An $\infty-\infty$ Integral

Put $x=\frac{1}{t} \Rightarrow dx=-\frac{1}{t^2} dt$

we have

$\large{\displaystyle \int^{1}_{0} \left( \frac{\arcsin (t)}{t^2}-\frac{1}{t} \right) dt}$

Using integration by parts taking $\arcsin(t)$ as first function we get

$\large{\left(-\frac{1}{t} \arcsin(t) \right)^{1}_{0}+\displaystyle \int^{1}_{0} \left( \frac{1}{t \sqrt{1-t^2} }dt \right) - \ln t|^{1}_{0} }$

Using $\large{\displaystyle \lim_{x \to 0} \frac{\arcsin(x)}{x}=1}$ and substituting $t=\sin \theta$ in the integral

$\large{-\frac{\pi}{2}+1+\displaystyle \int^{\frac{\pi}{2}}_{0} \csc \theta d \theta}$ $\large{+ \displaystyle \lim_{a \to 0} \ln (a)}$

$\large{\Rightarrow -\frac{\pi}{2}+1+\displaystyle \lim_{a \to 0} \left( \ln(\tan \left(\frac{\arcsin t}{2}\right)\right)^{1}_{a}+ \displaystyle \lim_{a \to 0} \ln (a)}$

$\large{\Rightarrow -\frac{\pi}{2}+1 - \ln \left( \tan \left(\frac{\arcsin a}{2}\right)\right)+\ln (a)}$

As $a \to 0~thus~ \arcsin(a)=a$

$\Large{\Rightarrow -\frac{\pi}{2}+1 - \displaystyle \lim_{a \to 0} \ln \left( \frac{\tan \left( \frac{a}{2} \right)}{2. \frac{a}{2}} \right)}$

$\displaystyle \lim_{x \to 0} \frac{\tan x}{x}=1$

$\large{=-\frac{\pi}{2}+1+\ln(2)}$

First, split the integral into two parts for it to become

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

Then, make the substitution $x=\csc \theta$ and $dx=-\cot(\theta)\csc(\theta)d\theta$ for the first integral to get

$\displaystyle -\int_{a}^{b}\theta\cot\theta\csc\theta d\theta-\ln(x)|_{1}^{\infty}$

NOTE: At the end, we will plug $x$ back into the integral because it is impossible to obtain an upper limit in the new integral.

Apply integration by parts to get $\displaystyle \theta\csc\theta|_{a}^{b}-\int_{a}^{b}\csc\theta d\theta-\ln(x)|_{1}^{\infty}$

Integrate to and substitute $x$ back to get $\displaystyle \bigg(x\arcsin\bigg( \frac{1}{x}\bigg)+\ln\bigg(x+\cot \bigg( \arcsin\bigg( \frac{1}{x}\bigg)\bigg)\bigg) -\ln(x)\bigg)\bigg|_{1}^{\infty}$

Combine the two natural logaritms and use $\cos(\arcsin(x))=\sqrt{1-x^{2}}$ to obtain

$\displaystyle \lim_{x\to\infty}\bigg(x\arcsin\frac{1}{x}\bigg)-\frac{\pi}{2}+\ln\bigg(1+\frac{x\sqrt{1-\frac{1}{x}^2}}{x}\bigg)\bigg|_{1}^{\infty}$

Using the identity $\lim_{x \to 0}\frac{\arcsin(x)}{x}=1$ and solving the logarithm, we have

$\displaystyle 1-\frac{\pi}{2}+\ln(1+1)-\ln(1)=\ln(2)-\frac{\pi}{2}+1$

Finally, $2+2+1=\boxed{5}$