# How can I solve this definite Integral?

I need to evaluate the integral

$$\displaystyle \int_0^{1} x^{-1/3}\left(1-x\right)^{-2/3}\left(1 + 2x\right)^{-1} dx$$

I don't know how can I simplify this to get in a proper form to integrate, I have tried various trigonometric substitutions but nothing seems to work over here

Since the limits are from $$0$$ to $$1$$. I strongly suspect this integral is related to beta function.

But I am not able to see how can I use this observation on this integral .

Can anyone please solve this integral ?

Thank you.

• WA says this here $$-\frac{15 x F_1\left(\frac{2}{3};\frac{1}{3},1;\frac{5}{3};x,-2 x\right)}{2 \sqrt[3]{-(x-1) x} (2 x+1) \left(-5 F_1\left(\frac{2}{3};\frac{1}{3},1;\frac{5}{3};x,-2 x\right)+6 x F_1\left(\frac{5}{3};\frac{1}{3},2;\frac{8}{3};x,-2 x\right)-x F_1\left(\frac{5}{3};\frac{4}{3},1;\frac{8}{3};x,-2 x\right)\right)}$$ – Dr. Sonnhard Graubner Oct 13 '19 at 13:30
• Hypergeometric form can be easily obtained by Euler's 2F1 representation. – MHZ Oct 13 '19 at 15:00

The beta function $$\mathrm{B}$$ is really involved.

The substitution $$x=\dfrac{y}{3+y}$$ yields $$\begin{gather*} \int_{0}^{1}x^{-1/3}(1-x)^{-2/3}(1+2x)^{-1}\, dx = 3^{-2/3}\int_{0}^{\infty}\dfrac{y^{2/3-1}}{(1+y)^{2/3+1/3}}\, dy = \\[2ex] 3^{-2/3}\mathrm{B} \left(\dfrac{2}{3},\dfrac{1}{3}\right) = 3^{-2/3} \dfrac{\Gamma\left(\dfrac{2}{3}\right)\Gamma\left(\dfrac{1}{3}\right)}{\Gamma(1)} = 3^{-2/3} \Gamma\left(1-\dfrac{1}{3}\right)\Gamma\left(\dfrac{1}{3}\right)=\\[2ex] 3^{-2/3}\dfrac{\pi}{\sin(\pi/3)} =\dfrac{2\pi\, 3^{5/6}}{9}. \end{gather*}$$

• Nice answer, My only question is :How did you think of that substitution ? – sat091 Oct 13 '19 at 17:25
• In a sense, I used trial and error. The beta function has a lot of properties e.g. \begin{equation*} \mathrm{B}(x,y) = \int_{0}^{\infty}\dfrac{t^{x-1}}{(1+t)^{x+y}}\, dt \end{equation*} In order to be able to integrate over $[0,\infty)$ I tested $y = \dfrac{x}{1-x}$ which I then modified to $y = \dfrac{3x}{1-x} \Leftrightarrow x = \dfrac{y}{y+3}$. The substitution $x = \dfrac{1}{3y+1}$ also works. – JanG Oct 13 '19 at 18:43

The integrand has an elementary primitive and here is a way to find it:

The curve $$y^3=x(1-x)^2$$ looks like this:

We note that $$(0,0)$$ belongs to the curve, and try to parametrize it with $$t$$ satisfying $$y=tx$$. This gives us $$x=\frac{1-t^{3/2}}{1-t^3},\quad y=t\frac{1-t^{3/2}}{1-t^3},\quad\text{and}\quad dx=-\frac{3\sqrt{t}}{2(1+t^{3/2})^2}\,dt.$$ The parameter $$t$$, being the direction coefficient from the line $$y=tx$$, will run from $$+\infty$$ to $$0$$ as $$x$$ runs from $$0$$ to $$1$$. A straight-forward calculation (in the last step we set $$t=u^2$$) gives \begin{aligned} \int_0^1\frac{1}{x^{1/3}(1-x)^{2/3}(1+2x)}\,dx &= \int_0^1\frac{1}{y(1+2x)}\,dx\\ &= \frac{3}{2}\int_0^{+\infty}\frac{1}{t^2+3\sqrt{t}}\,dt\\ &=3\int_0^{+\infty}\frac{1}{3+u^3}\,du. \end{aligned} The last integral (known by Euler, I've read somewhere) can be taken care of in the "usual" way with partial fraction decomposition and logarithms and inverse tangents will appear (Residue calculus could give a shorter/handier calculation).

I will not write details but just conclude that one could benefit much from knowing how to deal with the Beta function, as JanG demonstrates.