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chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 08 16 2014 on www.mathhelpforum.com by the user superzhangmch and not yet solved...

... prove...

$\displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n} = 0\ (1)$


...
or show it is not true...
From the expression of the function sin x as infinite product ...

$\displaystyle \sin x = x\ \prod_{k=1}^{\infty} (1 - \frac{x^{2}}{k^{2}\ \pi^{2}})\ (1)$

we derive...


$\displaystyle a_{n} = \frac{\ln |\sin n|}{n} = \frac{\ln n}{n} +\sum_{k=1}^{\infty} \frac{\ln |1 - \frac{n^{2}}{k^{2}\ \pi^{2}}|}{n}\ (2)$

The proposed question is not trivial since to show that $\displaystyle \lim_{n \rightarrow \infty} a_{n} = 0$ it is necessary to show that each term of the series (2) tends to zero as n tends to infinity. This can be critical when it is $\displaystyle \frac{n}{k} \sim \pi$, that is when $\displaystyle \frac{n}{k}$ is a 'good approximation' of $\pi$ since the logarithm can take negative values even higher. The workload needed for this investigation is not light but fortunately with a short research it has found a German text of the late nineteenth century, where are example values ​​$\displaystyle \frac{n}{k}$ 'good approximations' of $\pi$...

Archimedes,Huygens, Lambert, Legendre.

Vier Abhandlungen über die Kreismessung. Deutsh hrsg. und mit einerÜbersicht über die
Geschichte des Problemes von der Quadratur des Zirkels


Published 1892 by B.G. Teubner in Leipzig .

Written in German.



pages 146-147 has a table of rational approximations of pi...



1:3
7:22
106:333
113 : 355
33102: 103993
33215 : 104348
66317: 208341
99532 : 312689
265381: 833719
364913 : 1146408
1360120: 4272943
1725033 : 5419351
25510582: 80143857
52746197 : 165707065
78256779: 245850922
131002976 :411557987
340262713 :1068966896
811528438 : 2549491779
1963319607 : 6167950454
4738167652: 14885392687
6701487259 : 21053343141
567663097408 : 1783366216531
1142027682075 : 3587785776203
1709690779483 : 5371151992734
2851718461558 : 8958937768937
107223273857129 : 336851849443403
324521540032945 : 1019514486099146

The next job in the next posted ...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
From the expression of the function sin x as infinite product ...

$\displaystyle \sin x = x\ \prod_{k=1}^{\infty} (1 - \frac{x^{2}}{k^{2}\ \pi^{2}})\ (1)$

we derive...


$\displaystyle a_{n} = \frac{\ln |\sin n|}{n} = \frac{\ln n}{n} +\sum_{k=1}^{\infty} \frac{\ln |1 - \frac{n^{2}}{k^{2}\ \pi^{2}}|}{n}\ (2)$

The proposed question is not trivial since to show that $\displaystyle \lim_{n \rightarrow \infty} a_{n} = 0$ it is necessary to show that each term of the series (2) tends to zero as n tends to infinity. This can be critical when it is $\displaystyle \frac{n}{k} \sim \pi$, that is when $\displaystyle \frac{n}{k}$ is a 'good approximation' of $\pi$ since the logarithm can take negative values even higher. The workload needed for this investigation is not light but fortunately with a short research it has found a German text of the late nineteenth century, where are example values ​​$\displaystyle \frac{n}{k}$ 'good approximations' of $\pi$...

Archimedes,Huygens, Lambert, Legendre.

Vier Abhandlungen über die Kreismessung. Deutsh hrsg. und mit einerÜbersicht über die
Geschichte des Problemes von der Quadratur des Zirkels


Published 1892 by B.G. Teubner in Leipzig .

Written in German.



pages 146-147 has a table of rational approximations of pi...



