# Unsolved analysis and number theory from other sites...

#### chisigma

##### Well-known member
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
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
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
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
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
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
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
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
$\displaystyle \lim_{x\to \infty} \dfrac{\mathrm{li}(2)}{\frac{x}{\ln x}} = \lim_{x\to \infty} \mathrm{li}(2) \frac{\ln x}{x} = 0$
Typo 1 : You mean $1$ instead of $0$. We evidently don't want our estimates to be little-o! 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 #### Euge

##### MHB Global Moderator
Staff member
Typo 1 : You mean $1$ instead of $0$. We evidently don't want our estimates to be little-o! 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
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 ) 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
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! ...

Kind regards

$\chi$ $\sigma$

#### mathbalarka

##### Well-known member
MHB Math Helper
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
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
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
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
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
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
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
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

$\chi$ $\sigma$