For $x,y>0$, define two sequences $(x_n)$ and $(y_n)$ by $x_1=x,y_1=y$ and $x_{n+1}=(x_n+y_n)/2$ and $y_{n+1}=\sqrt{x_ny_n}$. Prove that $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n= \dfrac{\pi}{\int_0^\pi \dfrac{d\theta}{\sqrt{x^2 \cos^2\theta + y^2\sin^2\theta}}}.$
I think it might be easier to prove $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n.$ Let the LHS of this equation be denoted $L$ and let the RHS be denoted $M$. By the AM-GM inequality and induction, $x_{n}\leq y_n$ for all $n\ge 2$. It could be useful to define a new sequence with a limit that's easier to evaluate. We have $L\leq M$ by limit properties, so we just need to show $L\ge M$ to get $L=M$. Suppose for a contradiction that $L < M.$ Then by definition, there exists $N$ so that for all $n\ge N, x_n < \frac{L+M}2$ and $y_n > \dfrac{L+M}2.$ How can I proceed from here?
As for showing it equals an expression involving a given integral, I think the integral is actually fairly hard to compute explicitly, so one should use some properties of the sequences to show the desired equality.
Best Answer
I will handle the integral at the end using the transformation given by Gauss. An alternative transformation is available on my blog (linked in comments to question).
Let us write $$I(x, y) =\int_0^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}\tag{1}$$ for $x, y>0$ and we prove $$I(x, y) =I\left(\frac{x+y} {2},\sqrt {xy} \right) \tag{2}$$ Gauss used the substitution $$\sin t=\frac{2x\sin u} {x+y+(x-y) \sin^2u}\tag{3}$$ to establish $(2)$. The substitution maps the interval $[0,\pi/2]$ to $[0,\pi/2]$.
Let us first observe that $$\cos t=\frac{\sqrt{(x+y)^2+(x-y)^2\sin^4u-2(x^2+y^2)\sin^2u}} {x+y+(x-y) \sin^2u} $$ which can be rewritten as $$\frac{\sqrt{4a^2+4(a^2-b^2)\sin^4u-4(2a^2-b^2)\sin^2u}} {x+y+(x-y) \sin^2u} $$ where $2a=x+y,b^2=xy$. The above can be further simplified as $$\cos t=\frac{2\cos u\sqrt{a^2\cos^2u+b^2\sin^2u}}{x+y+(x-y)\sin^2u}\tag{4}$$ Next we have $$x^2\cos^2t+y^2\sin^2t=\frac{4x^2\cos^2u(a^2\cos^2u+b^2\sin^2u)+4x^2y^2\sin^2u}{(x+y+(x-y)\sin^2u)^2}$$ which can be rewritten as $$\frac{x^2[(x+y)^2\cos^4u+4xy\sin^2u\cos^2u+4y^2\sin^2u]}{(x+y+(x-y)\sin^2u)^2} $$ Replacing $\cos^2u$ with $1-\sin^2u$ we get $$\sqrt{x^2\cos^2t+y^2\sin^2t}=\frac{x(x+y-(x-y)\sin^2u)}{x+y+(x-y)\sin^2u}\tag{5}$$ Differentiating equation $(3)$ with respect to $u$ we get $$\cos t\cdot \frac{dt}{du} =\frac{2x\cos u(x+y-(x-y) \sin^2u)} {(x+y+(x-y)\sin^2u)^2} \tag{6}$$ Using $(4),(5),(6)$ we get $$\int_{0}^{\pi/2}\frac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}=\int_{0}^{\pi/2}\frac{du}{\sqrt{a^2\cos^2u+b^2\sin^2u}} $$ or $I(x,y) =I(a,b) $.
Thus we get $I(x, y) =I(x_n, y_n) $ and taking limits as $n\to\infty$ we get $$I(x, y) =\lim_{n\to\infty} I(x_n, y_n) =I(\lim_{n\to\infty} x_n, \lim_{n\to\infty} y_n) =I(L, L) $$ We have used the fact that $I(x, y) $ is a continuous function of $x, y$ and that the sequences $x_n, y_n$ tend to a common limit, say $L$.
Now $I(L, L) =\pi/(2L)$ it follows that $$L=\frac{\pi} {2I(x,y)}=\dfrac{\pi}{\displaystyle \int_0^{\pi}\dfrac{dt}{\sqrt{x^2\cos^2t+y^2\sin^2t}}}$$