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Δημοσιεύτηκε: **Σάβ Απρ 14, 2012 4:32 pm**

Κοτρώνης Αναστάσιος έγραψε: $\displaystyle{28)}$ Ας βάλουμε και κανένα γενικευμένο για επίλυση $\displaystyle{\int_{0}^{\pi/2}\frac{\tan^{-1}(a\sin x)}{\sin x}\,dx}$ όπου .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ
Από "Introduction to the theory of Fourier's series and integrals" Carslaw H.S. 1921.

Μια λύση, μιας και περασε καιρός.Iσχύει ότι $\displaystyle \arctan x=\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n x^{2n+1}}{2n+1}}, \left|x \right|\leqslant 1$ άρα $\displaystyle \frac{\arctan \left(a \sin x \right)}{\sin x}=\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n a^{2n+1} \sin ^{2n} x}{2n+1}}$ και άρα έχουμε $\displaystyle \int_{0}^{\pi /2}\frac{\arctan \left(a \sin x \right)}{\sin x}dx=\int_{0}^{\pi /2}\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n a^{2n} \sin ^{2n} x}{2n+1}} dx$ .

Μπορούμε να ενναλάξουμε ολοκληρώμα και σειρά.Αυτο διακιολογείται και από το θεώρημα Fubini αφού $\displaystyle \sum_{n=0}^{\infty}{\int_{0}^{\pi/2}\frac{\left|a \right|^{2n+1} \sin^{2n} x}{2n+1}}dx < \infty$ (*).

Έστω τώρα $\displaystyle I_{n}:=\int_{0}^{\pi /2}{\left(\sin x \right)^n dx}$ μετά από μια παραγοντική βρίσκουμε $\displaystyle \frac{I_{2k}}{I_{2k-2}}=\frac{k-1}{k}$ οπότε πολλαπλασιάζοντας έχουμε τηλεσκοπικά $\displaystyle I_{2n}=I_{0} \frac{\left(2n-1 \right)!!}{\left(2n \right)!!}=\frac{\pi}{2^{2n+1}}\binom{2n}{n}$ .

Άρα είναι εμφανές ότι η σειρά $\displaystyle \sum_{n=0}^{\infty}\frac{\pi \left|a \right|^{2n+1}}{2^{2n+1}}\binom{2n}{n}=\frac{\pi}{2} \sum_{n=0}^{\infty}{\left(\frac{a^2}{4} \right)^n \binom{2n}{n}}=\frac{\pi}{2\sqrt{1-a^2}}<\infty$ .

Άρα $\displaystyle \int_{0}^{\pi/2}{\frac{\arctan \left(a\sin x \right)}{\sin x}}dx=\frac{\pi}{2}\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n a^{2n+1}}{4^n \left(2n+1 \right)}}\binom{2n}{n}=\frac{\pi}{2}\sinh^{-1} a$ .

Μια απόδειξη για την τελευταία σειρά: Χρησιμοποιύμε το διωνυμικό θεώρημα για να δούμε ότι $\displaystyle \left(1+x^2 \right)^{-1/2} =\sum_{n=0}^{\infty}{\binom{2n}{n}\frac{\left(-1 \right)^n x^{2n}}{4^n}}, \left|x \right|<1$ . Ολοκληρώνοντας και τα δύο μέλη από 0 εως $\displaystyle z,\left|z \right|<1:$

$\displaystyle \int_{0}^{z}{\frac{dx}{\sqrt{1+x^2}}}=\int_{0}^{z}{\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n x^{2n}}{4^n}}}\binom{2n}{n} dx=\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n}{4^n}\binom{2n}{n}\int_{0}^{z}{x^{2n} dx}}$

$\displaystyle \Rightarrow \sinh^{-1} z =\sum_{n=0}^{\infty}{\frac{\left(-1 \right)^n}{4^n}\binom{2n}{n}\frac{z^{2n+1}}{2n+1}}$ .

Δημοσιεύτηκε: **Σάβ Απρ 14, 2012 10:22 pm**

Nebuchadnezzar έγραψε: 86) $\qquad \int_{-\infty}^{\infty} \cos(x^2) + \sin(x^2) dx$

Θεωρούμε την γραμμή που σχηματίζεται από το τμήμα $OA: z=x \; ,0\leq x\leq R$ , το τόξο $AB: \; z=Re^{it} \; , 0\leq t\leq \frac{\pi}{4}$
και το τμήμα $BO: z=re^{i\frac{\pi}{4}} \; , R\leq r\leq 0 .$

Η συνάρτηση $f(z)=e^{iz^2}$ είναι αναλυτική άρα:
$\displaystyle \int_{OA}f(z)dz +\int_{AB}f(z)dz +\int_{BO}f(z)dz = 0 \Rightarrow$

$\displaystyle \int_{0}^{R}e^{ix^2} dx + \int_{0}^{\pi /4}exp(iR^2e^{i2t})Rie^{it} dt + \int_{R}^{0}exp(ir^2e^{i\pi /2})e^{i\pi /4} dr = 0 \Rightarrow$

$\displaystyle \int_{0}^{R}e^{ix^2} dx = e^{i\pi /4}\int_{0}^{R}e^{-r^2} dr - \int_{0}^{\pi /4}exp(iR^2e^{i2t})Rie^{it} dt \quad (1)$

Είναι $\displaystyle \lim_{R\to \infty}\int_{0}^{R}e^{-r^2} dr = \int_{0}^{\infty}e^{-r^2} dr = \dfrac{\sqrt{\pi}}{2} .$

$\displaystyle \left| \int_{0}^{\pi /4}exp(iR^2e^{i2t})Rie^{it} dt\right| \leq R\int_{0}^{\pi /4}e^{-R^2\sin 2t} dt =$

$\displaystyle \dfrac{R}{2}\int_{0}^{\pi /2}e^{-R^2\sin u} du < \dfrac{R}{2}\cdot\dfrac{\pi}{2R^2} = \dfrac{\pi}{4R}\to 0$ καθώς $R\to\infty .$

Παίρνοντας όρια στην (1)

για $R\to\infty ,$ έχουμε: $\displaystyle \int_{0}^{\infty}\cos(x^2) dx + i \int_{0}^{\infty}\sin(x^2) dx = \dfrac{\sqrt{2\pi}}{4}+i\dfrac{\sqrt{2\pi}}{4} .$

Άρα $\displaystyle \int_{0}^{\infty}\cos(x^2) dx = \int_{0}^{\infty}\sin(x^2) dx = \dfrac{\sqrt{2\pi}}{4}$ οπότε $\displaystyle \int_{-\infty}^{\infty}\cos(x^2) dx = \int_{-\infty}^{\infty}\sin(x^2) dx = \dfrac{\sqrt{2\pi}}{2} .$

Επομένως, $\displaystyle \int_{-\infty}^{\infty}[\cos(x^2)+\sin(x^2)] dx = \sqrt{2\pi} .$

Δημοσιεύτηκε: **Δευ Απρ 16, 2012 10:54 am**

Κοτρώνης Αναστάσιος έγραψε: $\displaystyle{28)}$ Ας βάλουμε και κανένα γενικευμένο για επίλυση $\displaystyle{\int_{0}^{\pi/2}\frac{\tan^{-1}(a\sin x)}{\sin x}\,dx}$ όπου .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ
Από "Introduction to the theory of Fourier's series and integrals" Carslaw H.S. 1921.

