Limit of $\int_0^1\left(\frac {2}{\sqrt {(1-t^2)(1-xt^2)}}-\frac{x}{1-xt}\right)\,dt$ as $x\to 1^{-}$

While going through this question I was reminded of one of my earlier questions and I found that there is some unfinished business which needs some further exploration.

Let $$F(x) =\int_0^1\left(\frac{2}{\sqrt{(1-t^2)(1-xt^2)}} -\frac{x}{1-xt}\right)\, dt=\int_0^1 f(x, t) \, dt\tag{1}$$ for $x\in[0,1]$.

Let's observe that $$\lim_{x\to 1^-}f(x,t)=\frac{1}{1+t}\tag{2}$$ and hence it is natural to expect that $\lim_{x\to 1^-}F(x)$ should equal $\int_0^1 dt/(1+t)=\log 2$ but numerical evidence as well as some amount of elliptic function theory tells (see one of my questions linked earlier for details) us that this particular limit is $4\log 2$.

This suggests that there is some weird behavior of integrand as $x\to 1^{-}$ (in particular the convergence is not uniform).

I would like to have this limit evaluated using some analysis related to convergence of integrand as $x\to 1^-$. Any help in this direction would be appreciated.

Note: I have asked a new question instead of bumping an old one. The old question is more about solution verification and is related to elliptic integrals. I wanted to have a different perspective which involves general issues of uniform convergence to handle the limit of this integral.

The non-uniform convergence prevents the direct use of the given integrand.

However, the original problem is to compute $L=\lim\limits_{x\to1^-}\big(2K(\sqrt x)+\log(1-x)\big)$.

One may replace $x/(1-xt)$ with "something better" that approximates the original "elliptic" integrand uniformly (on $t\in(0,1)$ as $x\to1^-$) and is still elementarily integrable.

Let's replace $x$ by $x^2$ (to get rid of $\sqrt x$ everywhere), and consider $$F(x,t)=\frac1{\sqrt{(1-t^2)(1-x^2 t^2)}},\\G(x,t)=\frac1{a_x(1+t)\sqrt{(1-t)(1-xt)}},$$ where $a_x=\sqrt{(1+x)/2}$ is chosen to have $\lim\limits_{x\to1^-}\big(F(x,t)-G(x,t)\big)=0$ uniformly.

Now $\int_0^1 F(x,t)\,dt=K(x)$ and $\int_0^1 G(x,t)\,dt=a_x^{-2}\tanh^{-1}a_x$. Hence \begin{align*} L&=\lim_{x\to1^-}\big(2K(x)+\log(1-x^2)\big) \\&=\lim_{x\to1^-}\big(\frac{1}{a_x^2}\log\frac{1+a_x}{1-a_x}+\log(1-x^2)\big) \\&=\lim_{x\to1^-}\left[\log\left(\frac{1+a_x}{1-a_x}(1-x^2)\right)+\frac{1-x}{1+x}\log\frac{1+a_x}{1-a_x}\right] \\&=\lim_{x\to1^-}\log\left(\frac{(1+a_x)^2}{1-a_x^2}(1-x^2)\right) \\&=\lim_{x\to1^-}\log\big(2(1+x)(1+a_x)^2\big)=4\log 2. \end{align*}

Remark: @metamorphy gave a nice solution. I constructed something slightly different from @metamorphy's $G(x, t)$ due to the term $\sigma_x$. I hope my answer is helpful as a supplement.

