Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

4 Answers 4

up vote 3 down vote accepted

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

share|improve this answer

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 $\frac{\pi}{2} -1 - \ln2$

share|improve this answer
    
thanks, this was more than enough. –  aziiri Feb 10 '13 at 22:25
    
Sorry, how do you make hidden text in a textbox? –  Berci Feb 10 '13 at 22:51
    
Add >! to the beginning of the line. It makes the text that follows a spoiler. You can find all the ways to mark up your answers at math.stackexchange.com/editing-help –  ahruss Feb 11 '13 at 4:16

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, $$I=\frac{\pi}{2}-1-\ln(2)=-0.1223\ldots$$

share|improve this answer

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 $$ I=\frac\pi2-1-\log2\approx-0.12235. $$

share|improve this answer
    
I don't think that the first equality should be : $$ I=-\int_0^{\frac{\pi}{2}}(t-\sin t)\cos t\frac{\mathrm dt}{(\sin t)^2}$$ isn't it ? –  aziiri Feb 10 '13 at 22:13
    
Mathematica gives $(\pi - 2 - 2\log{2})/2 \approx -0.122$ –  Ron Gordon Feb 10 '13 at 22:15
    
@aziiri Thanks for the remark. Answer modified. –  Did Feb 10 '13 at 22:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.