Solving an integral using Dogbone contour

Prove that

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

$$\textit{proof}$$

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

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

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

Consider the indented circle around \( z=0 \)

$$\left| \int^{2\pi-\epsilon}_{\epsilon} e^{\frac{1}{2}\log(re^{i\theta}-r^2e^{2i\theta})} ir e^{i\theta}\,d\theta \right| \leq r \int^{2\pi-\epsilon}_{\epsilon} e^{\frac{1}{2}\log|re^{i\theta}-r^2e^{2i\theta}|} \,d\theta = r^{\frac{3}{2}} \int^{2\pi-\epsilon}_{\epsilon} e^{\frac{1}{2}\log|1-re^{i\theta}|} \,d\theta \to 0$$

Similarly for the contour around 1. Finally we get

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

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