Evaluate $\int_{0}^{1}K(x)^2\text{d}x -\int_{0}^{1} \frac{x\sqrt{1-x^2} }{2-x^2}K(x)^2\text{d}x$

Question

Recently, I found this identity on a mathematical site(seems true):

$$\int_{0}^{1}K(x)^2\text{d}x -\int_{0}^{1} \frac{x\sqrt{1-x^2} }{2-x^2}K(x)^2\text{d}x =\frac{\Gamma\left ( \frac{1}{4} \right )^4 }{64}$$ where $K(x)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}\sqrt{1-x^2t^2} }\text{d}t$.

A problem that resemble the above identity is here. Which has the power $3$. But so far, I still don't really know how to relate those two integrals.

Answer 1

To calculate this integral we can utilize the following Ramanujan identity:

$$\int_0^\sqrt{1-x^2} \frac{K(y)}{\sqrt{1-y^2}\sqrt{1-x^2-y^2}}dy=\frac{2}{1+x}K^2\left(\sqrt{\frac{1-x}{1+x}}\right)\tag{*}$$

This is found as eq. $(42)$ in this article by Yajun Zhou (and proven a bit later).

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In order to be able to apply the above identity, we'll start similarly to @Miracle Invoker's answer by applying some substitutions alongside Landen's Transformation (LT) to the second integral.

$$\int_0^1 \frac{x\sqrt{1-x^2}K^2(x)}{2-x^2}dx\overset{1-x^2\to x^2}=\int_0^1\frac{x^2}{1+x^2}K^2(\sqrt{1-x^2})dx$$

$$\overset{LT}=4\int_0^1 \frac{x^2K^2\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1+x)^2}dx\overset{\frac{1-x}{1+x}\to x}=\int_0^1 K^2(x)\left(1-\frac{2x}{1+x^2}\right)dx $$

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Plugging this into the original integral yields:

$$\int_0^1 K^2(x)dx-\int_0^1 \frac{x\sqrt{1-x^2}K^2(x)}{2-x^2}dx\overset{x^2\to x}=\int_0^1 \frac{K^2(\sqrt x)}{1+x}dx$$

$$\overset{\large x\to \frac{1-x}{1+x}}=\int_0^1 \frac{K^2\left(\sqrt{\frac{1-x}{1+x}}\right)}{1+x}dx\overset{\large (*)}=\frac12\int_0^1 \int_0^{\sqrt{1-x^2}}\frac{K(y)}{\sqrt{1-y^2} \sqrt{1-x^2-y^2}}dydx$$

$$=\frac12\int_0^1 \frac{K(y)}{\sqrt{1-y^2}} \int_0^\sqrt{1-y^2}\frac{1}{\sqrt{1-x^2-y^2}}dxdy=\frac{\pi}{4}\int_0^1 \frac{K(y)}{\sqrt{1-y^2}}dy =\boxed{\frac{\Gamma^4\left(\frac{1}{4}\right)}{64}}$$

Where the last integral can be found evaluated here as:

$$\int_0^1 \frac{K(y)}{\sqrt{1-y^2}}dy = \int_0^1 \frac{K(y)}{\sqrt y}dy =\frac12 \left(\int_0^1 \frac{dy}{\sqrt y\sqrt{1-y^2}}\right)^2=\frac{\Gamma^4\left(\frac{1}{4}\right)}{16\pi}$$

Answer 2

Not a complete answer.

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$$I=\int_0^1K^2dk-\int_0^1\frac{kk'}{2-k^2}K^2dk$$

Well as far as I have seen Closed Form of the First Integral is not known, so our first step should be to remove the same term from the second integral such that both cancel out.

Using $k\to k'$ on the Second Integral we have, $$\int_0^1\frac{kk'}{2-k^2}K^2dk=\int_0^1K'^2dk-\int_0^1\frac{K'^2}{1+k^2}dk$$

Using Landen's Transformation on the Integral of $K'^2$ we get,

$$K\left[\frac{1-k}{1+k}\right]=\left(\frac{1+k}{2}\right)K'$$

$$\int_0^1K'^2dk=2\int_0^1K^2dk$$

Use it again on the Second Integral we get,

$$\int_0^1\frac{K'^2}{1+k^2}dk=\int_0^1K^2dk+2\int_0^1\frac{kK^2}{1+k^2}dk$$

So finally we obtain,

$$I=2\int_0^1\frac{kK^2}{1+k^2}dk$$

Using Theorem 1 Equation $(15)$ from MOMENTS OF PRODUCTS OF ELLIPTIC INTEGRALS and setting $t=i$ results in,

$$\int_0^1\frac{kKK'}{1+k^2}dk=\frac{\Gamma^4(1/4)}{128}$$

Now we just need to prove that both of these Integrals are equal.

$$0=\int_0^1\frac{k}{1+k^2}[KK'-K^2]dk$$

I am having trouble in this part.