Let $E=E(k), K=K(k)$ be the complete elliptic integrals of the second and the first kind respectively. It is well-known that $\frac12\int_0^1K\,dk=G$ is the Catalan constant. We can also find, by switching integral trick that
\begin{align}
\int_0^1E\,dk
&=\int_0^{\frac\pi 2}\int_0^1\sqrt{1-k^2\sin^2\theta}d\theta\\
&=\int_0^{\frac\pi 2}\left(\frac12\cos\theta+\frac12\frac{\theta}{\sin\theta}\right)d\theta\\
&=\frac12+G
\end{align}
since $\int_0^{\pi/2}\frac{\theta}{\sin\theta}d\theta=2G$.
My question for EG: Is there an easier/another way for calculating $\int_0^1E\,dk$ without switching integrals trick?
Best Answer
We have $$\int_{0}^{1}(2E-K)\text{d}k =\int_0^1\left ( \frac{\mathrm{d} }{\mathrm{d} k} kE\right )\text{d}k=\left ( kE \right )\big|^1_0=1.$$ In the meantime, $\int_{0}^{1}K\text{d}k=2G$, which gives $$ \int_{0}^{1}E\text{d}k=G+\frac12. $$