Ολοκλήρωμα 32

china university · #1

For

, let (Για p>0

, έστω)

$g(x)=\begin{cases} p\left[\dfrac{x}{p}\right]+\dfrac{p}{2}& x\ge 0,\\ -g(-x)& x<0 \end{cases}$

Try to prove that for all $x \in \mathbb{R}$ (Προσπαθήστε να δείξετε ότι για κάθε $x \in \mathbb{R}$ )

$\dfrac{p}{2\pi}\displaystyle\int_{-\infty}^{+\infty}\displaystyle\sum_{n=-[x/p]}^{[x/p]}\dfrac{\sin{\left(n+\dfrac{1}{2}\right)pt}}{\sin{\dfrac{1}{2}pt}}\cdot\dfrac{\sin{xt}}{t}dt=\dfrac{1}{2}[g(x^{+})+g(x^{-})]$ .

I can usually solve similar integrals, but I cannot find the solution to this one. Your help would be appreciated. Thank you.
(Συνήθως μπορώ να υπολογίζω τέτοιου τύπου ολοκληρώματα αλλά δεν μπορώ να βρω τη λύση σε αυτό εδώ. Θα εκτιμούσα τη βοήθειά σας. Ευχαριστώ.)

#2

Θεωρούμε ότι στο θέμα εννοείται ότι : $\displaystyle{g\left( {{x^ + }} \right) = \mathop {\lim }\limits_{y \to {x^ + }} g\left( y \right)}$ και $\displaystyle{g\left( {{x^ - }} \right) = \mathop {\lim }\limits_{y \to {x^ - }} g\left( y \right)}$

Αν $\displaystyle{x < 0}$ θα ισχύει $\displaystyle{\left[ {\frac{x}{p}} \right] < 0}$ , οπότε το άθροισμα $\displaystyle{\sum\limits_{n = - \left[ {x/p} \right]}^{\left[ {x/p} \right]} {\sin \left( {n + \frac{1}{2}} \right)pt} }$ δεν έχει νόημα. Γι αυτό θεωρούμε ότι $\displaystyle{x \geqslant 0}$

Αν $\displaystyle{m \in \mathbb{N}}$ τότε $\displaystyle{\mathop {\lim }\limits_{x \to m \cdot {p^ + }} g\left( x \right) = \frac{p}{2} + m \cdot p}$ και $\displaystyle{\mathop {\lim }\limits_{x \to m \cdot {p^ - }} g\left( x \right) = \frac{p}{2} + \left( {m - 1} \right) \cdot p}$ , οπότε $\displaystyle{\frac{1}{2}\left( {g\left( {m \cdot {p^ + }} \right) + g\left( {m \cdot {p^ - }} \right)} \right) = \frac{p}{2} + \left( {\left[ {\frac{x}{p}} \right] - \frac{1}{2}} \right)p = \left[ {\frac{x}{p}} \right]p}$

Αν $\displaystyle{x \ne m \cdot p}$ τότε $\displaystyle{\frac{1}{2}\left( {g\left( {{x^ + }} \right) + g\left( {{x^ - }} \right)} \right) = \frac{p}{2} + \left[ {\frac{x}{p}} \right]p}$ (Μόνον οι τιμές $\displaystyle{x = m \cdot p}$ είναι θέσεις ασυνέχειας της συνάρτησης $\displaystyle{g\left( x \right)}$ )

$\displaystyle{I = \frac{p}{{2\pi }}\int\limits_{ - \infty }^{ + \infty } {\sum\limits_{n = - \left[ {x/p} \right]}^{\left[ {x/p} \right]} {\frac{{\sin \left( {n + \frac{1}{2}} \right)pt}}{{\sin \frac{1}{2}pt}} \cdot \frac{{\sin xt}}{t}\,} dt} = \frac{p}{\pi }\int\limits_0^\infty {\frac{{\sin \left( {\left[ {\frac{x}{p}} \right] + \frac{1}{2}} \right)pt}}{{\sin \frac{1}{2}pt}} \cdot \frac{{\sin xt}}{t}dt} = \mathop = \limits^{L:1} = \frac{p}{\pi }\int\limits_0^\infty {\left( { - 1 + 2\sum\limits_{n = 0}^{\left[ {x/p} \right]} {\cos npt} } \right) \cdot \frac{{\sin xt}}{t}dt} = }$