Μετά την όμορφη λύση με Σειρές, που μας χάρισε ο Κώστας παραπάνω .. και μια πιο στοιχειώδης ..

Έστω $\displaystyle{f\left( a \right) = \int\limits_0^{\pi /2} {\frac{{\arctan \left( {a\sin \left( x \right)} \right)}}{{\sin x}}dx} }$ .

Τότε $\displaystyle{f'\left( a \right) = \int\limits_0^{\pi /2} {\frac{{\sin x}}{{\sin x\left( {1 + {a^2}{{\sin }^2}x} \right)}}dx} = \int\limits_0^{\pi /2} {\frac{1}{{1 + {a^2}{{\sin }^2}x}}dx} = \int\limits_0^{\pi /2} {\frac{1}{{1 + {a^2}\dfrac{{{{\tan }^2}x}}{{1 + {{\tan }^2}x}}}}dx} = \mathop = \limits^{\tan x = y} = }$

$\displaystyle{\int\limits_0^\infty {\frac{1}{{1 + \left( {{a^2} + 1} \right){y^2}}}dy} = \mathop = \limits^{y = x/\sqrt {{a^2} + 1} } = \frac{1}{{\sqrt {{a^2} + 1} }}\int\limits_0^\infty {\frac{1}{{1 + {x^2}}}dx} = \frac{\pi }{{2\sqrt {{a^2} + 1} }}}$ .

Επομένως $\displaystyle{f\left( a \right) = \frac{\pi }{2}\int {\frac{1}{{\sqrt {{a^2} + 1} }}da} = \frac{\pi }{2}\ln \left( {a + \sqrt {{a^2} + 1} } \right) + c}$ .

Για $\displaystyle{a = 0:f\left( 0 \right) = 0 \Rightarrow c = 0}$ . Άρα $\displaystyle{\int\limits_0^{\pi /2} {\frac{{\arctan \left( {a\sin \left( x \right)} \right)}}{{\sin x}}dx} = \frac{\pi }{2}\ln \left( {a + \sqrt {{a^2} + 1} } \right)}$ .

Παρεμπιπτόντως $\displaystyle{\ln \left( {a + \sqrt {{a^2} + 1} } \right) = {\sinh ^{ - 1}}a}$ .

Δημοσιεύτηκε: **Δευ Απρ 23, 2012 8:24 pm**

Και ένα σχετικά απλό για να πάρουμε μια ανάσα :

101)

Για

και $n\geq0$ πραγματικούς, ας υπολογισθεί το $\displaystyle{\int_{0}^{1}x^n\ln\left(\sqrt{1+x^k}-\sqrt{1-x^k}\right)\,dx}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

Δημοσιεύτηκε: **Τετ Απρ 25, 2012 1:24 am**

Κοτρώνης Αναστάσιος έγραψε: Για και $n\geq0$ πραγματικούς, ας υπολογισθεί το $\displaystyle{\int_{0}^{1}x^n\ln\left(\sqrt{1+x^k}-\sqrt{1-x^k}\right)\,dx}$ .

$\displaystyle{\int_0^1 {{x^n}\ln \left( {\sqrt {1 + {x^k}} - \sqrt {1 - {x^k}} } \right)dx} = \frac{1}{{n + 1}}\int_0^1 {{{\left( {{x^{n + 1}}} \right)}{'}}\ln \left( {\sqrt {1 + {x^k}} - \sqrt {1 - {x^k}} } \right)dx} = }$

$\displaystyle{ = \frac{1}{{n + 1}}\left[ {{x^{n + 1}}\ln \left( {\sqrt {1 + {x^k}} - \sqrt {1 - {x^k}} } \right)} \right]_0^1 - \frac{1}{{n + 1}}\int_0^1 {{x^{n + 1}}\frac{{\dfrac{{k{x^{k - 1}}}}{{2\sqrt {1 + {x^k}} }} + \dfrac{{k{x^{k - 1}}}}{{2\sqrt {1 - {x^k}} }}}}{{\sqrt {1 + {x^k}} - \sqrt {1 - {x^k}} }}dx} = }$

$\displaystyle{ = \frac{{\ln 2}}{{2\left( {n + 1} \right)}} - \frac{k}{{2\left( {n + 1} \right)}}\int_0^1 {{x^n}\frac{{1 + \sqrt {1 - {x^{2k}}} }}{{\sqrt {1 - {x^{2k}}} }}dx} = \frac{{\ln 2}}{{2\left( {n + 1} \right)}} - \frac{k}{{2\left( {n + 1} \right)}}\int_0^1 {\left( {\frac{{{x^n}}}{{\sqrt {1 - {x^{2k}}} }} + {x^n}} \right)dx} = }$

$\displaystyle{ = \frac{{\left( {n + 1} \right)\ln 2 - k}}{{2{{\left( {n + 1} \right)}^2}}} - \frac{k}{{2\left( {n + 1} \right)}}\int_0^1 {\frac{{{x^n}}}{{\sqrt {1 - {x^{2k}}} }}dx} = \mathop = \limits^{{x^{2k}} = y} = \frac{{\left( {n + 1} \right)\ln 2 - k}}{{2{{\left( {n + 1} \right)}^2}}} - \frac{k}{{4k\left( {n + 1} \right)}}\int_0^1 {{y^\big{{\dfrac{{n + 1}}{{2k}} - 1}}}{{\left( {1 - y} \right)}^\big{{\dfrac{1}{2} - 1}}}dy} = }$

$\displaystyle{ = \frac{{\left( {n + 1} \right)\ln 2 - k}}{{2{{\left( {n + 1} \right)}^2}}} - \frac{k}{{4k\left( {n + 1} \right)}}B\left( {\frac{{n + 1}}{{2k}},\frac{1}{2}} \right) = \frac{{\left( {n + 1} \right)\ln 2 - k}}{{2{{\left( {n + 1} \right)}^2}}} - \frac{{k\sqrt \pi }}{{4k\left( {n + 1} \right)}}\frac{{\Gamma \left( {\dfrac{{n + 1}}{{2k}}} \right)}}{{\Gamma \left( {\dfrac{{n + k + 1}}{{2k}}} \right)}}}$ .

Κανα λαθάκι .. δεν αποκλείεται ..

Δημοσιεύτηκε: **Κυρ Μάιος 20, 2012 1:12 am**

Ας υπολογισθεί το $\displaystyle{\int_{0}^{\pi/2}\frac{\arctan^{2}(\sin^2t)}{\sin^2t}\,dt}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

Δημοσιεύτηκε: **Κυρ Ιούλ 01, 2012 11:38 pm**

Ας υπολογισθεί το $\displaystyle{\int_{-1}^{0} \frac{x^2+2x}{\ln(x+1)}\,dx}$ .

Δημοσιεύτηκε: **Δευ Ιούλ 02, 2012 10:27 am**

Κοτρώνης Αναστάσιος έγραψε: Ας υπολογισθεί το $\displaystyle{\int_{-1}^{0} \frac{x^2+2x}{\ln(x+1)}\,dx}$ .