Let (replace $x$ with $x^4$ for convenience) \begin{align*} &G(x)\\ =\ & \int_0^1 \left(\frac{2}{\sqrt{(1 - t^2)(1 - x^4 t^2)}} - \frac{x^4}{1 - x^4t}\right) \mathrm{d} t\\ =\ & \int_0^1 \left(\frac{2}{\sqrt{(1 + t)(1 + x^2t)}\sqrt{(1 - t)(1 - x^2 t)}} - \frac{x^4}{1 - x^4t}\right) \mathrm{d} t\\ =\ & \int_0^1 \left(\frac{2}{\sqrt{(1 + t)(1 + t)}\sqrt{(1 - t)(1 - x^2 t)}} - \frac{x^4}{1 - x^4t}\right) \mathrm{d} t\\ &\quad + \int_0^1 g(x, t)\mathrm{d} t \tag{1} \end{align*} where $$g(x, t) = \frac{2}{\sqrt{1 + t}\sqrt{(1 - t)(1 - x^2 t)}} \left(\frac{1}{\sqrt{1 + x^2t}} - \frac{1}{\sqrt{1 + t}}\right).$$

First, we deal with the second integral in (1), i.e., $\int_0^1 g(x, t)\mathrm{d} t$. Clearly, $g(x, t) \ge 0$ for all $x$ in $(0, 1)$ and all $t$ in $[0, 1)$. Also, we have, for all $x$ in $(0, 1)$ and all $t$ in $[0, 1)$, \begin{align*} &g(x, t) \\ \le\ & \frac{2}{\sqrt{(1 - t)(1 - x^2)}} \left(\frac{1}{\sqrt{1 + x^2t}} - \frac{1}{\sqrt{1 + t}}\right)\\ =\ & \frac{2}{\sqrt{(1 - t)(1 - x^2)}}\frac{(1 - x^2)t}{(1 + x^2t)(1 + t)}\Big(\frac{1}{\sqrt{1 + x^2t}} + \frac{1}{\sqrt{1 + t}}\Big)^{-1}\\ \le \ & \frac{2}{\sqrt{(1 - t)(1 - x^2)}}(1-x^2)\\ =\ & \frac{2\sqrt{1 - x^2}}{\sqrt{1 - t}}. \end{align*} Since $\int_0^1 \frac{2\sqrt{1 - x^2}}{\sqrt{1 - t}} \mathrm{d} t = 4\sqrt{1 - x^2}$, we have $\lim_{x\to 1^{-}} \int_0^1 g(x, t)\mathrm{d} t = 0$.

Second, we deal with the first integral in (1), denoted by $H(x)$. With the substitution $\frac{1 - x^2 t}{1 - t} = u^2$ (correspondingly $t = \frac{u^2 - 1}{u^2 - x^2}$), we have (after doing the partial fraction decomposition for the integrand) \begin{align*} &H(x)\\ =\ & \int_1^\infty \left(\frac{\frac{2}{\sqrt{2 + 2x^2}}}{u - \sqrt{\frac{1 + x^2}{2}}} - \frac{\frac{2}{\sqrt{2 + 2x^2}}}{u + \sqrt{\frac{1 + x^2}{2}}} + \frac{2(1 + x^2)u}{(1 + x^2)u^2 - x^2} - \frac{2u}{u^2 - x^2}\right) \mathrm{d} u\\ =\ &\left(\frac{2}{\sqrt{2 + 2x^2}} \ln \frac{u - \sqrt{\frac{1 + x^2}{2}}}{u + \sqrt{\frac{1 + x^2}{2}}} + \ln\frac{(1 + x^2)u^2 - x^2}{u^2 - x^2}\right)\Big\vert_1^\infty\\ =\ & \ln(1 + x^2) - \frac{2}{\sqrt{2 + 2x^2}} \ln \frac{1 - \sqrt{\frac{1 + x^2}{2}}}{1 + \sqrt{\frac{1 + x^2}{2}}} - \ln\frac{1}{1 - x^2}\\ =\ & \ln(1 + x^2) + \frac{2}{\sqrt{2 + 2x^2}}\ln 2 + \frac{4}{\sqrt{2 + 2x^2}} \ln \left(1 + \sqrt{\frac{1 + x^2}{2}}\right)\\ &\quad + \left(1 - \frac{2}{\sqrt{2 + 2x^2}}\right)\ln(1 - x^2). \end{align*} It is easy to obtain $\lim_{x\to 1^{-}} H(x) = 4\ln 2$.

Thus, we have $\lim_{x\to 1^{-}} G(x) = 4\ln 2$.

We are done.