Let $K(x)$ be the complete elliptic integral of the first kind
(the argument is the parameter $m = k^2$).


Let $$ A = \int\_0^1 \arcsin(K(x)) dx$$


With precision $1000$ decimal digits $\Re A = \frac{\pi}{2}$.



> 
> Is this true?
> 
> 
> 


According to Wolfram Alpha for the indefinite integral 
there is no result in terms of standard mathematical functions.


$A=1.570796326794\ldots - 1.285983901951989\ldots i$.



Conjectured closed form for definite integral
In the interval [0,1], K(x) takes real values greater than $\pi/2$. Hence $\arcsin(K(x))$ (with the appropriate choice of branch) is equal to $\pi/2$ plus something imaginary.


