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 $$(e^{-2i \pi /4}-1) \int_0^{+\infty} \frac{x^{1/4}}{x^3+1}dx= \int_{+\infty}^0 \frac{x^{1/4}}{x^3+1}+\int_0^{+\infty} \frac{(e^{-2i\pi} x)^{1/4}}{(e^{-2i\pi }x)^3+1}d(e^{-2i\pi }x) = \int_C \frac{z^{1/4}}{z^3+1}dz$$ $$=\int_{C\cup S} \frac{z^{1/4}}{z^3+1}dz = 2i\pi (Res(\frac{z^{1/4}}{z^3+1},e^{-i\pi})+Res(\frac{z^{1/4}}{z^3+1},e^{-i\pi /3})+Res(\frac{z^{1/4}}{z^3+1},e^{-5i\pi /3}))$$