I'm interested in the integral
$$ I = \int_{1}^{\infty}\left(\frac{1}{t}-\text{arcsin}\left(\frac{1}{t}\right)\right)\text{d}t $$
I've shown that this integral exists and I know that $I = \frac{\pi}{2} - \left(1+\ln 2\right)$. I've tried multiple change of variable, for example: $$ I = \int_{0}^{1}\frac{u - \text{arcsin}\left(u\right)}{u^2}\text{d} u $$
I've then tried to $u = \sin t$ and I've obtained an integral mixing $t$ and $\sin t$ : $$ I = \lim\limits_{\epsilon \rightarrow 0}\left(\int_{\epsilon}^{1}\frac{\cos t}{\sin t}\text{d}t + \int_{\epsilon}^{1}\frac{t \cos t}{\sin^2 t}\text{d}t\right) $$ What could I try ?
Integration by part: $$ I = \int_{0}^{1}\left(\arcsin(u)-u\right)d\left(\frac{1}u \right)=\arcsin(1)-1-\int_{0}^{1}\frac{\frac{1}{\sqrt{1-u^2}}-1}{u}du $$ We get: $$ I =\frac{\pi}2-1+\ln(1+\sqrt{1-u^2})\bigg|_0^1=\boxed{~\frac{\pi}2-1-\ln2~} $$