Τριγωνομετρικό με λογάριθμο

Tolaso J Kos · #1

Έστω $\gamma$ η σταθερά των Euler - Mascheroni και $n \in \mathbb{N}$ . Δειχθήτω:
$\displaystyle{\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}{\ln{x}} \, {\rm d}x ={12\gamma^2-\pi^2\over 2(4n)^2}}$

#2

Tolaso J Kos έγραψε:Δειχθήτω: $\displaystyle{\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}{\ln{x}} \, {\rm d}x ={12\gamma^2-\pi^2\over 2(4n)^2}}$

Σχόλιο 1: Μετασχηματισμοί Laplace $\displaystyle{\int\limits_0^\infty {{y^a} \cdot {e^{ - x \cdot y}}dy} = \frac{{\Gamma \left( {1 + a} \right)}}{{{x^{1 + a}}}}}$ και $\displaystyle{\int\limits_0^\infty {\cos y \cdot {e^{ - x \cdot y}}dy} = \frac{x}{{1 + {x^2}}}}$ (θεωρούνται γνωστοί).

Σχόλιο 2: Από εδώ http://mathworld.wolfram.com/GammaFunction.html (σχέσεις 39 και 35) γνωρίζουμε $\displaystyle{\Gamma \left( {1 + a} \right) = a \cdot \Gamma \left( a \right)}$ καθώς και

$\displaystyle{\frac{1}{{\Gamma \left( {1 + 2 \cdot m \cdot z} \right)}} = 1 + 2m\gamma \cdot z + \frac{{{m^2}}}{3}\left( {6{\gamma ^2} - {\pi ^2}} \right){z^2} + \frac{{2{m^3}}}{3}\left( {2{\gamma ^3} - \gamma {\pi ^2} + 4\zeta \left( 3 \right)} \right){z^3} + ..}$

Σχόλιο 3: Επειδή $\displaystyle{\Gamma \left( z \right) \cdot \Gamma \left( {1 - z} \right) = \frac{\pi }{{\sin \pi z}}}$ (σχέση 42 παραπάνω) προκύπτει (με ανάλυση Taylor) ότι

$\displaystyle{\Gamma \left( {1 + m \cdot z} \right)\Gamma \left( {1 - m \cdot z} \right) = 1 + \frac{{{m^2}{\pi ^2}}}{6}{z^2} + \frac{{7{m^4}{\pi ^4}}}{{360}}{z^4} + ..}$ και με γινόμενο σειρών ότι

$\displaystyle{\frac{{\Gamma \left( {1 + m \cdot z} \right)\Gamma \left( {1 - m \cdot z} \right)}}{{\Gamma \left( {1 + 2 \cdot m \cdot z} \right)}} = 1 + 2m\gamma \cdot z + {m^2}\left( {2{\gamma ^2} - \frac{{{\pi ^2}}}{6}} \right){z^2} + {m^3}\left( {\frac{{4{\gamma ^3}}}{3} - \frac{{\gamma {\pi ^2}}}{3} + \frac{8}{3}\zeta \left( 3 \right)} \right){z^3} + ..}$

Στο θέμα μας. $\displaystyle{\int\limits_0^\infty {\frac{{\cos {x^n} - \cos {x^{2n}}}}{x}\log x\;dx} \mathop { = = = }\limits^{{x^n} \to x} \frac{1}{{{n^2}}}\int\limits_0^\infty {\frac{{\cos x - \cos {x^2}}}{x}\log x\;dx} }$

Όμως $\displaystyle{\int\limits_0^\infty {\frac{{1 - \cos x}}{{{x^{a + 1}}}}\;dx} = \frac{1}{{\Gamma (a + 1)}}\int\limits_0^\infty {\left( {1 - \cos x} \right)\frac{{\Gamma (a + 1)}}{{{x^{a + 1}}}}\;dx} = \frac{1}{{\Gamma (a + 1)}}\int\limits_0^\infty {\left( {1 - \cos x} \right)\left( {\int\limits_0^\infty {{y^a}{e^{ - xy}}dy} } \right)\;dx} = }$

$\displaystyle{ = \frac{1}{{\Gamma (a + 1)}}\int\limits_0^\infty {{y^a}\left( {\int\limits_0^\infty {\left( {1 - \cos x} \right){e^{ - xy}}dx} } \right)\;dy} = \frac{1}{{\Gamma (a + 1)}}\int\limits_0^\infty {{y^a}\left( {\frac{1}{y} - \frac{y}{{1 + {y^2}}}} \right)\;dy} }$ $\displaystyle{ = \frac{1}{{\Gamma (a + 1)}}\int\limits_0^\infty {\frac{{{y^{a - 1}}}}{{1 + {y^2}}}\;dy} \mathop { = = = }\limits^{{y^2} = x} }$

