i'm just running through a problem, the example i have been given isn't extensive and so i'm trying to extrapolate from that. any help would be greatly appreciated.

Calculate $$g(z)=\int_0^\infty \frac{x^{\frac{1}{4}}}{x^3+1} \, dx$$

Now, i recognise this as an integral of the type $$\int_0^\infty x^{\alpha-1} f(x) \, dx$$ which we can move into the complex plane with $f$ is a rational function such that $f(\mathbb{R}) \subseteq \mathbb{R}$. $g$ is analytic on $\mathbb{C}$ with the exception of the finite points $\{-1, \frac{1-i\sqrt{3}}{2},\frac{1+i\sqrt{3}}{2}\}$ none of which are in $[0,\infty)$ and $a \in \mathbb{R} \setminus \mathbb{Z}$ (it's obvious to see that $a = \frac{5}{4}$ now we assume that there exists constants $R>0,M_1>0 \text{ and } \delta_1>0$ such that $$|f(z)z^{\alpha-1}|\leq \frac{M_1}{|z|^{1+\delta_1}},~|z|>R, \tag 1 $$ and constants $r>0,M_2>0 \text{ and } \delta_2>0$ such $$|f(z)z^{\alpha-1}| \leq M_2|z|^{\delta_2-1}, \tag 2$$ now we consider $$z^{\alpha - 1}=z^{\frac{1}{4}}$$ and $$f(z) = \frac{1}{z^3+1}$$

then the conditions of (1) and (2) works out to be, does $$\frac{z^{\frac{5}{4}}}{z^3+1} \longrightarrow 0$$ as $z \longrightarrow 0$ and $z \longrightarrow \infty$ obviously as $z$ tends to zero we have $0,$ and $z^3$ outpaces $z^{\frac{5}{4}}$ and so we have $0$ on both accounts.

now....on to the integral. $$\int_0^\infty \frac{z^{\frac{5}{4}}}{z^3+1} \, dz = \frac{2 \pi i}{1-e^{\frac{\pi i}{2} }}\left[ \operatorname{Res} \left(z^{\frac{1}{4}}f(z),-1\right) + \operatorname{Res} \left(z^{\frac{1}{4}}f(z),\frac{1-i\sqrt{3}}{2}\right) + \operatorname{Res} \left(z^{\frac{1}{4}}f(z),\frac{1+i\sqrt{3}}{2} \right)\right]-\frac{\left(\int_{S^{-}_{\epsilon}(0)} f(z) z^{\frac{1}{4}}+\int_{S^{+}_{\frac{1}{\epsilon}}(0)} f(z) z^{\frac{1}{4}}\right)}{1-e^{\frac{\pi i}{2}}}$$

now from my understanding this is the correct integral, i need to consider it as $\epsilon \longrightarrow 0$, now...one major issue i'm having is that i'm not understanding what the contour is in this instance, normally for $\pm \infty$ we just consider the upper half plane...are we considering only a single quarter in this instance?

I've seen solutions as both $$\frac{\pi}{3 \sin{(\frac{5 \pi}{12})}}$$ and $$\frac{1}{3}\Gamma\left(\frac{5}{12}\right)\Gamma\left(\frac{7}{12} \right)$$ where $$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x} \, dx$$ which i believe are equivalent.

Any and all help would be great.


Your question is unclear because the main step is to take the contour $C:+\infty \to +\infty $ enclosing $[0,\infty)$ clockwise and $S$ the counterclockwise circle $+\infty\to +\infty$ of infinite radius then $$(e^{-2i \pi /4}-1) \int_0^{+\infty} \frac{x^{1/4}}{x^3+1}dx= \int_{+\infty}^0 \frac{x^{1/4}}{x^3+1}+\int_0^{+\infty} \frac{(e^{-2i\pi} x)^{1/4}}{(e^{-2i\pi }x)^3+1}d(e^{-2i\pi }x) = \int_C \frac{z^{1/4}}{z^3+1}dz$$ $$=\int_{C\cup S} \frac{z^{1/4}}{z^3+1}dz = 2i\pi (Res(\frac{z^{1/4}}{z^3+1},e^{-i\pi})+Res(\frac{z^{1/4}}{z^3+1},e^{-i\pi /3})+Res(\frac{z^{1/4}}{z^3+1},e^{-5i\pi /3}))$$

  • 1
    $\begingroup$ Note Ninad's solution $\int_0^\infty \frac{x^{\frac{1}{4}}}{x^3+1}dx = \frac12 \int_{-\infty}^\infty \frac{4u^4}{u^{12}+1}du$ is easier but it works only for $x^{a/(2b)}$ whereas mine works for $x^\alpha,\alpha\in (-1,2)$ $\endgroup$ – reuns Nov 27 '19 at 6:20
  • $\begingroup$ sorry, just to make sure my maths is correct, isn't the poles of $1/(z^3+1)$, $$e^{i\pi}, e^{i \pi/3}~\&~e^{i5 \pi/3}$$ And so it'd be $\int_{C\cup S} \frac{z^{1/4}}{z^3+1}dz = 2i\pi (Res(\frac{z^{1/4}}{z^3+1},e^{i\pi})+Res(\frac{z^{1/4}}{z^3+1},e^{i\pi /3})+Res(\frac{z^{1/4}}{z^3+1},e^{5i\pi /3}))$ Of which they're all simple as well.. $\endgroup$ – Vaas Nov 27 '19 at 7:21
  • $\begingroup$ Another question is, do i need to calculate the residues explicitly and just do the algebra, or is there a method in which i can skip tedious algebra and go straight to the gamma functions? i understand where they come from in this instance but not whether its acase of just cracking on $\endgroup$ – Vaas Nov 27 '19 at 8:12
  • $\begingroup$ @reuns the problem with your solution (which is fixable) is that you use $e^{-i\pi/2}=(e^{-i2\pi})^{1/4}$ is not valid because the contour you choose (which seems like the quarter circle in the fourth quadrant) crosses the branch cut that makes that simplification valid. The solution is not as all encompassing as you say because of the shifting branch cuts of $z^{1/4}$ for each real value. $\endgroup$ – Ninad Munshi Nov 29 '19 at 6:21
  • $\begingroup$ hm, so then i need to consider a different branch in order to make this valid? $\endgroup$ – Vaas Nov 29 '19 at 8:51

First manipulate the real valued integral before you jump into the complex domain. Use the substitution $u^4=x$

$$\int_0^\infty \frac{x^{\frac{1}{4}}}{x^3+1}dx = \int_0^\infty \frac{4u^4}{u^{12}+1}du$$

Now there is no ambiguity with fractional powers.


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