 4 \end{align}$$

Finally, putting together $(3)$ and $(4)$ yields

$$\int_0^1 \sqrt{x(1-x)}\,dx=\frac{\pi}{8}$$

as expected.

NOTE: The expansion leading to $(4)$ is correct given the chosen branches of $\sqrt{z}$ and $\sqrt{1-z}$. Then,

$$\begin{align} \sqrt{z(1-z)}&=\sqrt{-z^2\left(1-\frac1z\right)}\\\\ &=e^{\frac12\log(-z^2)+\frac12\log\left(1-\frac1z\right)}\\\\ &=iz \sqrt{1-\frac1z} \end{align}$$

where we used $\log(-1)=i\pi$ and $\log(z^2)=2\log(z)$. Then, upon expanding $\sqrt{1-\frac1z}$ in its Laurent series in the annulus $1<z<\infty$, and setting $z=Re^{i\phi}$, we obtain the expansion used to arrive at $(4)$.

  • $\begingroup$ Hey Mark, I appreciate the effort. I still have some confusion you are saying that I have restricted the value of the argument but I know that any complex number argument should be restricted to a period of length $2\pi$ in order to make it single-valued. $\endgroup$ – Zaid Alyafeai Feb 14 '17 at 19:02
  • $\begingroup$ Zaid, I understand. In this case, we are mapping the $z$-plane to the $w$ plane. And we can see that $w$ gets covered twice. This is no different from the mapping of $w=z^2$ for which the upper half $z$-plane maps into the entire $w$ plane and the lower half $z$-plane does likewise. We really need $\sqrt{z(1-z)}$ to be single valued by selecting a branch- and it is single valued given the chosen branches for $\sqrt{z}$ and $\sqrt{1-z}$. Its argument is on the interval $(-\pi/2,3\pi/2]$. $\endgroup$ – Mark Viola Feb 14 '17 at 19:21
  • $\begingroup$ By way of enrichment I present a computation where only invertible logarithms of positive real numbers are used once the two branches for the integrand have been selected, consult my post below. This is computation-intensive but it produces the correct value. $\endgroup$ – Marko Riedel Feb 15 '17 at 3:25
  • $\begingroup$ Hey Mark , thanks. I marked your answer as accepted because of the thorough explanation. $\endgroup$ – Zaid Alyafeai Feb 15 '17 at 22:16
  • $\begingroup$ @ZaidAlyafeai Zaid, you're certainly welcome. It was my pleasure. And I really do hope that it helped resolve your concerns. -Mark $\endgroup$ – Mark Viola Feb 15 '17 at 23:47

1) When you have $\sqrt{z^2}$, technically we have $(e^{i\pi}z)^2$ by the branch cut, but obviously that cancels out to $z.$

2) Also the given assumption is valid here because of the expansion that has been chosen; the "circle" of radius of convergence for z in $\sqrt{1-\frac{1}{z}}$ contains $\infty$ (since the area of convergence is really outside the circle here).

  • $\begingroup$ How is the first line correct ? $\endgroup$ – Zaid Alyafeai Feb 13 '17 at 23:19
  • $\begingroup$ The $\sqrt{z-z^2}$ term simplifies as mentioned earlier. As for $\sqrt{1-\frac{1}{z}}$, note that we can use the binomial theorem since it's $(1-\frac{1}{z})^{\frac{1}{2}}.$ Or maybe I'm answering the wrong thing.. $\endgroup$ – pie314271 Feb 14 '17 at 4:58

We may use the method from the following MSE link for the residue at infinity. The introduction and the continuity argument can be copied verbatim and will not be repeated here. We use

$$f(z) = \exp(1/2\times \mathrm{LogA}(z)) \exp(1/2\times \mathrm{LogB}(1-z))$$

with the two logarithms defined as in the linked-to post. Note that this is not the choice Mathematica or Maple makes, but consistency is sufficient here and we can re-use a vetted computation.

The difference is that we have $f(z)\sim z$ at infinity so we need to determine the coefficients on $z,$ the constant coefficient and the one on $1/z.$ There was less work at the linked-to computation because the function was $\sim 1/z$ at infinity.