$\displaystyle{ = - \frac{p}{\pi }\int\limits_0^\infty {\frac{{\sin xt}}{t}dt} + \frac{{2p}}{\pi }\sum\limits_{n = 0}^{\left[ {x/p} \right]} {\int\limits_0^\infty {\cos npt \cdot \frac{{\sin xt}}{t}dt} } = \mathop = \limits^{L:2} = - \frac{p}{\pi } \cdot \frac{\pi }{2} + \frac{{2p}}{\pi }\sum\limits_{n = 0}^{\left[ {x/p} \right]} {\int\limits_0^\infty {\cos npt \cdot \frac{{\sin xt}}{t}dt} } = \mathop = \limits^{xt = y} = }$

$\displaystyle{ = - \frac{p}{2} + \frac{{2p}}{\pi }\sum\limits_{n = 0}^{\left[ {x/p} \right]} {\int\limits_0^\infty {\cos \frac{{np}}{x}y \cdot \frac{{\sin y}}{y}dy} } = - \frac{p}{2} + \frac{{2p}}{\pi }\sum\limits_{n = 0}^{\left[ {x/p} \right] - 1} {\int\limits_0^\infty {\cos \frac{{np}}{x}y \cdot \frac{{\sin y}}{y}dy} } + \frac{{2p}}{\pi }\int\limits_0^\infty {\cos \left[ {\frac{x}{p}} \right]\frac{p}{x}y \cdot \frac{{\sin y}}{y}dy} = \mathop = \limits^{L:3} = }$

$\displaystyle{ = - \frac{p}{2} + \frac{{2p}}{\pi }\left[ {\frac{x}{p}} \right]\frac{\pi }{2} + \frac{{2p}}{\pi }\int\limits_0^\infty {\cos \left[ {\frac{x}{p}} \right]\frac{p}{x}y \cdot \frac{{\sin y}}{y}dy} = - \frac{p}{2} + p\left[ {\frac{x}{p}} \right] + \frac{{2p}}{\pi }\int\limits_0^\infty {\cos \left[ {\frac{x}{p}} \right]\frac{p}{x}y \cdot \frac{{\sin y}}{y}dy} }$

Αν $\displaystyle{x \ne m \cdot p}$ τότε $\displaystyle{\left[ {\frac{x}{p}} \right]\frac{p}{x} < 1}$ , οπότε $\displaystyle{I = - \frac{p}{2} + p\left[ {\frac{x}{p}} \right] + \frac{{2p}}{\pi } \cdot \frac{\pi }{2} = p\left[ {\frac{x}{p}} \right] + \frac{p}{2} = \frac{1}{2}\left( {g\left( {{x^ + }} \right) + g\left( {{x^ - }} \right)} \right)}$

Αν $\displaystyle{x = m \cdot p}$ τότε $\displaystyle{\left[ {\frac{x}{p}} \right]\frac{p}{x} = 1}$ , οπότε $\displaystyle{I = - \frac{p}{2} + p\left[ {\frac{x}{p}} \right] + \frac{{2p}}{\pi } \cdot \frac{\pi }{4} = p\left[ {\frac{x}{p}} \right] = \frac{1}{2}\left( {g\left( {{x^ + }} \right) + g\left( {{x^ - }} \right)} \right)}$
Τελικά για $\displaystyle{x \geqslant 0}$ ισχύει $\displaystyle{\frac{p}{{2\pi }}\int\limits_{ - \infty }^{ + \infty } {\sum\limits_{n = - \left[ {x/p} \right]}^{\left[ {x/p} \right]} {\frac{{\sin \left( {n + \frac{1}{2}} \right)pt}}{{\sin \frac{1}{2}pt}} \cdot \frac{{\sin xt}}{t}\,} dt} = \frac{1}{2}\left( {g\left( {{x^ + }} \right) + g\left( {{x^ - }} \right)} \right)}$ (Για $\displaystyle{x = 0}$ η ισότητα ισχύει με τετριμμένο τρόπο)

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Λήμμα 1 $\displaystyle{L:1\quad \frac{{\sin \left( {N + \frac{1}{2}} \right)x}}{{\sin \frac{1}{2}x}} = - 1 + 2\sum\limits_{n = 0}^N {\cos nx} }$ διότι

