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

[chisigma]

Posted the 01 14 2014 on www.mathhelpforum.com by the user Latsabb and not yet solved... ... we have just started working on the Taylor series, and one of our problems is to find the MacLaurin series for $\displaystyle \frac{\cos (x^{3})}{x}$...

It is quite surprising that none of the experts of the site have considered that the function $\displaystyle f(x)= \frac{\cos (x^{3})}{x}$ in x=0 has a singularity and that means that the McLaurin series of f(x) [i.e. the Taylor series in x=0...] doesn't exist... the Laurent series exists and it is easily found as follows...

$\displaystyle \cos x = 1 - \frac{x^{2}}{2} + ... + (-1)^{n}\ \frac{x^{2\ n}}{(2 n)!} + ... $

$\displaystyle \cos (x^{3}) = 1 - \frac{x^{6}}{2} + ... + (-1)^{n}\ \frac{x^{6\ n}}{(2 n)!} + ... $

$\displaystyle \frac{\cos (x^{3})}{x} = \frac{1}{x} - \frac{x^{5}}{2} + ... + (-1)^{n}\ \frac{x^{6\ n- 1}}{(2 n)!} + ... \ (1)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 21 2014 on www.artofproblemsolving.com by the user aziiri and noy yet solved...

For $\displaystyle n \ge 1$ we define a sequence... $\displaystyle a_{n} = \int_{0}^{\infty} \frac{d x} {(1 + x^{2})^{n}}$

1) Prove that $\displaystyle a_{n} \rightarrow 0$

2) Find $\alpha$,$\beta$ such that... $\displaystyle a_{n} - \sqrt{\frac{\pi}{4\ n}} \sim \alpha\ n^{\beta}$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 01 21 2014 on www.artofproblemsolving.com by the user aziiri and noy yet solved...

For $\displaystyle n \ge 1$ we define a sequence... $\displaystyle a_{n} = \int_{0}^{\infty} \frac{d x} {(1 + x^{2})^{n}}$

1) Prove that $\displaystyle a_{n} \rightarrow 0$

2) Find $\alpha$,$\beta$ such that... $\displaystyle a_{n} - \sqrt{\frac{\pi}{4\ n}} \sim \alpha\ n^{\beta}$

Using the Cauchy integral theorem on the upper 'big half circle' in the complex plane we obtain directly for $n \ge 2$ ...$\displaystyle a_{n}= \int_{0}^{\infty} \frac{d x}{(1 + x^{2})^{n}} = \frac{2\ \pi\ i}{(n-1)!} \lim_{z \rightarrow i} \frac{d^{n-1}}{d z^{n-1}} \frac{(z-i)^{n}}{(1 + z^{2})^{n}} = \frac{\pi}{2^{n}}\ \frac{(2 n - 3)!}{(n-1)!}\ (1)$

... but that isn't what is requested and we will proceed from (1) on in next post... Kind regards $\chi$ $\sigma$

 

[alyafey22]

Another way is looking at the beta integral

\(\displaystyle a_n = \int^\infty_0 \frac{1}{(1+x^2)^n}\, dx =\frac{1}{2} \int^\infty_0 \frac{dx}{\sqrt{x}(1+x)^{n}}=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(n-\frac{1}{2} \right)}{\Gamma(n)}\)

Then the denominator goes faster using the Sitrling approximation.

 

[chisigma]

Posted the 01 21 2014 on www.artofproblemsolving.com by the user aziiri and noy yet solved...

