While trying to evaluate $\int_0^1 k^2K(k)dk$ related to elliptic integral of the first kind, by integral switching method, I reached the trigonometric integral
$$\int_0^{\frac\pi 2}\frac{\theta-\cos\theta\sin\theta}{2\sin^3\theta}d\theta$$
which is evaluated to $\frac{2C+1}4$ by Wolfram Alpha. Here, $C$ is the Catalan constant, sometimes denoted by $G$. This integral is complicated for me.
Are there another methods to evaluate the integrals $\int_0^1k^nK(k)dk$, $n\geq2$? Or can someone please evaluate the trigonometric integral above? Thank you.
Best Answer
With $K$ and $E$ as the Complete Elliptic Integral of the First and Second Kind respectively with parameter $k$.
Denote
$$K_n=\int_0^1k^nK\ dk$$
$$E_n=\int_0^1k^nE\ dk$$
then one can prove that,
$$n^2K_n-(n-1)^2K_{n-2}=1\tag{1}$$
with the initial values of
$$K_0=2G,\quad K_1=1$$
One may have the following closed forms by building upon the recurrence,
$$\int_{0}^{1}k^{2n+1}K\ dk=\frac{2^{4n}}{\left(2n+1\right)^{2}}\binom{2n}{n}^{-2}\sum_{k=0}^{n}\binom{2k}{k}^2\frac{1}{2^{4k}}$$
$$\int_{0}^{1}k^{2n}K\ dk=\frac{1}{2^{4n}}\binom{2n}{n}^2\left[2G+\frac{1}{4}\sum_{k=1}^n\frac{2^{4k}}{k^2\displaystyle\binom{2k}{k}^2}\right]$$
The proof of $(1)$ is as follows
Use IBP, Integrate $kK$ and differentiate $k^{n-1}$
Using the following result $$\int kK\ dk=E-(1-k^2)K$$
$$K_n=1-(n-1)\int_0^1k^{n-2}[E-(1-k^2)K]\ dk\tag{2}$$
Now we need relation between $E_n$ and $K_n$
One can use the Derivative of $E$ and obtain
$$[k^nE]'=k^{n-1}[E-K]+nk^{n-1}E$$
Then integrate to obtain $$1+K_n=(n+2)E_n\tag{3}$$
Using $(2)$ and $(3)$ the result in $(1)$ follows.