mathematica.gr

Δημοσιεύτηκε: **Τετ Αύγ 24, 2016 10:52 pm**

Δείξατε ότι:
$\displaystyle{\int_0^1 \frac{\log(1+x)}{(1+x)\sqrt{x}}\, {\rm d}x = \pi \log 2 - \mathcal{G}}$ όπου $\mathcal{G}$ η σταθερά Catalan.

Δημοσιεύτηκε: **Σάβ Αύγ 27, 2016 2:02 am**

Η αλλαγή μεταβλητής $\displaystyle x \mapsto \frac{1-x}{1+x}$ ,

$\begin{aligned} \int_0^1 \frac{\log (1+x)}{(1+x)\sqrt{x}}\,dx &\underbrace{=}_{x \mapsto \frac{1-x}{1+x}} 2\int_0^1 \frac{\log \left(1+\frac{1-x}{1+x}\right)}{\left(1+\frac{1-x}{1+x}\right)\sqrt{\frac{1-x}{1+x}}}\,\frac{dx}{(1+x)^2}\\&= \int_0^1 \frac{\log 2 - \log (1+x)}{\sqrt{1-x^2}}\,dx\\&= \pi\log 2 - 2\int_0^{\pi/2} \log (2\cos (\theta/2))\,d\theta \\&= \pi\log 2 - 2\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\int_0^{\pi/2}\cos (n\theta)\,d\theta\\&= \pi\log 2 - 2\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}\sin \left(\frac{n\pi}{2}\right)\\&= \pi \log 2 - 2G\end{aligned}$

Δημοσιεύτηκε: **Σάβ Αύγ 27, 2016 10:45 am**

Tolaso J Kos έγραψε:Δείξατε ότι: $\displaystyle{\int_0^1 \frac{\log(1+x)}{(1+x)\sqrt{x}}\, {\rm d}x = \pi \log 2 - \mathcal{G}}$

Περίπου .. αλλιώς (τέλος πάντων) ..

$\displaystyle{\int\limits_0^1 {\frac{{\log \left( {1 + x} \right)}}{{\left( {1 + x} \right)\sqrt x }}dx} \mathop { = = = = }\limits^{x \to {x^2}} 2 \cdot \int\limits_0^1 {\frac{{\log \left( {1 + {x^2}} \right)}}{{1 + {x^2}}}dx} \mathop { = = = = = = }\limits^{x \to \tan x} }$ $\displaystyle{ - 4 \cdot \int\limits_0^{\pi /4} {\log \left( {\cos x} \right)dx} = - 4 \cdot \int\limits_0^{\pi /4} {\left( {\log \left( {2\cos x} \right) - \log 2} \right)dx} = }$

$\displaystyle{ = \pi \log 2 - 4\int\limits_0^{\pi /2} {\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}\cos \left( {2nx} \right)}}{n}} } = \pi \log 2 - }$ $\displaystyle{4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}\int\limits_0^{\pi /4} {\cos \left( {2nx} \right)dx} } = \pi \log 2 - 4\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}\sin \left( {\frac{{n\pi }}{2}} \right)}}{{2{n^2}}}} = }$

$\displaystyle{ = \pi \log 2 - 2\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\frac{{\left( {2n - 1} \right)\pi }}{2}} \right)}}{{{{\left( {2n - 1} \right)}^2}}}} = \pi \log 2 - 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{{\left( {2n - 1} \right)}^2}}}} = \pi \log 2 - 2G}$

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Χρησιμοποιήθηκε η σειρά Fourier $\displaystyle{\log \left( {2\cos x} \right) = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n - 1}}\cos \left( {2nx} \right)}}{n}} }$

mathematica.gr

Ολοκλήρωμα με λογάριθμο

Ολοκλήρωμα με λογάριθμο

Re: Ολοκλήρωμα με λογάριθμο

Re: Ολοκλήρωμα με λογάριθμο