$\displaystyle{\int\limits_{ - 1}^0 {\frac{{{x^2} + 2x}}{{\ln \left( {x + 1} \right)}}dx} = \mathop = \limits^{x + 1 = u} = \int\limits_0^1 {\frac{{{u^2} - 1}}{{\ln u}}du} = \mathop = \limits^{\ln u = y} = \int\limits_{ - \infty }^0 {\frac{{{e^{2y}} - 1}}{y}{e^y}dy} = \mathop = \limits^{y = - x} = - \int\limits_0^\infty {\frac{{{e^{ - 2x}} - 1}}{x}{e^{ - x}}dx} = - \int\limits_0^\infty {\left( {{e^{ - 3x}} - {e^{ - x}}} \right)\left( {\int\limits_0^\infty {{e^{ - xy}}dy} } \right)dx} = }$

$\displaystyle{ = - \int\limits_0^\infty {\left( {\int\limits_0^\infty {\left( {{e^{ - \left( {y + 3} \right)x}} - {e^{ - \left( {y + 1} \right)x}}} \right)dx} } \right)dy} = - \int\limits_0^\infty {\left( {\frac{1}{{y + 3}} - \frac{1}{{y + 1}}} \right)dy} = - \left[ {\ln \frac{{y + 3}}{{y + 1}}} \right]_0^\infty = \ln 3}$

Δημοσιεύτηκε: **Παρ Ιούλ 20, 2012 8:20 pm**

Ας υπολογισθεί το $\displaystyle{\int_{0}^{\pi/2}\left(\cos x\ln(\cos^2 x)\right)^2\,dx}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

Δημοσιεύτηκε: **Παρ Αύγ 03, 2012 2:40 pm**

Δείξτε ότι $\displaystyle{\int_{0}^{\pi/2}x^2\ln^2(2\cos x)\,dx=\frac{1}{5}\left(\frac{\pi}{2}\right)^5+\pi\int_{0}^{+\infty}y\ln^2(1-e^{-2y})\,dy=\frac{11\pi^5}{1440}}$ και

106)

$\displaystyle{\int_{0}^{\pi/2}x^2\ln^2(2\cos x)\,dx=\frac{\pi^5}{160}-\frac{\pi}{24}\int_{0}^{+\infty}\ln^3(1-e^{-2y})\,dy}$

με μεθόδους πραγματικής ανάλυσης αν είναι δυνατόν.

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

Δημοσιεύτηκε: **Παρ Οκτ 19, 2012 1:10 am**

Ας υπολογισθεί το $\displaystyle{\int_{0}^{1}[-\ln x]^2\,dx}$ , όπου $[\,\cdot\,]$ το ακέραιο μέρος.

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ

Δημοσιεύτηκε: **Σάβ Ιαν 12, 2013 6:03 pm**

kwstas12345 έγραψε: $\displaystyle 72)$ Να υπολογιστεί το $\displaystyle \int_{-\infty}^{\infty}{\frac{\sinh ax}{\sinh bx}}dx,b>\left|a \right|$

$\displaystyle{b > \left| a \right|}$
$\displaystyle{ K\left( {a,b} \right) = \int\limits_{ - \infty }^{ + \infty } {\frac{{\sinh ax}} {{\sinh bx}}dx} = 2\int\limits_0^{ + \infty } {\frac{{e^{ax} - e^{ - ax} }} {{e^{bx} - e^{ - bx} }}dx} = 2\int\limits_0^{ + \infty } {\frac{{\left( {e^{ax} - e^{ - ax} } \right)e^{ - bx} }} {{1 - e^{ - 2bx} }}dx} }$

$\displaystyle{ = 2\int\limits_0^{ + \infty } {\left( {e^{ax} - e^{ - ax} } \right)e^{ - bx} \sum\limits_{k = 0}^{ + \infty } {e^{ - 2bxk} } dx} = 2\sum\limits_{k = 0}^{ + \infty } {\int\limits_0^{ + \infty } {\left( {e^{ - x\left( {2bk + b - a} \right)} - e^{ - x\left( {2bk + b + a} \right)} } \right)dx} } }$

$\displaystyle{ = 2\sum\limits_{k = 0}^{ + \infty } {\left( {\frac{1} {{2bk + b - a}} - \frac{1} {{2bk + b + a}}} \right)} = \frac{1} {b}\sum\limits_{k = 0}^{ + \infty } {\left( {\frac{1} {{k + \frac{{b - a}} {{2b}}}} - \frac{1} {{k + \frac{{b + a}} {{2b}}}}} \right)} }$

$\displaystyle{ = - \frac{1} {b}\sum\limits_{k = 0}^{ + \infty } {\left( {\frac{1} {{k + \frac{{b + a}} {{2b}}}} - \frac{1} {{k + \frac{{b - a}} {{2b}}}}} \right)} = - \frac{1} {b}\sum\limits_{k = 0}^{ + \infty } {\left( {\frac{1} {{k + \left( {\frac{{b + a}} {{2b}}} \right)}} - \frac{1} {{k + 1 - \frac{{b + a}} {{2b}}}}} \right)} }$

$\displaystyle{ = - \frac{1} {b}\left( {\sum\limits_{k = 0}^{ + \infty } {\frac{1} {{\frac{{b + a}} {{2b}} + k}}} + \sum\limits_{k = 0}^{ + \infty } {\frac{1} {{\frac{{b + a}} {{2b}} - \left( {k + 1} \right)}}} } \right) = - \frac{1} {b}\left( {\frac{1} {{\frac{{b + a}} {{2b}}}} + \sum\limits_{k = 1}^{ + \infty } {\frac{1} {{\frac{{b + a}} {{2b}} + k}}} + \sum\limits_{k = 1}^{ + \infty } {\frac{1} {{\frac{{b + a}} {{2b}} - k}}} } \right) }$

$\displaystyle{ = - \frac{\pi } {b} \cdot \frac{{\cos \left( {\frac{\pi } {2} + \frac{{\pi a}} {{2b}}} \right)}} {{\sin \left( {\frac{\pi } {2} + \frac{{\pi a}} {{2b}}} \right)}} = - \frac{\pi } {b} \cdot \frac{{\cos \left( {\frac{\pi } {2}} \right)\cos \left( {\frac{{\pi a}} {{2b}}} \right) - \sin \left( {\frac{\pi } {2}} \right)\sin \left( {\frac{{\pi a}} {{2b}}} \right)}} {{\sin \left( {\frac{\pi } {2}} \right)\cos \left( {\frac{{\pi a}} {{2b}}} \right) + \sin \left( {\frac{{\pi a}} {{2b}}} \right)\cos \left( {\frac{\pi } {2}} \right)}} = - \frac{\pi } {b} \cdot \frac{{ - \sin \frac{{\pi a}} {{2b}}}} {{\cos \frac{{\pi a}} {{2b}}}} = \frac{\pi } {b}\tan \frac{{\pi a}} {{2b}} }$

Δημοσιεύτηκε: **Δευ Ιαν 14, 2013 1:04 am**

erxmer έγραψε:Nα αποδειχθούν τα ακόλουθα:

43) $\boxed{\displaystyle{\int_{0}^{\infty}{\frac{dx}{(x^4+(1+\sqrt{2})x^2+1)(x^{100}-x^{98}+...+1)}=\frac{\pi}{2(1+\sqrt{2})}}}}$

