TITLE: Calculate : $\int_1^{\infty} \frac{1}{x} -\sin^{-1} \frac{1}{x}\ \mathrm{d}x $
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: 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,
$$I=\frac{\pi}{2}-1-\ln(2)=-0.1223\ldots$$

REPLY [3 votes]: $$
\begin{align}
&\int_1^\infty\left(\frac1x-\sin^{-1}\left(\frac1x\right)\right)\,\mathrm{d}x\\
&=-\int_0^{\pi/2}\left(\sin(t)-t\right)\,\mathrm{d}\csc(t)\\
&=\frac\pi2-1+\int_0^{\pi/2}\csc(t)(\cos(t)-1)\,\mathrm{d}t\\
&=\frac\pi2-1-\int_0^{\pi/2}\csc(t)(1-\cos(t))\frac{\sin^2(t)}{1-\cos^2(t)}\,\mathrm{d}t\\
&=\frac\pi2-1+\int_0^{\pi/2}\frac{\mathrm{d}\cos(t)}{1+\cos(t)}\\
&=\frac\pi2-1-\log(2)
\end{align}
$$