Estimate the integral of $(1+x^2)^{-\alpha}$, where $\alpha>1/2$

Question

I'm reading a proof of a theorem, and there's one step I couldn't understand why. It said that for all $a>0$ and $\alpha>1/2$, $$ \int_{a}^{\infty}(1+x^2)^{-\alpha} \ \mathrm dx \leq2^{2\alpha-1}\left(1+\frac{1}{2\alpha-1}\right)(1+a^2)^{-\alpha+1/2}. $$ Could anybody give me a hint about this inequality? Thanks a lot.

Answer 1

We have to estimate: $$\begin{eqnarray*}\int_{0}^{+\infty}\left(\frac{1+a^2}{1+(a+x)^2}\right)^{\alpha}\,dx&=&\int_{0}^{+\infty}\left(\frac{1}{1+\frac{2a}{1+a^2}x+\frac{1}{1+a^2}x^2}\right)^{\alpha}\,dx\\&=&\sqrt{1+a^2}\int_{0}^{+\infty}\left(\frac{1}{1+z^2+\frac{2a}{\sqrt{1+a^2}}z}\right)^{\alpha}\,dz\end{eqnarray*} $$ and assuming $a> 0$ and $\alpha>\frac{1}{2}$ the last integral is upper bounded by: $$ \int_{0}^{+\infty}\frac{dz}{(1+z^2)^\alpha}=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\alpha-\frac{1}{2}\right)}{\Gamma\left(\alpha\right)}\leq\frac{\sqrt{\pi\alpha}}{2\alpha-1},$$ much better. Putting all together, I get:

$$ \int_{a}^{+\infty}\frac{dx}{(1+x^2)^\alpha}\leq (1+a^2)^{\frac{1}{2}-\alpha}\frac{\sqrt{\pi\alpha}}{2\alpha-1},$$

tighter and simpler than your inequality.