For $\displaystyle n \ge 1$ we define a sequence... $\displaystyle a_{n} = \int_{0}^{\infty} \frac{d x} {(1 + x^{2})^{n}}$

1) Prove that $\displaystyle a_{n} \rightarrow 0$

2) Find $\alpha$,$\beta$ such that... $\displaystyle a_{n} - \sqrt{\frac{\pi}{4\ n}} \sim \alpha\ n^{\beta}$

Using the Cauchy integral theorem on the upper 'big half circle' in the complex plane we obtain directly for $n \ge 2$ ...$\displaystyle a_{n}= \int_{0}^{\infty} \frac{d x}{(1 + x^{2})^{n}} = \frac{2\ \pi\ i}{(n-1)!} \lim_{z \rightarrow i} \frac{d^{n-1}}{d z^{n-1}} \frac{(z-i)^{n}}{(1 + z^{2})^{n}} = \frac{\pi}{2^{n}}\ \frac{(2 n - 3)!}{(n-1)!}\ (1)$

... but that isn't what is requested and we will proceed from (1) on in next post...

If we write the (1) in a 'more rational form' we obtain... $\displaystyle a_{n}= \int_{0}^{\infty} \frac{d x}{(1 + x^{2})^{n}} = \frac{\pi}{2}\ \prod_{k=1}^{n-1} (1 - \frac{1}{2\ k})\ (1)$... and, because the series $\displaystyle - \sum_{k =1}^{\infty} \frac{1}{2\ k}$ is a divergent all negative terms series, the infinite product $\displaystyle \prod_{k=1}^{\infty} (1 - \frac{1}{2\ k})$ diverges to 0.

It is easy to see that the sequence $a_{n}$ is the solution of the difference equation...

$\displaystyle a_{n+1} = a_{n} (1 - \frac{1}{2\ n}),\ a_{1} = \frac{\pi}{2}\ (2)$

... so that is $\displaystyle \Delta_{n} = a_{n+1} - a_{n} = - \frac{a_{n}}{2\ n}$ and that leads to approximate (2) with the ODE...

$\displaystyle y^{\ '} = - \frac{y}{2\ x}\ (3)$

... the solution of which is $\displaystyle y= \frac{c}{\sqrt{x}}$ so that is $\displaystyle a_{n} \sim \frac{c}{\sqrt{n}}$... Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 02 27 2014 on www.artofproblemsolving.com by the user AndrewTom and not yet solved...

Prove that...

$\displaystyle \int_{0}^{1} \frac{x^{n} - 1}{\ln x}\ d x = \ln\ (n+1)$



Kind regards

$\chi$ $\sigma$

 

[Saitama]

Posted the 02 27 2014 on www.artofproblemsolving.com by the user AndrewTom and not yet solved...

Prove that...

$\displaystyle \int_{0}^{1} \frac{x^{n} - 1}{\ln x}\ d x = \ln\ (n+1)$



Kind regards

$\chi$ $\sigma$

The exact same integral is evaluated in Zaid's thread: http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233.html

 

[Saitama]

Posted the 01 03 2014 on www.artofproblemsolving.com by the user Pirkuliyev Rovsen and not yet solved... ... calculate $f^{(n)} (0)$ of $f(x)= (\sin^{-1} x)^{2}$... Kind regards $\chi$ $\sigma$

Hi chisigma!

Did you find a solution to this? I am confused by the notation. Does $f^2(x)$ means $f(f(x))$ or the second derivative $f''(x)$?

Thanks!

 

[chisigma]

Hi chisigma!

Did you find a solution to this? I am confused by the notation. Does $f^2(x)$ means $f(f(x))$ or the second derivative $f''(x)$?

Thanks!

The notation $f^n (x)$ means $f(x)$ eleved to the n-th power, so that $f^{2} (x)$ means $f(x)*f(x)$. The notation $f^{(n)} (x)$ means $\frac{d^{n}}{d x^{n}} f(x)$ so that $f^{(2)} (x)$ means $\frac{d^{2}}{d x^{2}} f(x)$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 03 02 2014 on www.artofproblemsolving.com by the user bobthesmartypants and not yet solved...

A sequence $\{a_{n}\}$ satisfies $\displaystyle a_{n+3} = \frac{a_{n+2} + a_{n+1} + a_{n}}{3}$. Given $a_{0},a_{1},a_{2}$ find $\displaystyle \lim_{n \rightarrow \infty} a_{n}$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 03 02 2014 on www.artofproblemsolving.com by the user bobthesmartypants and not yet solved...

