Inequality with Gamma function: how to prove it?

Question

Let $\alpha \in (0,1)$ and $\Gamma(\alpha) = \int_0^{\infty}s^{\alpha - 1}e^{-s}ds$. I would like to prove that $$\int_0^{\infty}\frac{s^{-\alpha}}{1 + s}ds \le \Gamma(1 - \alpha)\Gamma(\alpha).$$

Basically I know the following two facts, but I don't know if they are needed: $$1) \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} = \int_0^1s^{\alpha - 1}(1 - s)^{\beta - 1}ds$$ $$2)\ \Gamma(1 - \alpha)\Gamma(\alpha) = \frac{\pi}{\sin(\pi\alpha)}$$

Thanks for the help!

Answer 1

$$\begin{align} \int_{0}^{\infty} \frac{x^{-\alpha}}{1+x} \ dx &= \int_{0}^{\infty} \int_{0}^{\infty} x^{-\alpha} e^{-(1+x)t} \ dt \ dx \\ &= \int_{0}^{\infty} e^{-t} \int_{0}^{\infty} x^{-\alpha} e^{-tx} \ dx \ dt \\ &= \int_{0}^{\infty} e^{-t} \int_{0}^{\infty} \left( \frac{u}{t} \right)^{-\alpha} e^{-u} \ \frac{du}{t} \ dt \\ &=\int_{0}^{\infty} t^{\alpha-1} e^{-t} \ dt \int_{0}^{\infty} u^{(1-\alpha)-1}e^{-u} \ du \\ &= \Gamma(\alpha) \Gamma(1-\alpha) \end{align}$$

Answer 2

Your relation is actually an equality for the values of $\alpha$ you describe. (By the way, your value for $\Gamma(\alpha) \Gamma(1-\alpha)$ is the reciprocal of what you state.) In fact, you can show by the residue theorem that

$$\left ( 1-e^{-i 2 \pi \alpha}\right) \int_0^{\infty} ds \frac{s^{-\alpha}}{1+s} = i 2 \pi e^{-i \pi \alpha}$$

or,

$$\int_0^{\infty} ds \frac{s^{-\alpha}}{1+s} = \frac{\pi}{\sin{\pi \alpha}}$$