Calculate : $\int_1^{\infty} \frac{1}{x} -\sin^{-1} \frac{1}{x}\ \mathrm{d}x $ Find : $\displaystyle \int_1^{\infty} \frac{1}{x} -\sin^{-1} \frac{1}{x}\ \mathrm{d}x $.
I've done some work but I've got stuck, you may try to help me continue or give me another way , in both cases try to give me just hints (not the full solution), Thank you.

My  work :
setting : $x^{-1}=t$, we get : $\displaystyle \int_{0}^{1} \frac{t-\sin^{-1}t}{t^2}\ \mathrm{d}t$.
for $|t|\leq 1 $, we have : $\displaystyle \sin^{-1}t =\sum_{k=0}^{\infty}\frac{(2k)!z^{2k+1}}{4^k (k!)^2(2k+1)}.$
with some simplification :
$\displaystyle \int_{0}^{1} \frac{t-\sin^{-1}t}{t^2}\ \mathrm{d}t=-\sum_{k=1}^{\infty} \frac{(2k-1)!}{4^k (k!)^2(2k+1)}$.
The first problem is that I don't have any idea how to prove the Taylor series (in the general form ) for $\sin^{-1} t$ and it seem quite complicated, I just have it in my book.
The second problem is that I don't know how to evaluate the last sum.
I hope you can have an easier solution.
 A: Hint 1:
Don't try to use the Taylor series here. Try splitting the integral up into to two parts, like so: 
$$
\int_1^\infty\frac{dx}{x} - \int_1^\infty\ \arcsin\frac{1}{x}dx.
$$
Integrating the first term should be easy. For the second, the substitution you were trying to use was correct, but try using integration by parts.
If that's not enough of a hint, 

Hint 2:

Integrating by parts: 
Let $f(u) = \arcsin(u), d(g(u)) = \frac {1} {u^2}$. $\int fdg = f(u)g(u)du-\int g(u) d(f(u))$. This gives
$\int_0^1\frac{1}{u\sqrt{1-u^2}}du-\frac {\arcsin u} {u} + \int_1^\infty \frac{1} {x}dx.$ The first integral can be simplified through substitution.


Hint 3:

Substitute $s=\sqrt{1-u^2}$, $ds = -\frac {u} {\sqrt{1-u^2}}du$. This makes the first term $\int \frac {ds}{s^2-1}$, which you can integrate by using partial fraction decomposition.


Solution:

 Your final answer should come out to be $\frac{\pi}{2} -1 - \ln2$

A: Let $\frac{1}{x}=t$.
As you showed correctly, $$I=\int_{1}^{\infty}\frac{1}{x}-\sin^{-1}(\frac{1}{x})dx=\int_{0}^{1}\frac{t-\sin^{-1}(t)}{t^{2}}dt$$
Then we integrate by parts, 
$$\int_{0}^{1}\frac{t-\sin^{-1}(t)}{t^{2}}dt=\left[-\frac{t-\sin^{-1}(t)}{t}\right]^{1}_{0}+\int_{0}^{1}\frac{1-\frac{1}{\sqrt{1-t^{2}}}}{t}dt$$
The evaluation comes to $\pi/2-1$ in the case $t=1$ and, in the limit as $t \to 0$, $0$.
For this second integral, let $t=\sin(\theta)$
$$\int_{0}^{\pi/2}\frac{1-\frac{1}{\cos{\theta}}}{\sin(\theta)}\cos(\theta)d\theta=\int_{0}^{\pi/2}\frac{\cos(\theta)-1}{\sin(\theta)}d\theta=-\ln(2)$$
Hence,
$$I=\frac{\pi}{2}-1-\ln(2)=-0.1223\ldots$$
A: $$
\begin{align}
&\int_1^\infty\left(\frac1x-\sin^{-1}\left(\frac1x\right)\right)\,\mathrm{d}x\\
&=-\int_0^{\pi/2}\left(\sin(t)-t\right)\,\mathrm{d}\csc(t)\\
&=\frac\pi2-1+\int_0^{\pi/2}\csc(t)(\cos(t)-1)\,\mathrm{d}t\\
&=\frac\pi2-1-\int_0^{\pi/2}\csc(t)(1-\cos(t))\frac{\sin^2(t)}{1-\cos^2(t)}\,\mathrm{d}t\\
&=\frac\pi2-1+\int_0^{\pi/2}\frac{\mathrm{d}\cos(t)}{1+\cos(t)}\\
&=\frac\pi2-1-\log(2)
\end{align}
$$
A: First step: the change of variable $x=1/\sin t$ yields $\mathrm dx=-\cos t\mathrm dt/(\sin t)^2$ hence the integral to be computed is
$$
I=-\int_0^{\pi/2}(t-\sin t)\cos t\frac{\mathrm dt}{(\sin t)^2}=\int_0^{\pi/2}u(t)v'(t)\mathrm dt,
$$
with $u(t)=t-\sin t$ and $v(t)=1/\sin t$. 
Second step: integrate by parts, this yields a formula involving $u({\pi/2})v({\pi/2})={\pi/2}-1$, $u(0)v(0)=0$ and
$$
\int_0^{\pi/2}u'(t)v(t)\mathrm dt=\int_0^{\pi/2}(1-\cos t)\frac{\mathrm dt}{\sin t}.
$$
Third step: use standard techniques to compute the last integral, here $\frac{1-\cos t}{\sin t}=\tan(t/2)$ hence
$$
\int_0^{\pi/2}(1-\cos t)\frac{\mathrm dt}{\sin t}=\left[-2\log\cos(t/2)\right)_0^{\pi/2}=-2\log\cos(\pi/4)=\log2.
$$
The final formula might be
$$
I=\frac\pi2-1-\log2\approx-0.12235.
$$