το ολοκλήρωμα θα πρέπει να είναι
$\displaystyle{ I = \int\limits_0^{ + \infty } {\frac{{dx}} {{\left( {x^4 + \left( {1 + \sqrt 2 } \right)x^2 + 1} \right)\left( {x^{100} - x^{98} + ... + 1} \right)}}} = \frac{\pi } {{2\sqrt {3 + \sqrt 2 } }} }$

$\displaystyle{ I = \int\limits_0^{ + \infty } {\frac{{dx}} {{\left( {x^4 + \left( {1 + \sqrt 2 } \right)x^2 + 1} \right)\left( {x^{100} - x^{98} + ... + 1} \right)}}} \underbrace = _{x = \frac{1} {y}}\int\limits_0^{ + \infty } {\frac{{y^{102} }} {{\left( {y^4 + \left( {1 + \sqrt 2 } \right)y^2 + 1} \right)\left( {y^{100} - y^{98} + ... + 1} \right)}}dy} }$

$\displaystyle{ I = \frac{1} {2}\int\limits_0^{ + \infty } {\left( {\frac{1} {{\left( {x^4 + \left( {1 + \sqrt 2 } \right)x^2 + 1} \right)\left( {x^{100} - x^{98} + ... + 1} \right)}} + \frac{{x^{102} }} {{\left( {x^4 + \left( {1 + \sqrt 2 } \right)x^2 + 1} \right)\left( {x^{100} - x^{98} + ... + 1} \right)}}} \right)dx} }$

$\displaystyle{ = \frac{1} {2}\int\limits_0^{ + \infty } {\frac{{1 + x^{102} }} {{\left( {x^4 + \left( {1 + \sqrt 2 } \right)x^2 + 1} \right)\left( {x^{100} - x^{98} + ... + 1} \right)}}dx} }$

$\displaystyle{ x^{100} - x^{98} + ... + 1 = \frac{{1 + x^{2 \cdot 51} }} {{1 + x^2 }} = \frac{{1 + x^{102} }} {{1 + x^2 }} }$

$\displaystyle{ = \frac{1} {2}\int\limits_0^{ + \infty } {\frac{{1 + x^2 }} {{x^4 + \left( {1 + \sqrt 2 } \right)x^2 + 1}}dx} = \frac{1} {2}\int\limits_0^{ + \infty } {\frac{{1 + x^2 }} {{x^2 \left( {x^2 + \sqrt 2 } \right) + x^2 + 1}}dx} }$

$\displaystyle{ \underbrace = _{x^2 + 1 = u}\frac{1} {2}\int\limits_1^{ + \infty } {\frac{u} {{\left( {u - 1} \right)\left( {u - 1 + \sqrt 2 } \right) + u}}\frac{{du}} {{2\sqrt {u - 1} }}} \underbrace = _{u - 1 = y}\frac{1} {4}\int\limits_0^{ + \infty } {\frac{{y + 1}} {{y\left( {y + \sqrt 2 } \right) + y + 1}}\frac{{du}} {{\sqrt y }}} }$

$\displaystyle{ = \frac{1} {4}\int\limits_0^{ + \infty } {\frac{{y + 1}} {{y^2 + y\left( {1 + \sqrt 2 } \right) + 1}}\frac{{dy}} {{\sqrt y }}} }$

$\displaystyle{ \underbrace = _{y = \tan ^2 \theta }\frac{1} {2}\int\limits_0^{\frac{\pi } {2}} {\frac{{\tan ^2 \theta + 1}} {{\tan ^4 \theta + \tan ^2 \theta \left( {1 + \sqrt 2 } \right) + 1}}\frac{{\tan \theta \sec ^2 \theta d\theta }} {{\sqrt {\tan ^2 \theta } }}} }$

$\displaystyle{ = \frac{1} {2}\int\limits_0^{\frac{\pi } {2}} {\frac{{\sec ^4 \theta }} {{\sec ^2 \theta + \tan ^4 \theta \left( {1 + \frac{{\sqrt 2 }} {{\tan ^2 \theta }}} \right)}}d\theta } = \frac{1} {2}\int\limits_0^{\frac{\pi } {2}} {\frac{{d\theta }} {{\cos ^2 \theta + \sin ^4 \theta + \sqrt 2 \cos ^2 \theta \sin ^2 \theta }}} }$

$\displaystyle{ = \frac{1} {2}\int\limits_0^{\frac{\pi } {2}} {\frac{{d\theta }} {{1 + \left( {\sqrt 2 - 1} \right)\sin ^2 \theta + \left( {1 - \sqrt 2 } \right)\sin ^4 \theta }}} = \frac{1} {{2\left( {1 - \sqrt 2 } \right)}}\int\limits_0^{\frac{\pi } {2}} {\frac{{d\theta }} {{ - \left( {1 + \sqrt 2 } \right) - \sin ^2 \theta \left( {1 - \sin ^2 \theta } \right)}}} }$

$\displaystyle{ = 2\left( {\sqrt 2 + 1} \right)\int\limits_0^{\frac{\pi } {2}} {\frac{{d\theta }} {{4\left( {\sqrt 2 + 1} \right) + \sin ^2 2\theta }}} = \frac{1} {2}\int\limits_0^{\frac{\pi } {2}} {\frac{{d\theta }} {{1 + \frac{{\sin ^2 2\theta }} {{4\left( {\sqrt 2 + 1} \right)}}}}} = \frac{1} {2}\int\limits_0^{\frac{\pi } {2}} {\sum\limits_{k = 0}^{ + \infty } {\left( { - 1} \right)^k \left( {\frac{{\sin ^2 2\theta }} {{4\left( {\sqrt 2 + 1} \right)}}} \right)^k } } d\theta }$

$\displaystyle{ = \frac{1} {2}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{2^{2k} \left( {\sqrt 2 + 1} \right)^k }}} \int\limits_0^{\frac{\pi } {2}} {\sin ^{2k} 2\theta d\theta } = \frac{1} {2}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}} \int\limits_0^{\frac{\pi } {2}} {\sin ^{2k} \theta \cos ^{2k} \theta d\theta } }$

$\displaystyle{ = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}} 2\int\limits_0^{\frac{\pi } {2}} {\sin ^{2\left( {k + \frac{1} {2}} \right) - 1} \theta \cos ^{2\left( {k + \frac{1} {2}} \right) - 1} \theta d\theta } = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\beta \left( {k + \frac{1} {2},k + \frac{1} {2}} \right)} }$

$\displaystyle{ = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\frac{{\Gamma ^2 \left( {k + \frac{1} {2}} \right)}} {{\Gamma \left( {2k + 1} \right)}}} = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\frac{{\left( {\frac{{\sqrt \pi }} {{2^{2k - 1} }}\frac{{\Gamma \left( {2k} \right)}} {{\Gamma \left( k \right)}}} \right)^2 }} {{\Gamma \left( {2k + 1} \right)}}} }$

$\displaystyle{ = \frac{1} {4}\sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{\left( {\sqrt 2 + 1} \right)^k }}\frac{{\Gamma ^2 \left( {k + \frac{1} {2}} \right)}} {{\Gamma \left( {2k + 1} \right)}}} = \pi \sum\limits_{k = 0}^{ + \infty } {\frac{{\left( { - 1} \right)^k }} {{4^{2k} \left( {\sqrt 2 + 1} \right)^k }}\frac{{\Gamma ^2 \left( {2k} \right)}} {{\Gamma ^2 \left( k \right)\Gamma \left( {2k + 1} \right)}}} }$