1:3
7:22
106:333
113 : 355
33102: 103993
33215 : 104348
66317: 208341
99532 : 312689
265381: 833719
364913 : 1146408
1360120: 4272943
1725033 : 5419351
25510582: 80143857
52746197 : 165707065
78256779: 245850922
131002976 :411557987
340262713 :1068966896
811528438 : 2549491779
1963319607 : 6167950454
4738167652: 14885392687
6701487259 : 21053343141
567663097408 : 1783366216531
1142027682075 : 3587785776203
1709690779483 : 5371151992734
2851718461558 : 8958937768937
107223273857129 : 336851849443403
324521540032945 : 1019514486099146

The next job in the next posted ...
Using the calculator of windows and the table of 'best approximations' of $\pi$ found in the German text of the late nineteenth century, we have calculated the values ​​of ...

$\displaystyle p(\frac{k}{n}) = \frac{\ln |1 - \frac{n^{2}}{k^{2}\ \pi^{2}}|}{n}\ (1)$

The results are as follows ...

$\frac{k}{n}= \frac{1}{3} \implies p(\frac{k}{n}) = -2.429...$

$\frac{k}{n} = \frac{7}{22} \implies p(\frac{k}{n}) = -.3238...$

$\frac{k}{n}= \frac{106}{333} \implies p(\frac{k}{n}) = - .2956...$


$\frac{k}{n}= \frac{113}{355} \implies p(\frac{k}{n}) = -.0439...$

$\frac{k}{n}= \frac{33102}{103993} \implies p(\frac{k}{n}) = - 2.0889...\ 10^{-4}$

$\frac{k}{n}= \frac{33215}{104348} \implies p(\frac{k}{n}) = -2.135...\ 10^{-4}$

$\frac{k}{n}= \frac{66317}{208341} \implies p(\frac{k}{n}) = -1.117...\ 10^{-4}$

$\frac{k}{n}= \frac{99532}{312689} \implies p(\frac{k}{n}) = -7.902...\ 10^{-5}$

$\frac{k}{n}= \frac{265381}{833719} \implies p(\frac{k}{n}) = -3.108...\ 10^{-5}$

$\frac{k}{n}= \frac{364913}{1146408} \implies p(\frac{k}{n}) = -2.408...\ 10^{-5}$

$\frac{k}{n}= \frac{1360120}{4272943} \implies p(\frac{k}{n}) = -6.784...\ 10^{-6}$

$\frac{k}{n}= \frac{1725033}{5419351} \implies p(\frac{k}{n}) = -5.8849...\ 10^{-6}$

$\frac{k}{n}= \frac{25510582}{80143857} \implies p(\frac{k}{n}) = -4.4341...\ 10^{-7}$

$\frac{k}{n}= \frac{52746197}{165707065} \implies p(\frac{k}{n}) = -2.22...\ 10^{-7}$

$\frac{k}{n}= \frac{78256779}{245850922} \implies p(\frac{k}{n}) = -1.5269...\ 10^{-7}$

$\frac{k}{n}= \frac{31002976}{411557987} \implies p(\frac{k}{n}) = -9.4603...\ 10^{– 8}$

$\frac{k}{n}= \frac{340262713}{1068966896} \implies p(\frac{k}{n}) = -1.502...\ 10^{-8}$

$\frac{k}{n}= \frac{811528438}{2549491779} \implies p(\frac{k}{n}) = -1.6667...\ 10^{-8}$

$\frac{k}{n}= \frac{1963319607}{6167950454} \implies p(\frac{k}{n}) = -7.21...\ 10^{-9}$

$\frac{k}{n}= \frac{4738167652}{14885392687} \implies p(\frac{k}{n}) = -3.04758...\ 10^{-9}$

$\frac{k}{n}= \frac{6701487259}{21053343141} \implies p(\frac{k}{n}) = -2.38...\ 10^{-9}$

$\frac{k}{n}= \frac{567663097408}{1783366216531} \implies p(\frac{k}{n}) = -3.112...\ 10^{-11}$

$\frac{k}{n}= \frac{1142027682075}{3587785776203} \implies p(\frac{k}{n}) = -1.585...\ 10^{-11}$

$\frac{k}{n}= \frac{1709690779483}{5371151992734} \implies p(\frac{k}{n}) = -1.067...\ 10^{-11}$

