integral related to arc sine

We want to evaluate the integral

${\displaystyle {\int }_1^{\mathrm{\infty }}}\left(\arcsin {\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x}}\right)𝑑x.$(1)

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

$$I(t):={\int }_1^{\mathrm{\infty }}\left(\arcsin \frac{t}{x}-\frac{t}{x}\right)𝑑x,$$

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

$I^{\prime }(t)$$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_1^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{\sqrt{1-\frac{t^2}{x^2}}}}\cdot {\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x}}\right)𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_1^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{\sqrt{\frac{x^2}{t^2}-1}}}\cdot {\displaystyle \frac{1}{t}}-{\displaystyle \frac{1}{x}}\right)𝑑x$

$\mathrm{\hspace{0.33em}}=\underset{x=1}{\overset{\mathrm{\infty }}{/}}\left[\ln \left({\displaystyle \frac{x}{t}}+\sqrt{{\displaystyle \frac{x^2}{t^2}}-1}\right)-\ln x\right]$

$\mathrm{\hspace{0.33em}}=\underset{x=1}{\overset{\mathrm{\infty }}{/}}\ln {\displaystyle \frac{1+\sqrt{1-\frac{t^2}{x^2}}}{t}}$

$\mathrm{\hspace{0.33em}}=\ln {\displaystyle \frac{2}{t}}-\ln {\displaystyle \frac{1+\sqrt{1-t^2}}{t}}=\ln 2-\ln (1+\sqrt{1-t^2})$

The gotten expression implies, since  $I(0)={\int }_1^{\mathrm{\infty }}(\arcsin 0-0)𝑑x=0$,  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

$I(1)$$\mathrm{\hspace{0.33em}}=\ln 2-{\displaystyle {\int }_0^1}\ln (1+\sqrt{1-t^2})𝑑t=\ln 2-\underset{0}{\overset{1}{/}}t\ln (1+\sqrt{1-t^2})-{\displaystyle {\int }_0^1}{\displaystyle \frac{t^{2}dt}{(1+\sqrt{1-t^2})\sqrt{1-t^2}}}$

$\mathrm{\hspace{0.33em}}=\ln 2-{\displaystyle {\int }_0^1}{\displaystyle \frac{t^{2}dt}{1-t^2+\sqrt{1-t^2}}}.$

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

$${\int }_1^{\mathrm{\infty }}\left(\arcsin \frac{1}{x}-\frac{1}{x}\right)𝑑x=\mathrm{\hspace{0.33em}1}+\ln 2-\frac{\pi }{2}.$$

For the convergence, see the French version of http://en.wikipedia.org/wiki/Improper_integralthis article.