We use $f(z)/z^2$ for the first one and put $z = R\exp(i\theta).$ Now the modulus of $\mathrm{LogA}(z)$ is $\log R.$ We get for the modulus of $\mathrm{LogB}(1-z)$

$$\log\sqrt{(1-R\cos(\theta))^2 + R^2\sin(\theta)^2} \\ = \log\sqrt{1-2R\cos(\theta) + R^2}.$$

We are manipulating logarithms of positive real numbers using the real logarithm and may continue with

$$\log R + \log\sqrt{1-2\cos(\theta)/R+1/R^2}.$$

Using the leading term and the method from the link we immediately obtain $f(z) \sim -iz.$ For the constant coefficient we use $(f(z)+iz)/z.$ We have

$$\exp\left(\frac{1}{2}\log\sqrt{1-2\cos(\theta)/R+1/R^2}\right) = \sqrt[4]{1-2\cos(\theta)/R+1/R^2} \\ = 1 - \frac{1}{4} (2\cos(\theta)/R-1/R^2) - \frac{3}{32} (2\cos(\theta)/R-1/R^2)^2 -\cdots \\ = 1 - \frac{1}{2}\cos(\theta)\frac{1}{R} + \left(1/4 - \frac{3}{8}\cos(\theta)^2\right)\frac{1}{R^2} - \cdots$$

We also have

$$\arctan\left(\frac{-R\sin(\theta)}{1-R\cos(\theta)}\right) = \arctan\left(\frac{-\sin(\theta)}{1/R-\cos(\theta)}\right) \\ = \theta + \sin(\theta)\frac{1}{R} + \sin(\theta)\cos(\theta)\frac{1}{R^2} + \cdots$$

so that (this actually holds everywhere as long as $\theta$ matches the range of $\arg\mathrm{LogB}$)

$$\exp\left(\frac{1}{2}i\arg\mathrm{LogB(1-z)}\right) \\ = \exp\left(\frac{1}{2}i\theta\right) \left(1 + \frac{1}{2}i\sin(\theta)\frac{1}{R} + \frac{1}{2}i\sin(\theta)\cos(\theta)\frac{1}{R^2} - \frac{1}{8} \sin(\theta)^2\frac{1}{R^2} + \cdots\right).$$

Adding in $iz$ cancels the contribution from the first terms of these two expansions. We get from the second term and collecting the contributions from the lower and upper half plane

$$\int_0^{2\pi} \frac{1}{R\exp(i\theta)} \exp(\log R) \exp(i\theta+\pi i/2) \\ \times \left(\frac{1}{2}i\sin(\theta)\frac{1}{R} -\frac{1}{2}\cos(\theta)\frac{1}{R}\right) Ri\exp(i\theta) \; d\theta \\ = \int_0^{2\pi} \exp(\log R) \exp(i\theta+\pi i/2) i \left(-\frac{1}{2}\exp(-i\theta)\right) \frac{1}{R} \; d\theta \\ = \frac{1}{2} \int_0^{2\pi} \exp(i\theta)\exp(-i\theta) \; d\theta = \pi.$$

for a residue of $-(\pi)/(2\pi i) = i/2.$ This establishes $f(z) \sim -iz + i/2.$ We integrate $f(z) + iz - i/2$ to get the coefficient on $1/z$, obtaining

$$\int_0^{2\pi} \exp(\log R) \exp(i\theta+\pi i/2) \\ \times \frac{1}{R^2} \left(\frac{1}{4}-\frac{3}{8}\cos(\theta)^2 + \frac{1}{2} i\sin(\theta)\cos(\theta) - \frac{1}{8} \sin(\theta)^2 -\frac{1}{4}i\sin(\theta)\cos(\theta)\right) \\ \times Ri\exp(i\theta) \; d\theta$$

The inner term is

$$\frac{1}{8} - \frac{1}{4}\cos(\theta)^2 + \frac{1}{8}i\sin(2\theta) = - \frac{1}{8}\cos(2\theta) + \frac{1}{8}i\sin(2\theta) = - \frac{1}{8}\exp(-2i\theta).$$

which leaves for the integral

$$\frac{1}{8} \int_0^{2\pi} \exp(2i\theta)\exp(-2i\theta) \; d\theta = \frac{\pi}{4}.$$

We get the residue $-(\pi/4)/(2\pi i) = i/8.$ We have established that at infinity,

$$\bbox[5px,border:2px solid #00A000]{ f(z) \sim -iz + \frac{1}{2} i + \frac{1}{8} i \frac{1}{z}}$$

and hence $\mathrm{Res}_{z=\infty} f(z) = \frac{1}{8} i.$ Taking into account that the contour used with this branch produces twice the value of the integral we obtain

$$\frac{1}{2} \times -2\pi i \times \frac{1}{8} i$$

which is

$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{8}.}$$


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