# Definite Integral from 0 to 1 of Arcsine of x over x

## Theorem

$\displaystyle \int_0^1 \frac {\arcsin x} x = \frac \pi 2 \ln 2$

## Proof

Let:

$x = \sin \theta$

By Derivative of Sine Function, we have:

$\dfrac {\d x} {\d \theta} = \cos \theta$

We have, by Arcsine of Zero is Zero:

as $x \to 0$, $\theta \to \arcsin 0 = 0$.

By Arcsine of One is Half Pi, we have:

as $x \to 1$, $\theta \to \arcsin 1 = \dfrac \pi 2$.

We have:

 $\displaystyle \int_0^1 \frac {\arcsin x} x \rd x$ $=$ $\displaystyle \int_0^{\pi/2} \frac {\cos \theta \map \arcsin {\sin \theta} } {\sin \theta} \rd \theta$ substituting $x = \sin \theta$ $\displaystyle$ $=$ $\displaystyle \int_0^{\pi/2} \theta \cot \theta \rd \theta$ Definition of Arcsine, Definition of Real Cotangent Function
$\displaystyle \int \cot \theta \rd \theta = \map \ln {\sin \theta} + C$

So:

 $\displaystyle \int_0^{\pi/2} \theta \cot \theta \rd \theta$ $=$ $\displaystyle \intlimits {\theta \map \ln {\sin \theta} } 0 {\frac \pi 2} - \int_0^{\pi/2} \map \ln {\sin \theta} \rd \theta$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac \pi 2 \map \ln {\sin \frac \pi 2} - \lim_{\theta \to 0^+} \paren {\theta \map \ln {\sin \theta} } + \frac \pi 2 \ln 2$ Definite Integral from 0 to $\dfrac \pi 2$ of $\map \ln {\sin x}$ $\displaystyle$ $=$ $\displaystyle -\lim_{\theta \to 0^+} \paren {\theta \map \ln {\sin \theta} } + \frac \pi 2 \ln 2$ Sine of Right Angle, Natural Logarithm of 1 is 0

We have:

 $\displaystyle \lim_{\theta \to 0^+} \paren {\theta \map \ln {\sin \theta} }$ $=$ $\displaystyle \lim_{\theta \to 0^+} \paren {\theta \map \ln {\frac {\sin \theta} \theta \theta} }$ $\displaystyle$ $=$ $\displaystyle \lim_{\theta \to 0^+} \paren {\theta \map \ln {\frac {\sin \theta} \theta} } + \lim_{\theta \to 0^+} \theta \ln \theta$ Sum of Logarithms, Combination Theorem for Limits of Functions: Sum Rule $\displaystyle$ $=$ $\displaystyle \paren {\lim_{\theta \to 0^+} \theta} \paren {\map \ln {\lim_{\theta \to 0^+} \frac {\sin \theta} \theta} } + \lim_{\theta \to 0^+} \theta \ln \theta$ Combination Theorem for Limits of Functions: Product Rule $\displaystyle$ $=$ $\displaystyle 0 \ln 1 + 0$ Limit of $\dfrac {\sin x} x$, Limit of $x^n \paren {\ln x}^m$ $\displaystyle$ $=$ $\displaystyle 0$

giving:

$\displaystyle \int_0^{\pi/2} \theta \cot \theta \rd \theta = \frac \pi 2 \ln 2$

hence the result.

$\blacksquare$