How do I integrate these functions?

Solution 1:

First integral can be rewrite in the following form: $$I=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(x+1\right)x}\:dx$$ This view suggests the following change of variabel $u=x/(1-x)$, $dx=\dfrac{du}{(1+u)^2}$ $$I=\int _0^\infty\frac{ u^{-2/3}du}{(1+2u)}=\frac{1}{2^{1/3}}\int _0^\infty\frac{ u^{-2/3}du}{(1+u)}$$ The latter integral is the Beta function $$\frac{1}{2^{1/3}}B(1/3,2/3)=\frac{1}{2^{1/3}} \Gamma(1/3)\Gamma(2/3)=\frac{2^{2/3}\pi}{\sqrt{3}}\,.$$ Here I use the identity $\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin(\pi x)}$

To evaluate the secon integral I decompose the fraction $\dfrac{1}{1+x^2}=\frac{1}{2}\left(\dfrac{1}{1+i x}+\dfrac{1}{1-ix}\right)$ The integral $$I_a=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(1+a x\right)x}\:dx=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}$$ The integral $I_a$ is calculated absolutely analogously to the integral $I$. Note that this identity is valid not only for real positive $a$ but also for complex one. It is related wit the fact that the integral

$$I=\int _0^\infty\frac{ u^{-2/3}du}{(1+(1+a)u)}$$ can be evaluated for complex a using contour integration technique, but here I don't want touch it.

Thus we have $$ I_2=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(1+x^2\right)x}\:dx=\frac{2\pi}{\sqrt{3}}\frac{1}{2}\left(\frac{1}{(1-i)^{1/3}} + \frac{1}{(1+i)^{1/3}}\right)=\frac{2\pi}{\sqrt{3}} \frac{cos(\pi/12)}{2^{1/6}}=\frac{\pi}{2^{2/3}} \frac{\sqrt3+1}{\sqrt3} $$

EDIT: Let us show that $$I_a=\int _0^\infty\frac{ u^{-2/3}du}{(1+(1+a)u)}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$ The integrant has a cut in the complex plane from zero to infinity. We set this cut along posinive real axis. We will consider integral $$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}\,,$$ contour $\gamma$ is consists of two parts:

1) $z$ from $-i \epsilon +\infty$ to $-i\epsilon$ (against positive real axis a littel bit below cut

2) $z=\epsilon e^{-i \phi}$ $\phi$ from $\phi=\pi/2$ to $\phi=3\pi/2$ (semicircle)

3) $z$ from $i\epsilon$ to $i \epsilon +\infty$ (along positive real axis a littel bit higer cut

$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}=I_a(1-e^{2\pi i/3 })$$ In the other side this function has a pole in the poin $z=-1/(1+a)$ and we can calculate this integral by residue $$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}=\frac{-2\pi i e^{i \pi/3}}{(1+a)^{1/3}}\,. $$ Thus we obtain $$ I_a=\frac{\pi }{\sin(\pi/3)(1+a)^{1/3}}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$

Solution 2:

Let's avoid complex numbers if we don't need it. First, try the substitution $x = 1/z$, so $dx = -dz/z^2$. This changes the lower and upper limits of integration to $\infty$ and $1$ respectively, which we can flip and cancel the minus sign introduced by the differential substitution. The integrand, after some simple algebra, simplifies to $\frac{z^2}{(z+1)(z-1)^{1/3}}$. Thus, the original integral is equivalent to $\int_1^\infty \frac{dz}{(z+1)(z-1)^{1/3}}$. From here, we can use another substitution $u = (z-1)^{1/3}$ which changes the integral to $\int_0^\infty \frac{3udu}{u^3+2}$. The denominator can be factored using a sum of cubes formula, then partial fractions and splitting the numerator will give 3 terms. One will simplify to $\arctan$ after substitutions, while the other two will simplify to $\log$. Plug in your endpoints from the final substitution (or backsub to get an explicit indefinite integral in $x$) and you should get the answer marty cohen got from WolframAlpha.

Though I haven't tried it, I'm sure a similar approach can unpack the second integral as well, again avoiding complex numbers.

If you would like to use contour integration (which might make the computation easier), you can probably do it from the "$u$" equation with a clever choice of contour, but again I haven't tried it so no guarantees.