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

[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,

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

So for all sufficiently large $x$,

$$ |\int_0^x \frac{dt}{\ln t}| < \frac{3}{2} |\frac{x}{\ln x}|$$

Consequently,

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

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

$\chi$ $\sigma$