$\displaystyle{ = \frac{\pi } {4}\sum\limits_{k = 0}^{ + \infty } {\left( { - \frac{1} {{4^2 \left( {\sqrt 2 + 1} \right)}}} \right)^k \frac{{\left( {2k} \right)!}} {{\left( {k!} \right)^2 }}} = \frac{\pi } {4}\frac{1} {{\sqrt {1 - 4 \cdot - \frac{1} {{16\left( {\sqrt 2 + 1} \right)}}} }} = \frac{\pi } {4}\frac{1} {{\sqrt {1 + \frac{{\sqrt 2 - 1}} {4}} }} = \frac{\pi } {2}\frac{1} {{\sqrt {3 + \sqrt 2 } }} }$

Δημοσιεύτηκε: **Δευ Ιαν 14, 2013 4:59 am**

Κοτρώνης Αναστάσιος έγραψε: Ας υπολογισθεί το $\displaystyle{\int_{0}^{\pi/2}\left(\cos x\ln(\cos^2 x)\right)^2\,dx}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ
Paolo Perfetti

$\displaystyle{ K = \int\limits_0^{\frac{\pi } {2}} {\cos ^2 \theta \ln ^2 \left( {\cos ^2 \theta } \right)d\theta } }$

$\displaystyle{ \int\limits_0^{\frac{\pi } {2}} {\cos ^2 \theta \ln ^2 \left( {\cos ^2 \theta } \right)d\theta } \underbrace = _{\cos ^2 \theta = y}\frac{1} {2}\int\limits_0^1 {y\ln ^2 y\frac{{dy}} {{\sqrt y \sqrt {1 - y} }}} = \frac{1} {2}\int\limits_0^1 {\sqrt {\frac{y} {{1 - y}}} \ln ^2 ydy} }$

$\displaystyle{ = \frac{1} {2}\int\limits_0^1 {y^{\frac{1} {2}} \left( {1 - y} \right)^{ - \frac{1} {2}} \ln ^2 ydy} = \frac{1} {2}\int\limits_0^1 {y^{1 + \frac{1} {2} - 1} \left( {1 - y} \right)^{1 - \frac{1} {2} - 1} \ln ^2 ydy} }$

$\displaystyle{ = \frac{1} {2}\frac{{\partial ^2 \beta \left( {p,q} \right)}} {{\partial p^2 }}\left| {_{\left( {\frac{3} {2},\frac{1} {2}} \right)} } \right. }$

$\displaystyle{ \frac{{\partial ^2 \beta \left( {p,q} \right)}} {{\partial p^2 }} = \beta \left( {p,q} \right)\left( {\left( {\psi ^{\left( 0 \right)} \left( p \right) - \psi ^{\left( 0 \right)} \left( {p + q} \right)} \right)^2 + \psi ^{\left( 1 \right)} \left( p \right) - \psi ^{\left( 1 \right)} \left( {p + q} \right)} \right) }$

$\displaystyle{ = \frac{1} {2}\frac{{\partial ^2 \beta \left( {p,q} \right)}} {{\partial p^2 }}\left| {_{\left( {\frac{3} {2},\frac{1} {2}} \right)} } \right. = \frac{1} {2}\beta \left( {\frac{3} {2},\frac{1} {2}} \right)\left( {\left( {\psi ^{\left( 0 \right)} \left( {\frac{3} {2}} \right) - \psi ^{\left( 0 \right)} \left( {\frac{3} {2} + \frac{1} {2}} \right)} \right)^2 + \psi ^{\left( 1 \right)} \left( {\frac{3} {2}} \right) - \psi ^{\left( 1 \right)} \left( {\frac{3} {2} + \frac{1} {2}} \right)} \right) }$

$\displaystyle{ = \frac{1} {2}\frac{{\Gamma \left( {\frac{3} {2}} \right)\Gamma \left( {\frac{1} {2}} \right)}} {{\Gamma \left( {\frac{3} {2} + \frac{1} {2}} \right)}}\left( {\left( {\psi ^{\left( 0 \right)} \left( {\frac{3} {2}} \right) - \psi ^{\left( 0 \right)} \left( {\frac{3} {2} + \frac{1} {2}} \right)} \right)^2 + \psi ^{\left( 1 \right)} \left( {\frac{3} {2}} \right) - \psi ^{\left( 1 \right)} \left( {\frac{3} {2} + \frac{1} {2}} \right)} \right) }$

$\displaystyle{ = \frac{\pi } {4}\left( {\left( {1 - 2\ln 2} \right)^2 + \frac{{\pi ^2 }} {3} - 3} \right) }$

Δημοσιεύτηκε: **Δευ Ιαν 14, 2013 10:28 pm**

erxmer έγραψε: Nα αποδειχθεί οτι $\boxed{\displaystyle{\int_{0}^{+\infty}{\frac{cosx^2-cosx}{x}}dx=\frac{\gamma }{2}}}$

$\gamma = Euler \ constant$

$\displaystyle{ I\left( n \right) = \int\limits_0^{ + \infty } {\frac{{\cos \left( {x^2 } \right) - \cos x}} {{x^{n + 1} }}dx} }$

$\displaystyle{ \cos \left( {x^2 } \right) - \cos x = u \wedge \frac{{dx}} {{x^{n + 1} }} = dv }$

$\displaystyle{ \left( {\sin x - 2x\sin \left( {x^2 } \right)} \right)dx = du \wedge - \frac{1} {{nx^n }} = v }$

$\displaystyle{ I\left( n \right) = \int\limits_0^{ + \infty } {\frac{{\cos \left( {x^2 } \right) - \cos x}} {{x^{n + 1} }}dx} = \left( { - \frac{{\cos \left( {x^2 } \right) - \cos x}} {{nx^n }}} \right)\left| {_0^{ + \infty } } \right. + \frac{1} {n}\int\limits_0^{ + \infty } {\frac{{\sin x - 2x\sin \left( {x^2 } \right)}} {{x^n }}dx} }$

$\displaystyle{ I\left( n \right) = \frac{1} {n}\left( {\int\limits_0^{ + \infty } {\frac{{\sin x}} {{x^n }}dx} - \int\limits_0^{ + \infty } {\frac{{\sin \left( {x^2 } \right)}} {{x^n }} \cdot 2xdx} } \right) = \frac{1} {n}\left( {\int\limits_0^{ + \infty } {\frac{{\sin x}} {{x^n }}dx} - \int\limits_0^{ + \infty } {\frac{{\sin y}} {{y^{\frac{n} {2}} }}dy} } \right) }$

$\displaystyle{ = \frac{\Im } {n}\left( {\int\limits_0^{ + \infty } {x^{ - n} e^{ix} dx} - \int\limits_0^{ + \infty } {y^{ - \frac{n} {2}} e^{iy} dy} } \right) = \frac{\Im } {n}\left( { - \frac{1} {i}\int\limits_0^{ + \infty } {\left( {\frac{x} {i}} \right)^{ - n} e^{ - x} dx} + \frac{1} {i}\int\limits_0^{ + \infty } {\left( {\frac{x} {i}} \right)^{ - \frac{n} {2}} e^{ - x} dx} } \right) }$

