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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.

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

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1 Answer 1

up vote 10 down vote accepted

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.

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In other words, the real part of your integrand is the constant $\pi/2$, so of course the value of the integral has real part $\pi/2$. – Gerald Edgar May 25 '14 at 16:59

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