I was studying the property of Beta function and I encountered the following equality:
$$ \int_{0}^1 \lambda^{\alpha}(1-\lambda)^{\alpha-1}\frac{1}{1+e^{(2\lambda-1)k}}\text{d}\lambda = \text{B}(\alpha+1,\alpha+1), $$
where $\text{B}$ stands for the Beta function.
I can show that for every $\alpha>0$, there exists an unique $k \in (0,\infty)$ s.t. the above equality holds. What interets me is that when I plot the graph of $k$ in terms of $\alpha$ in Wolfram, it turns out that $k$ is actually a strictly-decreasing function w.r.t. $\alpha$.
I couldn't prove the above claim, but I do have some intuitions. Integrating by parts yields that the above equality is equivalent to:
$$ \int_{0}^1 \lambda^{\alpha}(1-\lambda)^{\alpha}[\frac{2k}{2+e^{(2\lambda-1)k}+e^{(1-2\lambda)k}}-1]\text{d}\lambda = 0. $$
So when $\alpha$ is large, the term $\lambda^{\alpha}(1-\lambda)^{\alpha}$ becomes dominated at $\lambda=1/2$. Therefore, $2k/4$ must stay close to $1$ as well. When $\alpha$ is small, $k$ must be significantly larger than $2$ to compensate the part where $\lambda$ stay away from $1/2$.
Any hints/suggestions are mostly appreciated.
Let $R\left(a,k\right)=\int_{0}^{1}\lambda^{a}\left(1-\lambda\right)^{a-1}\frac{1}{1+e^{\left(2\lambda-1\right)k}}-\text{Beta}\left(a+1,a+1\right)$. By the Implicit Function Theorem applied to $R\left(a,k\right)=0$ we have
$\frac{dk}{da}=-\frac{\frac{\partial R}{\partial a}}{\frac{\partial R}{\partial k}}<0$
because $\frac{\partial R}{\partial a}<0$ and $\frac{\partial R}{\partial k}<0$. Let me know if this is clear.