$\displaystyle{ = \frac{\Im } {n}\left( {i \cdot i^n \int\limits_0^{ + \infty } {x^{1 - n - 1} e^{ - x} dx} - i \cdot i^{\frac{n} {2}} \int\limits_0^{ + \infty } {x^{1 - \frac{n} {2} - 1} e^{ - x} dx} } \right) = \frac{\Im } {n}\left( {i \cdot i^n \Gamma \left( {1 - n} \right) - i \cdot i^{\frac{n} {2}} \Gamma \left( {1 - \frac{n} {2}} \right)} \right) }$

$\displaystyle{ i^n = \left( {\cos \frac{\pi } {2} + i\sin \frac{\pi } {2}} \right)^n = e^{i\frac{{n\pi }} {2}} = \cos \frac{{n\pi }} {2} + i\sin \frac{{n\pi }} {2} }$

$\displaystyle{ i^{\frac{n} {2}} = \left( {\cos \frac{\pi } {2} + i\sin \frac{\pi } {2}} \right)^{\frac{n} {2}} = e^{i\frac{{n\pi }} {4}} = \cos \frac{{n\pi }} {4} + i\sin \frac{{n\pi }} {4} }$

$\displaystyle{ = - \frac{\Im } {n}\left( {i\left( {\cos \frac{{n\pi }} {4} + i\sin \frac{{n\pi }} {4}} \right)\Gamma \left( {1 - \frac{n} {2}} \right) - i\left( {\cos \frac{{n\pi }} {2} + i\sin \frac{{n\pi }} {2}} \right)\Gamma \left( {1 - n} \right)} \right) }$

$\displaystyle{ = - \frac{\Im } {n}\left( {\sin \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right) - \sin \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right) + i\left( {\cos \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right) - \cos \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right)} \right)} \right) }$

$\displaystyle{ = \frac{{\cos \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right) - \cos \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right)}} {n} }$

$\displaystyle{ I\left( n \right) = \int\limits_0^{ + \infty } {\frac{{\cos \left( {x^2 } \right) - \cos x}} {{x^{n + 1} }}dx} = \frac{{\cos \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right) - \cos \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right)}} {n} }$

$\displaystyle{ \int\limits_0^{ + \infty } {\frac{{\cos \left( {x^2 } \right) - \cos x}} {x}dx} = \mathop {\lim }\limits_{n \to 0} \frac{{\cos \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right) - \cos \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right)}} {n} }$

$\displaystyle{ = \mathop {\lim }\limits_{n \to 0} \left( {\frac{\pi } {4}\sin \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right) + \frac{1} {2}\cos \frac{{n\pi }} {4}\Gamma \left( {1 - \frac{n} {2}} \right)\psi \left( {1 - \frac{n} {2}} \right) - \frac{\pi } {2}\sin \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right) - \cos \frac{{n\pi }} {2}\Gamma \left( {1 - n} \right)\psi \left( {1 - n} \right)} \right) }$

$\displaystyle{ = - \Gamma \left( 1 \right)\psi \left( 1 \right) + \frac{1} {2}\Gamma \left( 1 \right)\psi \left( 1 \right) }$

$\displaystyle{ = - \psi \left( 1 \right)\left( {1 - \frac{1} {2}} \right) }$

$\displaystyle{ = \frac{1} {2}\gamma }$

Δημοσιεύτηκε: **Κυρ Φεβ 24, 2013 4:38 pm**

Και ένα από το μέλος galactus: $\displaystyle{\displaystyle \int_{0}^{\frac{\pi}{4}}x\ln^{2}(\sin(x))dx}$ .

Δημοσιεύτηκε: **Σάβ Μαρ 02, 2013 9:45 am**

Κοτρώνης Αναστάσιος έγραψε: Ας υπολογισθεί το $\displaystyle{\int_{0}^{1}[-\ln x]^2\,dx}$ , όπου $[\,\cdot\,]$ το ακέραιο μέρος.

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ
Les Reid

Λήμμα Για $\displaystyle{\left| x \right| < 1}$ ισχύει $\displaystyle{\sum\limits_{n = 1}^\infty {{n^2}{x^n}} = \frac{{x\left( {1 + x} \right)}}{{{{\left( {1 - x} \right)}^3}}}}$ . Διότι
$\displaystyle{\sum\limits_{n = 1}^\infty {{x^n}} = \frac{x}{{1 - x}} \Rightarrow \sum\limits_{n = 1}^\infty {n{x^{n - 1}}} = \frac{d}{{dx}}\left( {\frac{x}{{1 - x}}} \right) \Rightarrow \sum\limits_{n = 1}^\infty {n{x^n}} = \frac{x}{{{{\left( {1 - x} \right)}^2}}} \Rightarrow \sum\limits_{n = 1}^\infty {{n^2}{x^{n - 1}}} = \frac{d}{{dx}}\left( {\frac{x}{{{{\left( {1 - x} \right)}^2}}}} \right) \Rightarrow \sum\limits_{n = 1}^\infty {{n^2}{x^n}} = \frac{{x\left( {1 + x} \right)}}{{{{\left( {1 - x} \right)}^3}}}}$ , $\displaystyle{(*)}$

Επίσης Για $\displaystyle{{e^{ - \left( {n + 1} \right)}} < x \leqslant {e^{ - n}} \Rightarrow - \left( {n + 1} \right) < \ln x \leqslant - n \Rightarrow n \leqslant - \ln x < \left( {n + 1} \right) \Rightarrow \left\lfloor { - \ln x} \right\rfloor = n}$ , $\displaystyle{(**)}$ . Τότε

$\displaystyle{\begin{gathered} \int\limits_0^1 {{{\left\lfloor { - \ln x} \right\rfloor }^2}dx} = \sum\limits_{n = 0}^\infty {\left( {\;\int\limits_{{e^{ - \left( {n + 1} \right)}}}^{{e^{ - n}}} {{{\left\lfloor { - \ln x} \right\rfloor }^2}dx} } \right)} = \mathop = \limits^{\left( * \right)} = \sum\limits_{n = 0}^\infty {\left( {\;\int\limits_{{e^{ - \left( {n + 1} \right)}}}^{{e^{ - n}}} {{n^2}dx} } \right)} = \sum\limits_{n = 0}^\infty {{n^2}\left( {{e^{ - n}} - {e^{ - \left( {n + 1} \right)}}} \right)} = \sum\limits_{n = 0}^\infty {{n^2}{e^{ - n}}\left( {1 - \frac{1}{e}} \right)} = \hfill \\ \frac{{e - 1}}{e}\sum\limits_{n = 0}^\infty {{n^2}{e^{ - n}}} = \mathop = \limits^{\left( {**} \right)} = \frac{{e - 1}}{e} \cdot \frac{{\frac{1}{e}\left( {1 + \frac{1}{e}} \right)}}{{{{\left( {1 - \frac{1}{e}} \right)}^3}}} \Rightarrow \boxed{\int\limits_0^1 {{{\left\lfloor { - \ln x} \right\rfloor }^2}dx} = \frac{{\left( {1 + e} \right)}}{{{{\left( {e - 1} \right)}^2}}}} \hfill \\ \end{gathered} }$

Δημοσιεύτηκε: **Δευ Απρ 08, 2013 2:51 am**

Δημοσιεύτηκε: **Δευ Σεπ 16, 2013 8:23 pm**

Κοτρώνης Αναστάσιος έγραψε:Mε μεθόδους πραγματικής ανάλυσης αν είναι δυνατόν.

