Τριγωνομετρικό ολοκλήρωμα με εκθετική

Tolaso J Kos · #1

Έστω $a \in \mathbb{R}$ . Υπολογισθήτω:
$\displaystyle{\mathcal{J} = \int_0^\infty \frac{\sin^2 ax}{x(1-e^x)} \, {\rm d}x}$

#2

Tolaso J Kos έγραψε: Δευ Δεκ 25, 2017 10:27 am Έστω $a \in \mathbb{R}$ . Υπολογισθήτω: $\displaystyle{\mathcal{J} = \int_0^\infty \frac{\sin^2 ax}{x(1-e^x)} \, {\rm d}x}$

$\displaystyle \int\limits_0^\infty {\frac{{{{\sin }^2}ax}}{{x\left( {1 - {e^x}} \right)}}dx} = - \frac{1}{2}\int\limits_0^\infty {\frac{{1 - \cos 2ax}}{{x\left( {{e^x} - 1} \right)}}dx} = - \frac{1}{2}\int\limits_0^\infty {\frac{{\left( {1 - \cos 2ax} \right){e^{ - x}}}}{{x\left( {1 - {e^{ - x}}} \right)}}dx} = - \frac{1}{2}\int\limits_0^\infty {\frac{{\left( {1 - \cos 2ax} \right){e^{ - x}}}}{x}\sum\limits_{n = 0}^\infty {{e^{ - nx}}} dx} =$

$\displaystyle = - \frac{1}{2}\sum\limits_{n = 0}^\infty {\int\limits_0^\infty {\frac{{\left( {1 - \cos 2ax} \right){e^{ - \left( {n + 1} \right)x}}}}{x}dx} } = - \frac{1}{2}\sum\limits_{n = 0}^\infty {\int\limits_0^\infty {\left( {1 - \cos 2ax} \right){e^{ - \left( {n + 1} \right)x}}\left( {\int\limits_0^\infty {{e^{ - yx}}dy} } \right)dx} } =$

$\displaystyle = - \frac{1}{2}\sum\limits_{n = 0}^\infty {\int\limits_0^\infty {\left( {\int\limits_0^\infty {\left( {1 - \cos 2ax} \right){e^{ - \left( {n + 1 + y} \right)x}}dx} } \right)dy} } = - \frac{1}{2}\sum\limits_{n = 0}^\infty {\int\limits_0^\infty {\left( {\frac{1}{{n + 1 + y}} - \frac{{n + 1 + y}}{{4{a^2} + {{\left( {n + 1 + y} \right)}^2}}}} \right)dy} } =$

$\displaystyle = - \frac{1}{4}\sum\limits_{n = 0}^\infty {\left[ {\log \frac{{{{\left( {n + 1 + y} \right)}^2}}}{{4{a^2} + {{\left( {n + 1 + y} \right)}^2}}}} \right]_0^\infty } = - \frac{1}{4}\sum\limits_{n = 1}^\infty {\log \frac{{{n^2}}}{{4{a^2} + {n^2}}}} = - \frac{1}{4}\log \mathop {\lim }\limits_{{\rm N} \to \infty } \frac{{{{\left( {{\rm N}!} \right)}^2}}}{{\prod\limits_{n = 1}^N {\left( { - 2ia + n} \right)} \prod\limits_{n = 1}^N {\left( {2ia + n} \right)} }} =$

$\displaystyle = - \frac{1}{4}\log \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{ - 2ia}} \cdot n! \cdot \left( { - 2ia} \right)}}{{\left( { - 2ia} \right)\left( { - 2ia + 1} \right)\left( { - 2ia + 2} \right)..\left( { - 2ia + n} \right)}} \cdot \frac{{{{\left( {n + 1} \right)}^{2ia}} \cdot n!\left( {2ia} \right)}}{{\left( {2ia} \right)\left( {2ia + 1} \right)\left( {2ia + 2} \right)..\left( {2ia + n} \right)}}} \right) =$

$\displaystyle = - \frac{1}{4}\log \left( { - 4 \cdot {i^2}{a^2}\Gamma \left( {2ia} \right)\Gamma \left( { - 2ia} \right)} \right) = - \frac{1}{4}\log \left( {4{a^2}\frac{{\Gamma \left( {2ia} \right)\Gamma \left( {1 - 2ia} \right)}}{{ - 2ia}}} \right) = - \frac{1}{4}\log \left( {4{a^2}\frac{\pi }{{ - 2ia \cdot \sin \left( {2\pi ia} \right)}}} \right) =$

$\displaystyle = - \frac{1}{4}\log \left( {2a\frac{\pi }{{ - i \cdot \dfrac{{{e^{ - 2\pi a}} - {e^{2\pi a}}}}{{2i}}}}} \right) \Rightarrow \int\limits_0^\infty {\frac{{{{\sin }^2}ax}}{{x\left( {1 - {e^x}} \right)}}dx} = \log \sqrt[{^4}]{{\frac{{{e^{2\pi a}} - {e^{ - 2\pi a}}}}{{4a\pi }}}}$

Λόγω του τετραγώνου, μπορούμε να θεωρήσουμε $\displaystyle a \ge 0$

Χρησιμοποιήθηκαν οι τύποι $\displaystyle \int\limits_0^\infty {{e^{ - ax}}dx} = \frac{1}{a}$ και $\displaystyle \int\limits_0^\infty {\cos bx \cdot {e^{ - ax}}dx} = \frac{a}{{{b^2} + {a^2}}}$ (στοιχειώδεις παραγοντικές ολοκληρώσεις)
καθώς και $\displaystyle \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^z} \cdot n!}}{{z\left( {z + 1} \right)\left( {z + 2} \right)..\left( {z + n} \right)}}} \right) = \Gamma \left( z \right)$ , $\displaystyle \Gamma \left( {1 + z} \right) = z \cdot \Gamma \left( z \right)$ και $\displaystyle \Gamma \left( z \right) \cdot \Gamma \left( {1 - z} \right) = \frac{\pi }{{\sin \left( {\pi \cdot z} \right)}}$ από εδώ http://mathworld.wolfram.com/GammaFunction.html

Τριγωνομετρικό ολοκλήρωμα με εκθετική

Τριγωνομετρικό ολοκλήρωμα με εκθετική

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