integral related to arc sine

We want to evaluate the integral

∫1∞(arcsin⁑1x-1x)⁒𝑑x. (1)

Therefore we put an extra variable t to the integrand and thus get the function


and in to obtain a simpler integral, we differentiate it under the integral sign (, then integrate:

I′⁒(t)  =∫1∞(11-t2x2β‹…1x-1x)⁒𝑑x

The gotten expression implies, since  I⁒(0)=∫1∞(arcsin⁑0-0)⁒𝑑x=0,  that


and consequently

I⁒(1)  =ln⁑2-∫01ln⁑(1+1-t2)⁒𝑑t=ln⁑2-/01⁑t⁒ln⁑(1+1-t2)-∫01t2⁒d⁒t(1+1-t2)⁒1-t2

Here, the substitution (  t=sin⁑u  helps, yielding


Accordingly, we have the result

∫1∞(arcsin⁑1x-1x)⁒𝑑x= 1+ln⁑2-Ο€2.

For the convergence, see the French version of article.

Title integral related to arc sine
Canonical name IntegralRelatedToArcSine
Date of creation 2013-03-22 18:44:58
Last modified on 2013-03-22 18:44:58
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 8
Author pahio (2872)
Entry type Example
Classification msc 26A09
Related topic SubstitutionNotation
Related topic ArcSine
Related topic Arcosh
Related topic MethodsOfEvaluatingImproperIntegrals
Related topic CyclometricFunctions