# integral related to arc sine

We want to evaluate the integral

 $\displaystyle\int_{1}^{\infty}\!\left(\arcsin\frac{1}{x}-\frac{1}{x}\right)dx.$ (1)

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

 $I(t)\;:=\;\int_{1}^{\infty}\!\left(\arcsin\frac{t}{x}-\frac{t}{x}\right)dx,$

and in to obtain a simpler integral, we differentiate it under the integral sign (http://planetmath.org/DifferentiationUnderIntegralSign), then integrate:

 $\displaystyle I^{\prime}(t)$ $\displaystyle\;=\;\int_{1}^{\infty}\!\left(\frac{1}{\sqrt{1\!-\!\frac{t^{2}}{x% ^{2}}}}\cdot\frac{1}{x}-\frac{1}{x}\right)dx$ $\displaystyle\;=\;\int_{1}^{\infty}\left(\frac{1}{\sqrt{\frac{x^{2}}{t^{2}}\!-% \!1}}\cdot\frac{1}{t}-\frac{1}{x}\right)dx$ $\displaystyle\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!x=1}^{\,\quad\infty}% \!\left[\ln\left(\frac{x}{t}+\sqrt{\frac{x^{2}}{t^{2}}\!-\!1}\right)-\ln{x}\right]$ $\displaystyle\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!x=1}^{\,\quad\infty}% \!\ln\frac{1+\sqrt{1\!-\!\frac{t^{2}}{x^{2}}}}{t}$ $\displaystyle\;=\;\ln\frac{2}{t}-\ln\frac{1+\sqrt{1\!-\!t^{2}}}{t}\;=\;\ln{2}-% \ln(1+\sqrt{1\!-\!t^{2}})$

The gotten expression implies, since  $I(0)=\int_{1}^{\infty}(\arcsin{0}-0)dx=0$,  that

 $I(t)\;=\;\int_{0}^{t}[\ln{2}-\ln(1+\sqrt{1\!-\!t^{2}})]\,dt\;=\;t\ln{2}-\int_{% 0}^{t}\ln(1+\sqrt{1\!-\!t^{2}})\,dt,$

and consequently

 $\displaystyle I(1)$ $\displaystyle\;=\;\ln{2}-\!\int_{0}^{1}\ln(1+\sqrt{1\!-\!t^{2}})\,dt\;=\;\ln{2% }-\!\operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\quad 1}\!t\ln(1+\sqrt{1\!-% \!t^{2}})-\!\int_{0}^{1}\frac{t^{2}\,dt}{(1+\sqrt{1\!-\!t^{2}})\sqrt{1\!-\!t^{% 2}}}$ $\displaystyle\;=\;\ln{2}-\!\int_{0}^{1}\frac{t^{2}\,dt}{1\!-\!t^{2}+\sqrt{1\!-% \!t^{2}}}.$

Here, the substitution (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)  $t=\sin{u}$  helps, yielding

 $I(1)\;=\;\ln{2}-\int_{0}^{\frac{\pi}{2}}\!(1-\cos{u})\,du\;=\;\ln{2}-\frac{\pi% }{2}+1.$

Accordingly, we have the result

 $\int_{1}^{\infty}\!\left(\arcsin\frac{1}{x}-\frac{1}{x}\right)dx\;=\;1+\ln{2}-% \frac{\pi}{2}.$