Integral involving Elliptic Integral

$\displaystyle \begin{aligned} \int_0^1 \left(E(k)-\frac{\pi}{2}\right)\frac{dk}{k} &=\int_0^1 \left(\int_0^{\pi/2} \sqrt{1-k^2\sin^2\theta}\,d\theta-\frac{\pi}{2}\right)\,\frac{dk}{k} \\ &= \int_0^1 \int_0^{\pi/2} \frac{\sqrt{1-k^2\sin^2\theta}-1}{k}\,d\theta\,dk\\ &=\int_0^{\pi/2} \int_0^1 \frac{\sqrt{1-k^2\sin^2\theta}-1}{k}\,dk\,d\theta\\ \end{aligned}$

It is easy to show that:

$\displaystyle \begin{aligned} \int_0^1 \frac{\sqrt{1-k^2\sin^2\theta}-1}{k}\,dk &=\left(\sqrt{1-k^2\sin^2\theta}-\ln\left(\sqrt{1-k^2\sin^2\theta}+1\right)\right|_0^1 \\ &= \ln 2-1+\cos\theta-\ln\left(1+\cos\theta\right) \\ \end{aligned}$

Hence,

$\displaystyle \begin{aligned} \int_0^{\pi/2} \int_0^1 \frac{\sqrt{1-k^2\sin^2\theta}-1}{k}\,dk\,d\theta &=\int_0^{\pi/2} \left(\ln 2-1+\cos\theta-\ln\left(1+\cos\theta\right)\right)\,d\theta \\ &=\frac{\pi}{2}\left(\ln 2-1\right)+1-\int_0^{\pi/2}\ln(1+\cos\theta)\,d\theta \\ &=\frac{\pi\ln2}{2}-\frac{\pi}{2}+1-\int_0^{\pi/2} \ln\left(2\cos^2\frac{\theta}{2}\right) \\ &=1-\frac{\pi}{2}+\frac{\pi \ln2}{2}-\int_0^{\pi/2} \ln2 \,d\theta-2\int_0^{\pi/2} \ln\left(\cos\frac{\theta}{2}\right) \,d\theta \\ &=1-\frac{\pi}{2}-4\int_0^{\pi/4} \ln(\cos\theta)\,d\theta \\ \end{aligned}$

So $a=1,b=-2$ and $c=-4$ i.e $|a|+|b|+|c|=7$ .

We need to find the value of:

$\displaystyle\int_{0}^{1}\bigg( \displaystyle\int_{0}^{\frac{\pi}{2}} \sqrt{1-k^{2}\sin^{2} \theta}\text{d}\theta-\dfrac{\pi}{2}\bigg )\frac{\text{d}k}{k}$

First let us simplify the expression enclosed in the parentheses (let us call it $\mathfrak{I}$ ):

$\mathfrak{I}=\displaystyle\int_{0}^{\frac{\pi}{2}} \sqrt{1-k^{2}\sin^{2} \theta}\text{d}\theta-\dfrac{\pi}{2}$

$= \displaystyle\int_{0}^{\frac{\pi}{2}} \sqrt{1-k^{2}\cos^{2} \theta}\text{d}\theta-\dfrac{\pi}{2}$ (As $\displaystyle\int_{a}^{b} f(x)\text{d}x=\displaystyle\int_{a}^{b} f(a+b-x)\text{d}x$ )

Now using integration by parts:

$\mathfrak{I}=\left.\theta\sqrt{1-k^{2}\cos^{2} \theta}\right|_{0}^{\frac{\pi}{2}}-\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{k^{2}\theta\sin \theta\cos \theta}{\sqrt{1-k^{2}\cos^{2} \theta}}\text{d}\theta-\dfrac{\pi}{2}$

$=-\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{k^{2}\theta\sin \theta\cos \theta}{\sqrt{1-k^{2}\cos^{2} \theta}}\text{d}\theta$

Therefore the required answer is:

$\displaystyle\int_{0}^{1}\!\!\!\!\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{-k^{2}\theta\sin \theta\cos \theta}{\sqrt{1-k^{2}\cos^{2} \theta}}\text{d}\theta\frac{\text{d}k}{k}$

$=\displaystyle\int_{0}^{1}\!\!\!\!\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{-k\theta\sin \theta\cos \theta}{\sqrt{1-k^{2}\cos^{2} \theta}}\text{d}k\text{ }\text{d}\theta$ (As we can integrate w.r.t. either $k$ or $\theta$ first.)

Now the integral w.r.t. $k$ is easy. Just substitute $1-k^{2}\cos^{2} \theta=t$ and integrate to obtain:

$\displaystyle\int_{0}^{\frac{\pi}{2}} \theta\tan \theta(\sin \theta-1)\text{d}\theta$

This too is easily managed by integration by parts. Thus the value of the integral is:

$1-\dfrac{\pi}{2}-4\displaystyle\int_{0}^{\frac{\pi}{4}} \ln (\cos x)\text{d}x$

Thus the required answer is $|1|+|-2|+|-4|=\boxed{7}$