- Thread starter
- #1
- Jan 17, 2013
- 1,667
Prove that
\(\displaystyle \int^1_0 \frac{dx}{\sqrt[3]{x^2-x^3}} = \frac{2\pi }{\sqrt{3}}\)
\(\displaystyle \int^1_0 \frac{dx}{\sqrt[3]{x^2-x^3}} = \frac{2\pi }{\sqrt{3}}\)
Welcome to our community
Be a part of something great, join today!
An interesting integral
- Thread starter ZaidAlyafey
- Start date
- Jan 29, 2012
- 661
Hint: Use the Beta function and the complement formula
$$\int_0^1\frac{dx}{\sqrt[3]{x^2-x^3}}=\int_0^1x^{-2/3}(1-x)^{-1/3}dx=B\left(\frac{1}{3},\frac{2}{3}\right)= \frac{\Gamma\left( \frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{3}+\frac{2}{3}\right)}=\frac{\pi\;/\sin (\pi/3)}{1}=\frac{2\pi}{\sqrt{3}}$$
- Thread starter
- #3
- Jan 17, 2013
- 1,667
Yep , still there is another way . Possibly more complicated .
- Thread starter
- #4
- Jan 17, 2013
- 1,667
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 .
\(\displaystyle \oint_{\Gamma_2} f(z) \, dz +\oint_{\Gamma_4} f(z) \, dz+ \int_{\Gamma_3} f(z) \, dz + \int_{\Gamma_1} f(z) \, dz = 2\pi i \text{Res}(f(z) ; -1) \)
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 \(\displaystyle \square\) .
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 .
\(\displaystyle \oint_{\Gamma_2} f(z) \, dz +\oint_{\Gamma_4} f(z) \, dz+ \int_{\Gamma_3} f(z) \, dz + \int_{\Gamma_1} f(z) \, dz = 2\pi i \text{Res}(f(z) ; -1) \)
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 \(\displaystyle \square\) .