$\displaystyle{ = \frac{1}{{2\Gamma (a + 1)}}\int\limits_0^\infty {\frac{{{x^{\frac{a}{2} - 1}}}}{{1 + x}}\;dx} = \frac{1}{{2\Gamma (a + 1)}}\int\limits_0^\infty {{x^{\frac{a}{2} - 1}}\int\limits_0^\infty {{e^{ - xw}}{e^{ - w}}dw} \;dx} = }$ $\displaystyle{\frac{1}{{2\Gamma (a + 1)}}\int\limits_0^\infty {{e^{ - w}}\int\limits_0^\infty {{x^{\frac{a}{2} - 1}}{e^{ - xw}}dx} \;dw} = }$

$\displaystyle{ = \frac{{\Gamma \left( {\frac{a}{2}} \right)}}{{2\Gamma (a + 1)}}\int\limits_0^\infty {{e^{ - w}}\frac{1}{{{w^{\frac{a}{2}}}}}\;dw = } \frac{{\Gamma \left( {\frac{a}{2}} \right)\Gamma \left( {1 - \frac{a}{2}} \right)}}{{2\Gamma (a + 1)}} \Rightarrow \int\limits_0^\infty {\frac{{1 - \cos x}}{{{x^{2a + 1}}}}\;dx} = \frac{{\Gamma \left( a \right)\Gamma \left( {1 - a} \right)}}{{2\Gamma \left( {2a + 1} \right)}} = \frac{{\Gamma \left( {1 + a} \right)\Gamma \left( {1 - a} \right)}}{{2a \cdot \Gamma \left( {2a + 1} \right)}}}$

Δηλαδή $\displaystyle{\int\limits_0^\infty {\frac{{1 - \cos x}}{{{x^{2a + 1}}}}\;dx} = \frac{1}{{2a}} + \gamma + \left( {{\gamma ^2} - \frac{{{\pi ^2}}}{{12}}} \right)a + \left( {\frac{{2{\gamma ^3}}}{3} - \frac{{\gamma {\pi ^2}}}{6} + \frac{4}{3}\zeta \left( 3 \right)} \right){a^2} + ..}$

Παραγωγίζοντας ως προς $\displaystyle{a}$ προκύπτει $\displaystyle{{\int\limits_0^\infty {\frac{{1 - \cos x}}{{{x^{2a + 1}}}}\log x\;dx} = \frac{1}{{4{a^2}}} - \frac{1}{2}\left( {{\gamma ^2} - \frac{{{\pi ^2}}}{{12}}} \right) - \left( {\frac{{2{\gamma ^3}}}{3} - \frac{{\gamma {\pi ^2}}}{6} + \frac{4}{3}\zeta \left( 3 \right)} \right)a + ..}}$

Επίσης $\displaystyle{\int\limits_0^\infty {\frac{{1 - \cos {x^2}}}{{{x^{2a + 1}}}}\log x\;dx} \mathop { = = = }\limits^{{x^2} = y} \frac{1}{4}\int\limits_0^\infty {\frac{{1 - \cos y}}{{{y^{a + 1}}}}\log y\;dx} = \frac{1}{{4{a^2}}} - \frac{1}{8}\left( {{\gamma ^2} - \frac{{{\pi ^2}}}{{12}}} \right)}$ $\displaystyle{ - \left( {\frac{{2{\gamma ^3}}}{3} - \frac{{\gamma {\pi ^2}}}{6} + \frac{4}{3}\zeta \left( 3 \right)} \right)\frac{a}{8} + ..}$

Αφαιρώντας $\displaystyle{\int\limits_0^\infty {\frac{{\cos x - \cos {x^2}}}{{{x^{2a + 1}}}}\log x\;dx} = \frac{3}{8}\left( {{\gamma ^2} - \frac{{{\pi ^2}}}{{12}}} \right) + \frac{7}{8}\left( {\frac{{2{\gamma ^3}}}{3} - \frac{{\gamma {\pi ^2}}}{6} + \frac{4}{3}\zeta \left( 3 \right)} \right)a + ..}$

Παίρνοντας όριο $\displaystyle{a \to 0}$ προκύπτει $\displaystyle{\int\limits_0^\infty {\frac{{\cos x - \cos {x^2}}}{x}\log x\;dx} = \frac{{12{\gamma ^2} - {\pi ^2}}}{{32}}}$ και τελικά $\displaystyle{\int\limits_0^\infty {\frac{{\cos {x^n} - \cos {x^{2n}}}}{x}\log x\;dx} = \frac{{12{\gamma ^2} - {\pi ^2}}}{{32 \cdot {n^2}}}}$

Ενδιαφέρον παρουσιάζει το όριο $\displaystyle{\mathop {\lim }\limits_{a \to 0} \dfrac{{\int\limits_0^\infty {\dfrac{{\cos x - \cos {x^2}}}{{{x^{2a + 1}}}}\log x\;dx} - \dfrac{{12{\gamma ^2} - {\pi ^2}}}{{32}}}}{a} = \frac{{7\left( {4{\gamma ^3} - \gamma {\pi ^2} + 8\zeta \left( 3 \right)} \right)}}{{48}}}$

Τριγωνομετρικό με λογάριθμο

Τριγωνομετρικό με λογάριθμο

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