Another (perhaps tricky) integral.

Question

While solving my Math paper, I came across this integral, and I can't see any way to solve it. At least, any easy way. The integral is-

$$ \int{x^{2} \over 1 + x^{5}}\,\mathrm{d}x $$

I'm not even sure it exists, by the looks of it. Any help ?.

Answer 1

We have $$x^5+1 = (x+1)(x^4-x^3+x^2-x+1) = (x+1)(x^2+ax+1)(x^2-bx+1)$$ where $a-b=-1$ and $ab=1$.

Now use partial fractions and proceed.

Answer 2

I have the following rather lengthy approach: Initially, we note that the integral does not exist due to the pole at x=-1. However, $\int_0^\infty x^2/x^5 dx$ exists since $|x|^2/|x|^5 \le |x|^2$ for sufficiently large $|x| > 0$. I think, we can deal with this integral using the Residue theorem from function theory: f(z):= z^2/(1+z^5) is holomorphic on $\mathbb{C}$ except for the roots of $(1+z^5)$ which are given as $$ a_k = e^{\pi i/5} e^{2\pi i k/5} \qquad k=0,...,4.$$ Consider the path $$\gamma_R:= \alpha_R * \beta_R * \tilde\alpha_R$$ (first go $\alpha_R$ then $\beta_R$...) with $$\alpha_R(t) = R t \qquad t \in [0,1]$$ $$\beta_R(t) = Re^{it} \qquad t \in [0, 2\pi/5]$$ $$\tilde\alpha_R(t) = (1-t)Re^{2\pi i/5} \qquad t\in [0,1]$$ for $R>1$. $\gamma_R$ goes a around a segment of B_R(0) which only encloses $a_0$. Therefore the Residue theorem yields $$2\pi i Res_{a_0}f = \int_{\gamma_R}f(z)\, dz = \int_{\alpha_R}f + \int_{\beta_R}f + \int_{\tilde \alpha_R}f$$ Using Cauchys inequality from function theory the middle integral goes to 0 for $R\to \infty$. Now it can be shown that (found in most textbooks) $$Res_{a_0}f = a_0^2/(5a_0^4) = 1/5 a_0^{-2} = 1/5 e^{-2/5 \pi i}.$$ and then $$2/5 \pi i e^{-2/5 \pi i} = \int_0^R x^2/(1+x^5) dx - \int_0^R e^{2/5\pi i}\frac{(te^{2/5\pi i})^2}{1+(te^{2/5\pi i})^5} dt \\ = (1-e^{6/5\pi i})\int_0^R x^2/(1+x^5) dx \to (1-e^{6/5\pi i})\int_0^\infty x^2/(1+x^5) dx$$ for $R\to \infty$. Therefore $$ \int_0^\infty x^2/(1+x^5) dx = \frac{2/5\pi i e^{-2/5\pi i}}{(1- e^{6/5\pi i})} = \frac{\pi/5}{1/(2i)(e^{2/5\pi i} - e^{8/5\pi i})} = \frac{\pi/5}{\sin(2/5 \pi)}$$ I have compared this to the result form wolfram alpha and it checks out. I hope I could help.

BS