A sequence $\{a_{n}\}$ satisfies $\displaystyle a_{n+3} = \frac{a_{n+2} + a_{n+1} + a_{n}}{3}$. Given $a_{0},a_{1},a_{2}$ find $\displaystyle \lim_{n \rightarrow \infty} a_{n}$...

The characteristic equation related to the difference equation is...

$\displaystyle x^{3} - \frac{x^{2}}{3} - \frac{x}{3} - \frac{1}{3} = 0\ (1)$

... the solutions of which are $x_{1} = 1,\ x_{2,3} = - \frac{1}{3} \pm i\ \frac{\sqrt{2}}{3}$, so that the general solution is... $\displaystyle a_{n} = c_{1} + (-\frac{1}{3})^{n}\ \{c_{2}\ \cos (n\ \frac{\sqrt{2}}{3}) + c_{3}\ \sin (n\ \frac{\sqrt{2}}{3})\}\ (2)$

... and...

$\displaystyle \lim_{n \rightarrow \infty} a_{n} = c_{1}\ (3)$

The constant $c_{1}$ can be derived from the initial conditions solving the following linear system...

$\displaystyle c_{1} + c_{2} = a_{0}$

$\displaystyle c_{1} - \frac{1}{3}\ \cos \frac{\sqrt{2}}{3}\ c_{2} - \frac{1}{3}\ \sin \frac{\sqrt{2}}{3}\ c_{3} = a_{1}$

$\displaystyle c_{1} + \frac{1}{9}\ \cos \frac{2\ \sqrt{2}}{3}\ c_{2} + \frac{1}{9}\ \sin \frac{2\ \sqrt{2}}{3}\ c_{3} = a_{2}\ (4)$

Kind regards $\chi$ $\sigma$

 

[chisigma]

Posted the 03 19 2014 on www.artofproblemsolving by the user james digol and not jet solved...

Solve the differential equation...

$\displaystyle y^{\ ''} + \frac{1}{x}\ y^{\ '} = c\ (1)$

... where c is a constant...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 03 19 2014 on www.artofproblemsolving by the user james digol and not jet solved...

Solve the differential equation...

$\displaystyle y^{\ ''} + \frac{1}{x}\ y^{\ '} = c\ (1)$

... where c is a constant...

With the substitution y ' = z the DE becomes...

$\displaystyle z^{\ '} + \frac{z}{x} = c\ (1)$

... and with standard procedure we arrive to...

$\displaystyle z + e^{- \int \frac{d x}{x}}\ (\int c\ e^{\int \frac{d x}{x}}\ d x + c_{1} ) = c\ \frac{x}{2} + \frac{c_{1}}{x}\ (2)$

... and with another integration we arrive at...

$\displaystyle y = c\ \frac{x^{2}}{2} + c_{1}\ \ln |x| + c_{2}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Posted the 05 17 2014 on www.artofproblemsolving.com by the user uniquesaylor and not yet solved...

Evaluate PV of the following improper integral...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{\cos (\alpha\ t)}{1 + t^{3}}\ dt\ (1)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

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...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

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]

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]

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]

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]

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]

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]

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]

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 $\displaystyle x \to \infty$.

 

[mathbalarka]

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

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 (Lipssealed)

 

[Euge]

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]

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 $o(x^\epsilon)$.

No.

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

 

[chisigma]

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]

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.

 

[I like Serena]

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]

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]

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]

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

$\chi$ $\sigma$

 

[chisigma]

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-mathematics-set-theory-logic-15/difference-equation-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

$\chi$ $\sigma$

 

[chisigma]

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]

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...

http://d2b4wtkw5si7f7.cloudfront.net/96/93/i97162134._szw380h285_.jpg

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

$\chi$ $\sigma$