$\int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}$

Question

I want to evaluate following integral \begin{align} \int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)} \end{align}

Simple try on this integral is using branch cut and apply residue theorem.

Usual procedure gives for $0 < \alpha < 1$, with $Q(x)$ deg $n$ and $P(x)$ deg $m$, for $x>0$, $Q(x) \neq 0$

\begin{align} \int_0^{\infty} \frac{x^\alpha P(x)}{Q(x)} dx = \frac{2\pi i}{1- e^{i\alpha 2 \pi}} \sum_j Res[\frac{z^\alpha P(z)}{Q(z)} , z_j] \end{align}

where $z_j$ are poles which does not make $\frac{P}{Q}$ be zero.
This formula comes from Mathews and Howell's complex analysis textbook. And this is nothing but applying branch cut to make $x^{\frac{1}{4}}$ singled valued function. I think this formula works for above improper integral but results seems different.

Apply $\alpha=\frac{1}{4}$ and take poles $z_0=-1$, $z_1 = e^{\frac{i \pi}{3}}$ , $z_2 = e^{\frac{i5 \pi}{3}}$, i got different things.

Am i doing right?

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\begin{align} \frac{2\pi i}{1-i}\frac{1}{3} \left( e^{\frac{1}{4} \pi i} + e^{-\frac{7}{12}\pi i} + e^{-\frac{35}{12} \pi i}\right) \end{align}

Answer 1

You may simply remove the branch cut by setting $x=z^4$: $$ I = 4 \int_{0}^{+\infty}\frac{z^4\,dz}{1+z^{12}} = 2\int_{-\infty}^{+\infty}\frac{z^4}{1+z^{12}} \tag{1}$$ and by evaluating the residues at the roots of $1+z^{12}$ in the upper half-plane, $$ I = \frac{2\pi}{3\sqrt{2+\sqrt{3}}}=\frac{\pi}{3}\left(\sqrt{6}-\sqrt{2}\right)\tag{2} $$ follows. By setting $\frac{1}{1+z^{12}}=u$, the integral $(1)$ can also be evaluated through Euler's Beta function and the reflection formula for the $\Gamma$ function, since: $$ I = \frac{1}{3}\int_{0}^{1}u^{-5/12}(1-u)^{-7/12}\,du = \frac{1}{3}\,\Gamma\left(\frac{5}{12}\right)\,\Gamma\left(\frac{7}{12}\right)=\frac{\pi}{3\sin\frac{5\pi}{12}}.\tag{3}$$

Answer 2

Letting $y=\frac{1}{1+x^3}$ yields $$ \begin{aligned} I &=\int_{1}^{0} y\left(\frac{1}{y}-1\right)^{\frac{1}{12}} \frac{1}{3}\left(\frac{y}{1-y}\right)^{\frac{2}{3}} \frac{d y}{-y^{2}} \\ &=\frac{1}{3} \int_{0}^{1} y^{\frac{1}{12}-1}(1-y)^{\frac{5}{12}-1} d y\\&= \frac{1}{3} B\left(\frac{7}{12}, \frac{5}{12}\right) \end{aligned} $$ Using the property of Beta function$$ B(x, 1-x)=\pi \csc (\pi x), \quad \textrm{ where } x \notin \mathbb{Z}, $$ we can now conclude that $$ \boxed{I=\frac{\pi}{3} \csc \frac{5 \pi}{12}} $$