The arithmetic-geometric mean$^{[1]}$$\!^{[2]}$ of positive numbers $a$ and $b$ is denoted $\operatorname{agm}(a,b)$ and defined as follows: $$\text{Let}\quad a_0=a,\quad b_0=b,\quad a_{n+1}=\frac{a_n+b_n}2,\quad b_{n+1}=\sqrt{a_n b_n}.$$ $$\text{Then}\quad\operatorname{agm}(a,\,b)=\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n.\tag1$$ It can be expressed$^{[3]}$ in terms of the complete elliptic integral of the $1^{\text{st}}$ kind$^{[4]}$$\!^{[5]}$: $$\operatorname{agm}(a,\,b)=\frac\pi4\cdot\frac{a+b}{{\bf K}\!\left(\frac{a+b}{a-b}\right)}.\tag2$$

It appears that $$\int_0^1\frac{x^z}{\operatorname{agm}(1,\,x)}dx\stackrel{\color{gray}?}=\frac{\Gamma\!\left(\frac z2+\frac12\right)}{2\,\Gamma\!\left(\frac z2+1\right)},\quad\forall z\in\mathbb C,\,\Re(z)>-1.\tag3$$ How can we prove it?


$$\int_{0}^{+\infty}\frac{dt}{\sqrt{(1+t^2)(1+x^2 t^2)}}=\frac{\pi}{2\,\text{agm}(1,x)}\tag{1}$$ hence $$ I(z)=\int_{0}^{1}\frac{x^z}{\text{agm}(1,x)}\,dx = \frac{2}{\pi}\int_{0}^{1}\int_{0}^{+\infty}\frac{x^z}{\sqrt{1+x^2\sinh^2(t)}}\,dt\,dx\tag{2}$$ that can be managed through integration by parts, getting the same recurrence relation of the Euler Beta function involved in the claim. That, plus the logarithmic convexity of $I(z)$ (that follows from the integral form of the Cauchy-Schwarz inequality applied to the original definition of $I(z)$) is enough to prove the given conjecture.

For instance, by exchanging the order of integration we have: $$ I(0) = \frac{2}{\pi}\int_{0}^{+\infty}\frac{t}{\sinh(t)}\,dt = \frac{4}{\pi}\eta(2) = \frac{\pi}{2}\tag{3}$$ as well as: $$ I(1) = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\cosh(t)-1}{\sinh(t)^2}\,dt = \frac{2}{\pi}.\tag{4}$$

I also think that your conjecture directly follows from the Legendre differential equation for $K(k)$ and $K(k')$, coupled with $(3)$ and $(4)$.

Nice catch, Vladimir!


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