Από εδώ viewtopic.php?f=9&t=27979 θα χρησιμοποιηθούν: το λήμμα 6, δηλαδή $\displaystyle{\left[ {L6} \right]:\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}} = \frac{{{\pi ^4}}}{{72}}}$ , το λήμμα 1, δηλαδή $\displaystyle{ - \frac{{\log \left( {1 - x} \right)}}{{1 - x}} = \sum\limits_{n = 1}^\infty {{H_n}{x^n}} }$

και το τελικό αποτέλεσμα, δηλαδή ότι $\displaystyle{64\int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right)dx} = \frac{{22{\pi ^5}}}{{45}} \Rightarrow \int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right)dx} = \frac{{11{\pi ^5}}}{{1440}}\;\;\left[ * \right]}$ ,

καθώς και ότι $\displaystyle{\left[ {L1} \right]:\int\limits_0^1 {{x^n}{{\log }^2}x\,dx} = \frac{2}{{{{\left( {n + 1} \right)}^3}}}}$ και $\displaystyle{\left[ {L2} \right]:\int\limits_0^1 {{x^n}{{\log }^3}x\,dx} = \frac{{ - 6}}{{{{\left( {n + 1} \right)}^4}}}}$ (απλές παραγοντικές ολοκληρώσεις).

Κοτρώνης Αναστάσιος έγραψε: Δείξτε ότι $\displaystyle{\int_{0}^{\pi/2}x^2\ln^2(2\cos x)\,dx=\frac{1}{5}\left(\frac{\pi}{2}\right)^5+\pi\int_{0}^{+\infty}y\ln^2(1-e^{-2y})\,dy=\frac{11\pi^5}{1440}}$

$\displaystyle{\int\limits_0^\infty {y{{\log }^2}\left( {1 - {e^{ - 2y}}} \right)dy} = \mathop {\mathop = \limits_{y = - \log x} }\limits^{{e^{ - y}} = x} = - \int\limits_0^1 {\frac{{\log x}}{x}{{\log }^2}\left( {1 - {x^2}} \right)dx} = \mathop = \limits^{x = \sqrt y } = - \frac{1}{4}\int\limits_0^1 {\frac{{\log y}}{y}{{\log }^2}\left( {1 - y} \right)dy} = \mathop = \limits^{y = 1 - x} = }$

$\displaystyle{ = - \frac{1}{4}\int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{{1 - x}}{{\log }^2}x\,dx} = \frac{1}{4}\int\limits_0^1 {{{\log }^2}x\sum\limits_{n = 1}^\infty {{H_n}{x^n}} \,dx} = \frac{1}{4}\sum\limits_{n = 1}^\infty {{H_n}\int\limits_0^1 {{x^n}{{\log }^2}x\,dx} } = \mathop = \limits^{\left[ {L1} \right]} = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{{\left( {n + 1} \right)}^3}}}} = }$

$\displaystyle{ = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{H_{n + 1}} - 1/\left( {n + 1} \right)}}{{{{\left( {n + 1} \right)}^3}}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{{H_{n + 1}}}}{{{{\left( {n + 1} \right)}^3}}}} - \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^4}}}} = \mathop = \limits^{\left[ {L6} \right]} = \frac{1}{2}\left( {\frac{{{\pi ^4}}}{{72}} - \frac{{{\pi ^4}}}{{90}}} \right) = \frac{{{\pi ^4}}}{{720}}}$

Τότε $\displaystyle{\frac{1}{5}{\left( {\frac{\pi }{2}} \right)^5} + \pi \int\limits_0^{ + \infty } {y{{\log }^2}(1 - {e^{ - 2y}}){\kern 1pt} dy} = \frac{{{\pi ^5}}}{{160}} + \frac{{{\pi ^5}}}{{720}} = \frac{{11{\pi ^5}}}{{1440}} = \mathop = \limits^{\left[ * \right]} = \int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right){\kern 1pt} dx} }$

Κοτρώνης Αναστάσιος έγραψε: $\displaystyle{\int_{0}^{\pi/2}x^2\ln^2(2\cos x)\,dx=\frac{\pi^5}{160}-\frac{\pi}{24}\int_{0}^{+\infty}\ln^3(1-e^{-2y})\,dy}$

$\displaystyle{\int\limits_0^\infty {{{\log }^3}\left( {1 - {e^{ - 2y}}} \right)dy} = \mathop = \limits^{{e^{ - 2y}} = x} = \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^3}\left( {1 - x} \right)}}{x}dx} = \mathop = \limits^{x \to 1 - x} = \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^3}x}}{{1 - x}}dx} = \frac{1}{2}\int\limits_0^1 {{{\log }^3}x\sum\limits_{n = 0}^\infty {{x^n}} dx} = }$

$\displaystyle{ = \frac{1}{2}\sum\limits_{n = 0}^\infty {\int\limits_0^1 {{x^n}{{\log }^3}x\,dx} } = \mathop = \limits^{\left[ {L2} \right]} = - 3\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^4}}}} = - \frac{{{\pi ^4}}}{{30}}}$

Τότε $\displaystyle{\frac{{{\pi ^5}}}{{160}} - \frac{\pi }{{24}}\int\limits_0^{ + \infty } {{{\log }^3}\left( {1 - {e^{ - 2y}}} \right)dy} = \frac{{{\pi ^5}}}{{160}} + \frac{{{\pi ^5}}}{{720}} = \frac{{11{\pi ^5}}}{{1440}} = \mathop = \limits^{\left[ * \right]} = \int\limits_0^{\pi /2} {{x^2}{{\log }^2}\left( {2\cos x} \right){\kern 1pt} dx} }$

Κοτρώνης Αναστάσιος έγραψε:
ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ
Cody Thompson

Δημοσιεύτηκε: **Τρί Οκτ 29, 2013 12:47 am**

Κοτρώνης Αναστάσιος έγραψε:Και το γουρλίδικο το κατοστάρι, (όσο σάπιο γίνεται) αφιερωμένο στο φίλο μου το Σεραφείμ που με παίζει.

$\displaystyle{100)}$ Αν $a\in[-1,1)$ , δείξτε ότι $\displaystyle{\int_{0}^{1}\frac{(a-t)\ln(1-t)}{1-2at+t^2}\,dt=\frac{\pi^2}{12}-\frac{(\arccos a-\pi)^2}{8}-\frac{1}{8}\ln^2(2-2a)}$ .

ΕΜΦΑΝΙΣΗ ΚΕΙΜΕΝΟΥ
Omran Kouba, Syria. (...Βάστα φίλε...)

Μετά από ενάμιση χρόνο .. καλά είναι ..