$\frac{k}{n}= \frac{2851718461558}{8958937768937} \implies p(\frac{k}{n}) = -6.76...\ 10^{-12}$


$\frac{k}{n}= \frac{107223273857129}{336851849443403} \implies p(\frac{k}{n}) = -1.81...\ 10^{-13}$


At this point I have to stop because The subsequent step provides a value $\frac{n}{k} = \frac{1019514486099146}{324521540032945}$ that coincides to the twenty-four decimal approximation of $\pi$ of the calculator I used. It is clear that even with a more powerful calculation tool is difficult, following this way, to give a definite answer to the proposed question and a different way must be tried ... always that what has been already done by someone in the past ...

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 8 18 2014 on www.artofproblemsolving.com by the user Kid_Dynamite and not jet solved...

How to prove that $\displaystyle \int_{0}^{x} \frac{dt} {\ln t} = \mathcal{O}\ \{\frac{x}{\ln x}\}$ ?...

Kind regards

$\chi$ $\sigma$
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,905
Posted the 8 18 2014 on www.artofproblemsolving.com by the user Kid_Dynamite and not jet solved...

How to prove that $\displaystyle \int_{0}^{x} \frac{dt} {\ln t} = \mathcal{O}\ \{\frac{x}{\ln x}\}$ ?...

Kind regards

$\chi$ $\sigma$
Does Kid_Dynamite want this to be proved for $x \to 0$ or $x \to \infty$?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Does Kid_Dynamite want this to be proved for $x \to 0$ or $x \to \infty$?
I stronghly suppose $\displaystyle x \rightarrow \infty$...

Kind regards

$\chi$ $\sigma$
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,905
I stronghly suppose $\displaystyle x \rightarrow \infty$...

Kind regards

$\chi$ $\sigma$
By L'hospital's rule,

\(\displaystyle \lim_{x\to \infty} \dfrac{\int_0^x \frac{dt}{\ln t}}{\frac{x}{\ln x}} = \lim_{x\to \infty} \dfrac{\frac{1}{\ln x}}{\frac{\ln x - 1}{\ln^2 x}} = \lim_{x \to \infty} \frac{\ln x}{\ln x - 1} = 1\).

So for all sufficiently large $x$,

\(\displaystyle |\int_0^x \frac{dt}{\ln t}| < \frac{3}{2} |\frac{x}{\ln x}|\)

Consequently,

\(\displaystyle \int_0^x \frac{dt}{\ln t} = \mathcal{O}\left(\frac{x}{\ln x}\right)\) as \(\displaystyle x\to \infty\).
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Euge, can you elaborate on how you're getting the inequality at the second step from the first.

Also, I might as well mention that

$$\int_0^x \frac{dt}{\log(t)}$$

Doesn't quite make sense, as the usual integral blows up at $t = 1$. One usually makes an indentation around the contour of the integral around the pole, to have

$$\int_0^{1-\epsilon} \frac{dt}{\log(t)} + \int_{1+\epsilon}^x \frac{dt}{\log(t)}$$
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,905
Euge, can you elaborate on how you're getting the inequality at the second step from the first.

Also, I might as well mention that

$$\int_0^x \frac{dt}{\log(t)}$$

Doesn't quite make sense, as the usual integral blows up at $t = 1$. One usually makes an indentation around the contour of the integral around the pole, to have

$$\int_0^{1-\epsilon} \frac{dt}{\log(t)} + \int_{1+\epsilon}^x \frac{dt}{\log(t)}$$
This logarithmic integral is a special function, usually denoted $\text{li}(x)$. It is defined for all positive $x$ different from 1, and when $x > 1$, the integral is to interpreted as the Cauchy principal value. In any case, I argued thinking of $\text{Li}(x)$ (where 2 is the lower limit) instead of $\text{li}(x)$, so I must approach this differently. I'll come back later.

