$\int\limits_0^1{\frac{{E\left( {\sqrt{\frac{t}{{1+t}}}}\right)-K\left({\sqrt{\frac{t}{{1+t}}}}\right)}}{{t\sqrt {1+t} }}}\ln \left( {1 - t}\right)dt$

Question

I found this integral from a Facebook page : Find a closed form for this integral $$\int\limits_0^1{\frac{{E\left( {\sqrt{\frac{t}{{1+t}}}}\right)-K\left({\sqrt{\frac{t}{{1+t}}}}\right)}}{{t\sqrt {1+t} }}}\ln \left( {1 - t}\right)dt$$ $${\text{Where}}:K\left( k \right),{\text{ }}E\left( k \right){\text{ are in order the complete elliptic integral of the first and second kind}}{\text{.}}$$ $${\text{And}}:K\left( k \right) = \int\limits_0^1 {\frac{1}{{\sqrt {\left( {1 - {x^2}} \right)\left( {1 - {k^2}{x^2}} \right)} }}dx} ,E\left( k \right) = \int\limits_0^1 {\frac{{\sqrt {1 - {k^2}{x^2}} }}{{\sqrt {1 - {x^2}} }}dx} $$

I really don't know how to express a relation between $E(k)$ and $K(k)$ to reduce the problem. I think this is an intriguing integral, but I can't evaluate it yet. May you guys please help me with this? Thank you very much.

Answer 1

Using Mathematica notation for the elliptic integrals, we want to compute $$I=\int_0^1 \frac{E\left(\frac{t}{t+1}\right)-K\left(\frac{t}{t+1}\right)} {t \sqrt{t+1}}\,\, \log(1-t) \, dt$$ which did resist to the few $CAS$ I tried.

May be a series solution would work. But, we need to decide if we expand the whole integrand or just its denominator (what I would prefer).

We have $$E\left(\frac{t}{t+1}\right)-K\left(\frac{t}{t+1}\right)=\frac \pi 4 \sum_{n=1}^\infty (-1)^n \, \frac { a_n}{b_n}\, t^n$$ where the $b_n$ form sequence $A278145$ in $OEIS$ and the first $a_n$ are $$\{1,5,31,417,5853,42159,309035,9175801,275032529,2076150633,\cdots\}$$ and $$I_n=\int_0^1 \frac {t^n}{t \sqrt{t+1}}\log(1-t) \, dt=- \sqrt \pi \,\frac{\Gamma (n)}{\Gamma \left(n+\frac{1}{2}\right)}\left(H_{n-\frac{1}{2}}+2\log (2)\right) $$

Now, "just" compute !

Answer 2

The following properties are key to this problem,

$$K(ik)=\frac{1}{\sqrt{1+k^2}}K\left[\frac{k}{\sqrt{1+k^2}}\right]$$

$$E(ik)=\sqrt{1+k^2}E\left(\frac{k}{\sqrt{1+k^2}}\right)$$

Use the substitution $t\to t^2$,

$$I=2\int_0^1\frac{\ln(1-t^2)}{t}\left[\frac{E(t/\sqrt{1+t^2})-K(t/\sqrt{1+t^2})}{\sqrt{1+t^2}}\right]dt$$

Or, $$I=2\int_0^1\frac{\ln(1-t^2)}{t}\left[\frac{E(it)}{1+t^2}-K(it)\right]dt$$

Recall the Derivatives,

$$k\frac{d}{dk}K(ik)=\frac{E(ik)}{1+k^2}-K(ik)$$

Therefore we have after the application of the following integral,

$$\int_0^1t^{2n-1}\ln(1-t^2)dt=-\frac{H_n}{2n}$$

$$I=-\pi\sum_{n=1}^{\infty}(-1)^n\binom{2n}{n}^2\frac{H_n}{2^{4n}}$$

One can make use of their favorite methods to evaluate Harmonic Series from here on.

Or we can use the following,

$$\frac{\pi}{2}\sum_{n=0}^{\infty}x^{2n}\binom{2n}{n}^2\frac{H_n}{2^{4n}}=-2\int_0^{\pi/2}\frac{\ln\left[2\sin (t)\sqrt{1-x^2\sin^2t}\right]}{\sqrt{1-x^2\sin^2t}}dt$$

So,

$$I=4\ln(2)\int_0^{\pi/2}\frac{1}{\sqrt{1+\sin^2t}}dt+4\int_0^{\pi/2}\frac{\ln(\sin t)}{\sqrt{1+\sin^2t}}+2\int_0^{\pi/2}\frac{\ln(1+\sin^2t)}{\sqrt{1+\sin^2t}}dt$$

Third Integral,

$$\int_0^1\frac{\ln(1+t^2)}{\sqrt{1-t^4}}dt=\frac{\Gamma^2(1/4)\ln 2}{8\sqrt{2\pi}}$$

Second Integral

First Integral is just $K(i)=\Gamma^2(1/4)/(4\sqrt{2\pi})$

Combining all gives,

$$I=\left[\frac{5\ln2-\pi}{4\sqrt{2\pi}}\right]\Gamma^2(1/4)$$

Which is the result mentioned by @User Setness Ramesory.

Addendum: I am curious about the closed form of the following sum too but no matter how I attacked it no answer came. (I think I did see this on the site before but can't find it)

$$\sum_{n=1}^{\infty}(-1)^n\binom{2n}{n}^2\frac{H_n}{n2^{4n}}$$