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.

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


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.

| cite | improve this answer | |
  • 4
    $\begingroup$ 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 $\pi/2$. $\endgroup$ – Gerald Edgar May 25 '14 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.