Edit: For $x \ge 2$,

$\displaystyle \mathrm{li}(x) = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \varepsilon}^x \frac{dt}{2}\right) = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \epsilon}^2 \frac{dt}{\ln t} + \int_2^x \frac{dt}{\ln t}\right)$
$\displaystyle = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \epsilon}^2 \frac{dt}{\ln t}\right) + \mathrm{Li}(x) =\mathrm{li}(2) + \mathrm{Li}(x).$

In the argument I gave earlier, the lower limit of $0$ is to be replaced with $2$. Then the argment is valid and $\mathrm{Li}(x) = \mathcal{O}(x/\ln x)$ as $x \to \infty$. Since

$\displaystyle \lim_{x\to \infty} \dfrac{\mathrm{li}(2)}{\frac{x}{\ln x}} = \lim_{x\to \infty} \mathrm{li}(2) \frac{\ln x}{x} = 0$,

it follows that $\mathrm{li}(2) = \mathcal{O}(x/\ln x)$ and thus

$\displaystyle \mathrm{li}(x) = \mathcal{O}\left(\frac{x}{\ln x}\right)$ as $\displaystyle x \to \infty$.
 
Last edited:

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
$\displaystyle \lim_{x\to \infty} \dfrac{\mathrm{li}(2)}{\frac{x}{\ln x}} = \lim_{x\to \infty} \mathrm{li}(2) \frac{\ln x}{x} = 0$
Typo 1 : You mean $1$ instead of $0$. We evidently don't want our estimates to be little-o! :eek:

it follows that $\mathrm{li}(2) = \mathcal{O}(x/\ln x)$ and thus ...
Typo-2 : You meant $\mathrm{li}(2) = O(1)$.

OK, now I'll just zip the lip (Lipssealed)
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,905
Typo 1 : You mean $1$ instead of $0$. We evidently don't want our estimates to be little-o! :eek:
No, it's 0 by L'hospital's rule. Also, a function that is $o(g)$ is also $O(g)$ (although the converse does not hold).



Typo-2 : You meant $\mathrm{li}(2) = O(1)$.
No. Since $\mathrm{li}(2)/(x/\ln x) \to 0$ as $x\to \infty$, there exists a positive integer $x_0$ such that $|\mathrm{li}(2)| < |x/\ln x|$ for all $x \ge x_0$. Thus $\mathrm{li}(2) = \mathcal{O}(x/\ln x)$.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
No, it's 0 by L'hospital's rule.
Then I am sure you have made an error. Natural asymptotic expansions (in my case PNT :p) shows that $\mathrm{li}(x) \sim x/\log(x)$, i.e., $\mathrm{li}(x) = x/\log(x) + o(x/\log(x))$ so you're almost definitely wrong.

EDIT : Ah, I misread. I thought you were trying to prove $\li(x) = o(x/\log(x))$. Yes, you are indeed right, and one can do it much easier that L'Hopital : note that $\log(x)$ grows like $o(x^\epsilon)$.

Well, potato pohtato. $\mathrm{li}(2)$ is a constant hence $O(1)$ and anything $O(1)$ is automatically $O(x/\log(x))$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
By L'hospital's rule,

\(\displaystyle \lim_{x\to \infty} \dfrac{\int_0^x \frac{dt}{\ln t}}{\frac{x}{\ln x}} = \lim_{x\to \infty} \dfrac{\frac{1}{\ln x}}{\frac{\ln x - 1}{\ln^2 x}} = \lim_{x \to \infty} \frac{\ln x}{\ln x - 1} = 1\).

So for all sufficiently large $x$,

\(\displaystyle |\int_0^x \frac{dt}{\ln t}| < \frac{3}{2} |\frac{x}{\ln x}|\)

Consequently,

\(\displaystyle \int_0^x \frac{dt}{\ln t} = \mathcal{O}\left(\frac{x}{\ln x}\right)\) as \(\displaystyle x\to \infty\).
I must confess that when I proposed this problem taken from another site, I assumed that the solution You were to go through the prime number theorem ... Euge instead found a brilliant application of the l'Hopital rule that greatly simplifies the job... .. excellent! (Yes)...

