How to Prove this Interesting Integral Equation?

[alyafey22]

Prove that

\(\displaystyle \int^1_0 \frac{dx}{\sqrt[3]{x^2-x^3}} = \frac{2\pi }{\sqrt{3}}\)

 

[Fernando Revilla]

Hint: Use the Beta function and the complement formula

Spoiler

$$\int_0^1\frac{dx}{\sqrt[3]{x^2-x^3}}=\int_0^1x^{-2/3}(1-x)^{-1/3}dx=B\left(\frac{1}{3},\frac{2}{3}\right)= \frac{\Gamma\left( \frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{3}+\frac{2}{3}\right)}=\frac{\pi\;/\sin (\pi/3)}{1}=\frac{2\pi}{\sqrt{3}}$$

 

[alyafey22]

Yep , still there is another way . Possibly more complicated .

 

[alyafey22]

It can be solved using contour integration . The approach is quite long I might post it later .

On the other hand we can transform to another contour integration using a substitution .

\(\displaystyle t = \frac{1}{x}-1 \,\,\, dt = - \frac{1}{x^2}dx\)

\(\displaystyle \int^{\infty}_0 \frac{dt}{ \sqrt[3]{t}(t+1)}\)

we can use the following function to integrate

\(\displaystyle f(z) = \frac{e^{\frac{-1}{3}\text{Log}_0(z)}}{z+1}\)

Notice we are choosing a branch cut along the positive x-axis .

We will use a key-hole contour as indicated in the picture .

View attachment 909

\(\displaystyle \oint_{\Gamma_2} f(z) \, dz +\oint_{\Gamma_4} f(z) \, dz+ \int_{\Gamma_3} f(z) \, dz + \int_{\Gamma_1} f(z) \, dz = 2\pi i \text{Res}(f(z) ; -1) \)

Integration along the big circle :

\(\displaystyle \oint_{\Gamma_2}\frac{e^{\frac{-1}{3}\text{Log}_0(z)}}{z+1}\, dz \)

We will use the following paramaterization \(\displaystyle z = Re^{it}\,\,\, 0< t < 2\pi \)

\(\displaystyle iR\int_{0}^{2\pi }\frac{e^{it}\, e^{\frac{-1}{3}\text{Log}_{0}(Re^{it})}}{Re^{it}+1}\, dt \leq 2 \pi \frac{R^{1-\frac{1}{3}}}{R-1} \)

Now take $R$ to be arbitrarily large

\(\displaystyle \lim_{R \to \infty}2 \pi \frac{R^{\frac{2}{3}}}{R-1} = 0 \)

Similarity the integration along the smaller circle goes to $0$ as \(\displaystyle r \to 0\) .

Integration along the x-axis :

\(\displaystyle \text{P.V} \int^{\infty}_{0} \frac{e^{\frac{-1}{3}\text{Log}_0(z)}}{z+1}\, dz \)

Along the positive x-axis we get the following

\(\displaystyle \text{P.V} \int^{\infty}_{0} \frac{e^{\frac{-1}{3}\ln(x)}}{x+1}\, dx\)

For the other integral on the opposite direction

\(\displaystyle \text{P.V} \int^{0}_{\infty} \frac{e^{\frac{-1}{3}(\ln(x)+2\pi i)}}{x+1}\, dx\)

The residue at \(\displaystyle -1\)

Since we have a simple pole

\(\displaystyle \text{Res}(f;-1)= e^{-\frac{1}{3}\text{Log}_0(-1)} = e^{-\frac{\pi}{3} i} \)

Final step

\(\displaystyle \text{P.V}(1-e^{-\frac{2\pi}{3} i } )\int^{\infty}_{0} \frac{1}{\sqrt[3]{x}(x+1)}\, dx = 2\pi i e^{-\frac{\pi}{3} i} \)

\(\displaystyle \text{P.V}\int^{\infty}_{0} \frac{1}{\sqrt[3]{x}(x+1)}\, dx = 2\pi i \frac{e^{-\frac{\pi}{3} i}}{1-e^{-\frac{2\pi}{3} i } } \)

We can easily prove that \(\displaystyle 2\pi i \frac{e^{-\frac{\pi}{3} i}}{1-e^{-\frac{2\pi}{3} i } }=\frac{\pi }{\sin \left( \frac{\pi}{3}\right)}=\frac{2\pi }{\sqrt{3}}\)

Hence the result

\(\displaystyle \text{P.V}\int^{\infty}_{0} \frac{1}{\sqrt[3]{x}(x+1)}\, dx = \frac{2\pi }{\sqrt{3}} \)

We can easily prove that the integral converges hence the principle value is equal to the integral \(\displaystyle \square\) .

 

[CellCoree]

This is a fascinating integral that involves a rational function with a fractional exponent. To prove the given statement, we can use the substitution u = x^2, which transforms the integral into:

\int^1_0 \frac{dx}{\sqrt[3]{x^2-x^3}} = \int^1_0 \frac{du}{2\sqrt[3]{u-u^2}}

Next, we can use the substitution v = u - \frac{1}{2} to simplify the integrand:

\int^1_0 \frac{du}{2\sqrt[3]{u-u^2}} = \int^{\frac{1}{2}}_{-\frac{1}{2}} \frac{dv}{\sqrt[3]{v(1-v)}}

Using the trigonometric substitution v = \sin^2\theta, we can further simplify the integral to:

\int^{\frac{1}{2}}_{-\frac{1}{2}} \frac{dv}{\sqrt[3]{v(1-v)}} = \int^{\frac{\pi}{6}}_{-\frac{\pi}{6}} \frac{2\sin\theta \cos\theta d\theta}{\sqrt[3]{\sin^2\theta \cos^2\theta}} = \int^{\frac{\pi}{6}}_{-\frac{\pi}{6}} 2\sqrt[3]{\cot^2\theta \csc^2\theta} d\theta

Next, we can use the identity \cot^2\theta = \csc^2\theta - 1 to simplify the integrand:

\int^{\frac{\pi}{6}}_{-\frac{\pi}{6}} 2\sqrt[3]{\cot^2\theta \csc^2\theta} d\theta = \int^{\frac{\pi}{6}}_{-\frac{\pi}{6}} 2\sqrt[3]{\csc^2\theta(\csc^2\theta - 1)} d\theta

Using the substitution w = \csc\theta, we can further simplify the integral to:

\int^{\frac{\pi}{6}}_{-\frac{\pi}{6}} 2\sqrt[3]{\csc^2\theta(\csc^2\theta - 1