# How do I prove that $\alpha=\beta$ in this case?

Define $$f_\alpha(x):=\frac{\Gamma(\frac{\alpha+1}{2})}{\sqrt{\alpha \pi} \Gamma(\frac{\alpha}{2})}(1+\frac{x^2}{\alpha})^{-\frac{\alpha+1}{2}}$$ for $$x\in \mathbb{R}$$ and $$\alpha >0$$

Let $$m$$ be the Lebesgue measure. If $$f_\alpha=f_\beta$$ $$m$$-a.e., then how do I prove that $$\alpha=\beta$$?

• Suppose that $\alpha \neq \beta$ and prove that $f_\alpha$ and $f_\beta$ differ on a set of Lebesgue measure > 0 – windircurse Dec 7 at 13:41
• @windircurse Isn't that just a restatement of my question? How do I find such set? – Rubertos Dec 7 at 13:42
• prove that the function $\alpha \to f_\alpha$ is decreasing – windircurse Dec 7 at 13:48
• Could you be more specific? And the map $\alpha\to f_\alpha$ is not decreasing, as you can see here : en.m.wikipedia.org/wiki/Student%27s_t-distribution – Rubertos Dec 7 at 13:53