How can I convert this tricky complex number into a real number: $\int_0^{\infty}\frac{x^α}{x^3+1}dx$?

Question

The problem statement is:

$$\int_0^{\infty}\frac{x^α}{x^3+1}dx$$ for α in the range −1<α<2.

$$\huge \frac{2\pi i}{1-e^{\frac{i2\pi (\alpha+1)}{3}}} \frac {e^{\frac{i \pi \alpha}{3}}} { 3e^{\frac{2\pi i}{3}}}$$

$\alpha$ is some constant between -1 and 2.

Thanks,

Answer 1

Multiply by $\frac{e^{\frac{-i\pi(a+1)}3}}{e^{\frac{-i\pi(a+1)}3}}$: $$ \begin{align} \frac{2\pi i}{1-e^{\frac{i2\pi(a+1)}3}}\frac{e^{\frac{i\pi a}3}} {3e^{\frac{2\pi i}3}} &=\frac{2\pi i}{e^{-\frac{i\pi(a+1)}3}-e^{\frac{i\pi(a+1)}3}}\frac{e^{-\frac{i\pi}3}} {3e^{\frac{2\pi i}3}}\\[6pt] &=\frac{\frac\pi3}{-\sin\left(\frac\pi3(a+1)\right)}e^{-i\pi}\\[6pt] &=\frac{\frac\pi3}{\sin\left(\frac\pi3(a+1)\right)} \end{align} $$ since $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$.

Answer 2

HINT:

Assuming $x\in\mathbb{R}$:

$$\frac{2\pi i}{1-e^{\frac{i2\pi x}{3}}} \frac {e^{\frac{i \pi x}{3}}} { 3e^{\frac{2\pi i}{3}}}=$$ $$\frac{\left(2\pi i\right)\cdot\left(e^{\frac{x\pi i}{3}}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{2\pi i e^{\frac{x\pi i}{3}}}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{2\pi i\left(\cos\left(\frac{\pi x}{3}\right)+\sin\left(\frac{\pi x}{3}\right)i\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{-2\pi\sin\left(\frac{\pi x}{3}\right)+2i\pi\cos\left(\frac{\pi x}{3}\right)}{\left(1-e^{\frac{i2\pi x}{3}}\right)\cdot\left(3e^{\frac{2\pi i}{3}}\right)}=$$ $$\frac{-2\pi\sin\left(\frac{\pi x}{3}\right)+2i\pi\cos\left(\frac{\pi x}{3}\right)}{3(i+\sqrt{3})e^{\frac{i\pi x}{3}}\sin\left(\frac{\pi x}{3}\right)}$$