Kind regards

$\chi$ $\sigma$
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Prime number theorem is just overkill. The proof I had in mind was by Steepest descent, but evidently it was unnecessary too.

Well done, Euge.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,146
Beautiful!

However, the original problem didn't specify a $\text{li}$ function, nor did it specify a Cauchy principal value for the integral.
So as I see it, $\int_0^x \frac{dt}{\ln t}$ is undefined for $x\to\infty$.

Perhaps Kid_Dynamite wanted it for $x \to 0$ after all...
 

chisigma

Well-known member
Feb 13, 2012
1,704
Beautiful!

However, the original problem didn't specify a $\text{li}$ function, nor did it specify a Cauchy principal value for the integral.
So as I see it, $\int_0^x \frac{dt}{\ln t}$ is undefined for $x\to\infty$.

Perhaps Kid_Dynamite wanted it for $x \to 0$ after all...
Because for $\displaystyle x \ge \mu = 1.4513692348...$ is $\displaystyle \text{li}\ (x) = \int_{\mu}^{x} \frac{dt}{\ln t}$ and $\displaystyle \text{Li}\ (x) = \int_{2}^{x} \frac{dt} {\ln t}$ is also $\displaystyle \text{li}\ (x) - \text{Li}\ (x) = \text{li}\ (2) = 1.04516378...$, i.e. the difference between the two functions is a constant and it has no effect on the behavior for $\displaystyle x \to \infty$. For more details see...

Logarithmic Integral -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,905
Beautiful!

However, the original problem didn't specify a $\text{li}$ function, nor did it specify a Cauchy principal value for the integral.
So as I see it, $\int_0^x \frac{dt}{\ln t}$ is undefined for $x\to\infty$.

Perhaps Kid_Dynamite wanted it for $x \to 0$ after all...
I used the notation $\mathrm{li}(x)$ since there was some confusion as to the meaning of $\int_0^x \frac{dt}{\ln t}$, but like I've said before, this integral (for $x > 1$) is to be understood in the principal value sense:

$\displaystyle \mathrm{li}(x) := \int_0^x \frac{dt}{\ln t} = \lim_{\varepsilon \to 0^+} \left(\int_0^{1 - \varepsilon} \frac{dt}{\ln t} + \int_{1 + \varepsilon}^x \frac{dt}{\ln t}\right)$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 03 27 2014 on www.artofproblemsolving.com by the user TheCaffeinheartmachine and not yet solved...

For $\alpha> 2$ find $\displaystyle \int_{0}^{\infty} \frac{x-1}{x^{\alpha} - 1}\ dx$...

Kind regards

$\chi$ $\sigma$
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 03 27 2014 on www.artofproblemsolving.com by the user TheCaffeinheartmachine and not yet solved...

For $\alpha> 2$ find $\displaystyle \int_{0}^{\infty} \frac{x-1}{x^{\alpha} - 1}\ dx$...
The way to solve this integral is the formula obtained in...

http://mathhelpboards.com/discrete-...ation-tutorial-draft-part-i-426.html#post2494

$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{(n + a) (n+b)} = \frac{\phi(b) - \phi(a)}{b - a}\ (1)$

... where...

$\displaystyle \phi(x) = \frac{d}{d x} \ln x!\ (2)$

First step is to separate the integral in two parts...

$\displaystyle \int_{0}^{\infty} \frac{1 - x}{1 - x^{\alpha}}\ d x = \int_{0}^{1} \frac{1 - x}{1 - x^{\alpha}}\ dx + \int_{0}^{1} x^{\alpha - 3}\ \frac{1 - x}{1 - x^{\alpha}}\ dx\ (3)$

For the first integral, using the (1), we find ...

$\displaystyle \frac{1 - x}{1 - x^{\alpha}} = 1 - x + x^{\alpha} - x^{\alpha + 1} + ... + x^{n\ \alpha} - x^{n\ \alpha+1} + ... \implies \int_{0}^{1} \frac{1 - x}{1 - x^{\alpha}} = \sum_{n = 0}^{\infty} \frac{1}{(n\ \alpha + 1)(n\ \alpha + 2)} = \frac{1}{2} + \frac{\phi(\frac{2}{\alpha}) - \phi(\frac{1}{\alpha})}{\alpha} \ (4) $

... and for the second...