$\displaystyle{\int\limits_0^1 {\frac{{\left( {a - t} \right)\log \left( {1 - t} \right)}}{{1 - 2at + {t^2}}}dt = } - \int\limits_0^1 {\frac{{\left( {t - a} \right)\log \left( {1 - t} \right)}}{{{{\left( {t - a} \right)}^2} + 1 - {a^2}}}dt} = \mathop {\mathop = \limits_{\big{{z_2} = a - i\sqrt {1 - {a^2}} }} }\limits^{\big{{z_1} = a + i\sqrt {1 - {a^2}}} } = - \int\limits_0^1 {\frac{{\left( {t - a} \right)\log \left( {1 - t} \right)}}{{\left( {t - {z_1}} \right)\left( {t - {z_2}} \right)}}dt} = }$

$\displaystyle{ = - \frac{1}{2}\int\limits_0^1 {\log \left( {1 - t} \right)\left( {\frac{1}{{t - {z_1}}} + \frac{1}{{t - {z_2}}}} \right)dt} = \frac{1}{2}\int\limits_0^1 {\log \left( {1 - t} \right)\left( {\frac{1}{{{z_1}}}\sum\limits_{n = 0}^\infty {\frac{{{t^n}}}{{z_1^n}}} + \frac{1}{{{z_2}}}\sum\limits_{n = 0}^\infty {\frac{{{t^n}}}{{z_2^n}}} } \right)dt} = \frac{1}{{2{z_1}}}\sum\limits_{n = 0}^\infty {\frac{1}{{z_1^n}}\int\limits_0^1 {\log \left( {1 - t} \right){t^n}dt} } + }$

$\displaystyle{ + \frac{1}{{2{z_2}}}\sum\limits_{n = 0}^\infty {\frac{1}{{z_2^n}}\int\limits_0^1 {\log \left( {1 - t} \right){t^n}dt} } = - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{{H_{n + 1}}}}{{n + 1}}\frac{1}{{z_1^{n + 1}}}} - \frac{1}{2}\sum\limits_{n = 0}^\infty {\frac{{{H_{n + 1}}}}{{n + 1}}\frac{1}{{z_2^{n + 1}}}} = - \frac{1}{2}\left( {\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\frac{1}{{z_1^n}}} + \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{n}\frac{1}{{z_2^n}}} } \right) = }$

$\displaystyle{ = - \frac{1}{2}\left( {\frac{1}{2}{{\log }^2}\left( {1 - {z_1}} \right) + L{i_2}\left( {{z_1}} \right) + \frac{1}{2}{{\log }^2}\left( {1 - {z_2}} \right) + L{i_2}\left( {{z_2}} \right)} \right) = - \frac{1}{4}\left( {{{\log }^2}\left( {1 - {z_1}} \right) + {{\log }^2}\left( {1 - {z_2}} \right)} \right) - }$

$\displaystyle{ - \frac{1}{2}\left( {L{i_2}\left( {{z_1}} \right) + L{i_2}\left( {\frac{1}{{{z_1}}}} \right)} \right) = - \frac{1}{4}\left( {{{\log }^2}\left( {1 - a - i\sqrt {1 - {a^2}} } \right) + {{\log }^2}\left( {1 - a + i\sqrt {1 - {a^2}} } \right)} \right) - \frac{1}{2}\left( { - \frac{{{\pi ^2}}}{6} - \frac{1}{2}{{\log }^2}\left( { - {z_1}} \right)} \right) = }$

$\displaystyle{ = - \frac{1}{4}\left( {{{\log }^2}\left( {\sqrt {2 - 2a} \cdot {e^{\big{ - i \cdot \arctan \sqrt {\frac{{1 + a}}{{1 - a}}}} }}} \right) + {{\log }^2}\left( {\sqrt {2 - 2a} \cdot {e^{\big{i \cdot \arctan \sqrt {\frac{{1 + a}}{{1 - a}}} }}} \right)} \right) + \frac{1}{2}\left( {\frac{{{\pi ^2}}}{6} + \frac{1}{2}{{\log }^2}\left( { - a - i\sqrt {1 - {a^2}} } \right)} \right) = }$

$\displaystyle{ = - \frac{1}{4}\left( {{{\left( {\log \sqrt {2 - 2a} - i \cdot \arctan \sqrt {\frac{{1 + a}}{{1 - a}}}} } \right)}^2} + {{\left( {\log \sqrt {2 - 2a} - i \cdot \arctan \sqrt {\frac{{1 + a}}{{1 - a}}} } \right)}^2}} \right) + \frac{{{\pi ^2}}}{{12}} + \frac{1}{4}{\log ^2}{e^{\big{i\left( {\arctan \frac{{\sqrt {1 - {a^2}} }}{a} - \pi }} \right)}} = }$

$\displaystyle{ = \frac{{{\pi ^2}}}{{12}} - \frac{1}{4}\left( {2{{\log }^2}\sqrt {2 - 2a} - 2{{\arctan }^2}\sqrt {\frac{{1 + a}}{{1 - a}}} } \right) - \frac{1}{4}{\left( {\arctan \frac{{\sqrt {1 - {a^2}} }}{a} - \pi } \right)^2} = }$

$\displaystyle{ = \frac{{{\pi ^2}}}{{12}} - \frac{1}{8}{\log ^2}\left( {2 - 2a} \right) + \frac{1}{2}{\arctan ^2}\sqrt {\frac{{1 + a}}{{1 - a}}} - \frac{1}{4}{\left( {\arctan \frac{{\sqrt {1 - {a^2}} }}{a} - \pi } \right)^2}}$

Όμως $\displaystyle{\arctan \sqrt {\frac{{1 + a}}{{1 - a}}} = \frac{{\pi - \arccos a}}{2}}$ και $\displaystyle{\arctan \frac{{\sqrt {1 - {a^2}} }}{a} = \arccos a}$ (προκύπτουν στοιχειωδώς). Οπότε

$\displaystyle{\int\limits_0^1 {\frac{{\left( {a - t} \right)\log \left( {1 - t} \right)}}{{1 - 2at + {t^2}}}dt = } \frac{{{\pi ^2}}}{{12}} - \frac{1}{8}{\log ^2}\left( {2 - 2a} \right) + \frac{1}{2}{\arctan ^2}\sqrt {\frac{{1 + a}}{{1 - a}}} - \frac{1}{4}{\left( {\arctan \frac{{\sqrt {1 - {a^2}} }}{a} - \pi } \right)^2} = }$

$\displaystyle{ = \frac{{{\pi ^2}}}{{12}} - \frac{1}{8}{\log ^2}\left( {2 - 2a} \right) + \frac{1}{8}{\left( {\pi - \arccos a} \right)^2} - \frac{1}{4}{\left( {\pi - \arccos a} \right)^2} = \frac{{{\pi ^2}}}{{12}} - \frac{{{{\log }^2}\left( {2 - 2a} \right)}}{8} - \frac{{{{\left( {\pi - \arccos a} \right)}^2}}}{8}}$

Σχόλιο 1 : Χρησιμοποιήθηκε το Λήμμα $\displaystyle{2}$ από εδώ viewtopic.php?f=9&t=27979 με κάποια επεξεργασία και η ιδιότητα $\displaystyle{1}$ των διλογαριθμικών συναρτήσεων από εδώ viewtopic.php?f=9&t=12954&hilit=otto (μάλλον θα επανέλθω για διευκρινήσεις).

Σχόλιο 2 : Πράγματι .. όσο σάπιο γίνεται ..

mathematica.gr

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