$\displaystyle{\sum\limits_{n = 0}^N {\cos nx} = \frac{1}{2}\sum\limits_{n = 0}^N {\left( {{e^{inx}} + {e^{ - inx}}} \right)} = \frac{1}{2}\left( {\sum\limits_{n = 0}^N {{e^{inx}}} + \sum\limits_{n = 0}^N {{e^{ - inx}}} } \right) = \frac{1}{2}\left( {\frac{{1 - {e^{i\left( {N + 1} \right)x}}}}{{1 - {e^{ix}}}} + \frac{{1 - {e^{ - i\left( {N + 1} \right)x}}}}{{1 - {e^{ - ix}}}}} \right) = }$

$\displaystyle{ = \frac{1}{2} \cdot \frac{{\left( {1 - {e^{i\left( {N + 1} \right)x}}} \right)\left( {1 - {e^{ - ix}}} \right) + \left( {1 - {e^{ - i\left( {N + 1} \right)x}}} \right)\left( {1 - {e^{ix}}} \right)}}{{\left( {1 - {e^{ix}}} \right)\left( {1 - {e^{ - ix}}} \right)}} = \frac{1}{2} \cdot \frac{{2 - 2\cos x - 2\cos \left( {N + 1} \right)x + + 2\cos Nx}}{{2 - 2\cos x}}}$

$\displaystyle{ = \frac{1}{2}\left( {1 + \frac{{2\sin \left( {N + \frac{1}{2}} \right)x \cdot \sin \frac{1}{2}x}}{{2{{\sin }^2}\frac{1}{2}x}}} \right) = \frac{1}{2}\left( {1 + \frac{{\sin \left( {N + \frac{1}{2}} \right)x}}{{\sin \frac{1}{2}x}}} \right) \Rightarrow \frac{{\sin \left( {N + \frac{1}{2}} \right)x}}{{\sin \frac{1}{2}x}} = - 1 + 2\sum\limits_{n = 0}^N {\cos nx} }$

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Λήμμα 2 $\displaystyle{L:2\quad }$ Για $\displaystyle{x > 0}$ τότε $\displaystyle{\int\limits_0^\infty {\frac{{\sin xt}}{t}dt} = \frac{\pi }{2}}$ διότι $\displaystyle{\int\limits_0^\infty {\frac{{\sin xt}}{t}dt} = \mathop = \limits^{xt = y} = \int\limits_0^\infty {\frac{{\sin y}}{y}dy} = \frac{\pi }{2}}$ (θεωρείται γνωστό)

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Λήμμα 3 $\displaystyle{L:3\quad \int\limits_0^\infty {\frac{{\cos at \cdot \sin t}}{t}dt} = \left\{ {\begin{array}{*{20}{c}} {0\;,\;\left| a \right| > 1} \\ {\frac{\pi }{2}\;,\;\left| a \right| < 1} \\ {\frac{\pi }{4}\;,\;\left| a \right| = 1} \\ \end{array} } \right\}}$ διότι

$\displaystyle{\int\limits_0^\infty {\frac{{\cos at \cdot \sin t}}{t}dt} = \int\limits_0^\infty {\frac{{\cos \left| a \right|t \cdot \sin t}}{t}dt} = \frac{1}{2}\int\limits_0^\infty {\frac{{\sin \left( {1 + \left| a \right|} \right)t + \sin \left( {1 - \left| a \right|} \right)t}}{t}dt} = \frac{1}{2}\left( {\int\limits_0^\infty {\frac{{\sin \left( {1 + \left| a \right|} \right)t}}{t}dt} + \int\limits_0^\infty {\frac{{\sin \left( {1 - \left| a \right|} \right)t}}{t}dt} } \right) = }$

$\displaystyle{ = \frac{\pi }{4} + \frac{1}{2}\int\limits_0^\infty {\frac{{\sin \left( {1 - \left| a \right|} \right)t}}{t}dt} = \left\{ {\begin{array}{*{20}{c}} {0\;,\;\left| a \right| > 1} \\ {\frac{\pi }{2}\;,\;\left| a \right| < 1} \\ {\frac{\pi }{4}\;,\;\left| a \right| = 1} \\ \end{array} } \right\}}$

Ολοκλήρωμα 32

Ολοκλήρωμα 32

Re: Ολοκλήρωμα 32

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