$\displaystyle x^{\alpha - 3}\ \frac{1 - x}{1 - x^{\alpha}} = x^{\alpha - 3} - x^{\alpha - 2} + x^{2\ \alpha - 3} - x^{2\ \alpha - 2} + ... + x^{(n + 1)\ \alpha - 3} - x^{(n+1)\ \alpha - 2} + ... \implies $

$\displaystyle \implies \int_{0}^{1} x^{\alpha - 3}\ \frac{1 - x}{1 - x^{\alpha}}\ d x = \sum_{n=1}^{\infty} \frac{1}{(n\ \alpha - 2)\ (n\ \alpha - 1)} = \frac{\phi(- \frac{1}{\alpha}) - \phi(- \frac{2}{\alpha})}{\alpha}\ (5)$

... so that is...

$\displaystyle \int_{0}^{\infty} \frac{1 - x}{1 - x^{\alpha}}\ dx = \frac{1}{2} + \frac{\phi(\frac{2}{\alpha}) - \phi(\frac{1}{\alpha})}{\alpha} + \frac{\phi(- \frac{1}{\alpha}) - \phi(- \frac{2}{\alpha})}{\alpha}\ (6)$

Kind regards

$\chi$ $\sigma$
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 10 24 2014 on www.artofproblemsolving.com by the user TheChainheartMachine and not yet solved...

Evaluate $\displaystyle \int_{0}^{\infty} \frac{3\ x\ \sin x - \cos x + 1}{x^{2}}\ dx$

Kind regards

$\chi$ $\sigma$


 

chisigma

Well-known member
Feb 13, 2012
1,704
Posted the 10 24 2014 on www.artofproblemsolving.com by the user TheChainheartMachine and not yet solved...

Evaluate $\displaystyle \int_{0}^{\infty} \frac{3\ x\ \sin x - \cos x + 1}{x^{2}}\ dx$

It is well known that...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x}\ dx = \frac{\pi}{2}\ (1)$

... so that the problem is to compute...

$\displaystyle \int_{0}^{\infty} \frac{1 - \cos x}{x^{2}}\ dx\ (2)$

The integral (2) can be found integrating the function $\displaystyle f(z) = \frac{1 - e^{i\ z}}{z^{2}}$ along the path C shown in the figure...



Is...

$\displaystyle \int_{- R}^{- r} \frac{1 - e^{i\ x}}{x^{2}}\ dx + \int_{\gamma} \frac{1 - e^{i\ z}}{z^{2}}\ dz + \int_{r}^{R} \frac {1 - e^{i\ x}}{x^{2}}\ dx + \int_{\Gamma} \frac{1 - e^{i\ z}}{z^{2}}\ dz = 0\ (3)$

... where we indicated with $\gamma$ the 'small half circle' and with $\Gamma$ the 'big half circle'. If R tends to infinity the fourth integral tends to 0 and for the second integral is...

$\displaystyle \lim_{r \rightarrow 0} \int_{\gamma} \frac{1 - e^{i\ z}}{z^{2}}\ dz = \lim_{r \rightarrow 0} - i\ \int_{0}^{\pi} \frac{1 - e^{i\ r\ e^{i\ \theta}}}{r} e^{- i\ \theta}\ d \theta = - \int_{0}^{\pi} d \theta = - \pi\ (4)$

Combining (3) and (4), if R tends to infinity and r tends to 0, we found that...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{1 - \cos x}{x^{2}}\ d x = \pi\ (5)$

... so that, taking into account (1), we arrive to write...

$\displaystyle \int_{0}^{\infty} \frac{3\ x\ \sin x - \cos x + 1}{x^{2}}\ d x = \frac{3\ \pi}{2} + \frac{\pi}{2} = 2\ \pi\ (6)$

Kind regards

$\chi$ $\sigma$