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Гамма-функция
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Euler's reflection formula

\[\Gamma \left( z \right)\Gamma \left( {1 - z} \right) = \frac{\pi }{{\sin \left( {\pi z} \right)}},{\text{ }}z \notin \mathbb{Z}\]
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\[{\Gamma ^{\left( n \right)}}\left( x \right) = \int\limits_0^{ + \infty } {{t^{x - 1}}{e^{ - t}}{{\left( {\ln t} \right)}^n}dt} \]
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\[\begin{array}{l} \Gamma \left( {z + 1} \right) = z\Gamma \left( z \right) \hfill \\ \Gamma \left( z \right) = \int\limits_0^{ + \infty } {{e^{ - t}}{t^{z - 1}}dt} \hfill \\ {\text{Integration by parts}}: \hfill \\ \Gamma \left( {z + 1} \right) = \int\limits_0^{ + \infty } {{e^{ - t}}{t^z}dt} = - \int\limits_0^{ + \infty } {{t^z}d\left( {{e^{ - t}}} \right)} = \hfill \\ - \left( {\left. {\left( {{t^z}{e^{ - t}}} \right)} \right|_0^{ + \infty } - \int\limits_0^{ + \infty } {{e^{ - t}}d\left( {{t^z}} \right)} } \right) = z\int\limits_0^{ + \infty } {{e^{ - t}}{t^{z - 1}}dt} = z\Gamma \left( z \right) \hfill \\ \end{array} \]
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Разложение Куммера
$$\ln \Gamma \left( x \right) = \ln \sqrt {2\pi } + \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{2n}}\cos \left( {2\pi nx} \right)} + \sum\limits_{n = 1}^{ + \infty } {\frac{{\gamma + \ln \left( {2\pi n} \right)}}{{\pi n}}\sin \left( {2\pi nx} \right)} ,{\text{ }}0 < x < 1.$$
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$$\Gamma \left( x \right)\Gamma \left( {x + \frac{1}{2}} \right) = {2^{1 - 2x}}\sqrt \pi \Gamma \left( {2x} \right)$$
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$$\int\limits_0^{ + \infty } {{e^{ - st}}{t^{x - 1}}dt} = \frac{{\Gamma \left( x \right)}}{{{s^x}}}$$
\[\begin{array}{l} {\text{Докажите}}{\text{, что}} \hfill \\ \int\limits_0^{ + \infty } {\frac{1}{{1 + {x^{2n}}}}dx} = \frac{\pi }{{2n \cdot \sin \frac{\pi }{{2n}}}}. \hfill \\ n \in \mathbb{N} \hfill \\ \end{array}\]
\[\int\limits_0^{ + \infty } {{e^{ - {x^{\frac{1}{k}}}}}dx} = k \cdot \Gamma \left( k \right)\]
\[\begin{array}{l} N = \frac{{\sqrt 2 {\pi ^{\frac{3}{2}}}}}{{4\Gamma {{\left( {\frac{3}{4}} \right)}^2}}} = \frac{{\pi \cdot G}}{2} = \frac{1}{4}{\rm B}\left( {\frac{1}{4},\frac{1}{2}} \right). \hfill \\ \Gamma {\text{ - гамма - функция;}} \hfill \\ G{\text{ - постоянная Гаусса;}} \hfill \\ {\rm B}{\text{ - бета - функция}}{\text{.}} \hfill \\ {\text{Докажите}}{\text{, что }}1 + \frac{{1 + \frac{{1 + \frac{{1 + \frac{{1 + ...}}{{2 + 3/7}}}}{{2 + 3/5}}}}{{2 + 3/3}}}}{{2 + 3/1}} = N. \hfill \\ {a_k} = 2 + \frac{3}{{2k - 1}}. \hfill \\ \end{array}\]
\[{\text{Докажите}}{\text{, что }}\prod\limits_{k = 1}^{2n} {\Gamma \left( {\frac{k}{{2n + 1}}} \right)} = \frac{{{{\left( {2\pi } \right)}^n}}}{{\sqrt {2n + 1} }}.\]
\[\int\limits_0^1 {\Gamma \left( {2 - x} \right)\Gamma \left( {2 + x} \right)dx} = \frac{{21\zeta \left( 3 \right)}}{{2{\pi ^2}}}.\]
\[\int\limits_0^{\frac{1}{2}} {\frac{{\Gamma \left( {1 - x} \right)\Gamma \left( {1 + x} \right)}}{{\Gamma \left( {1 - \frac{x}{2}} \right)\Gamma \left( {1 + \frac{x}{2}} \right)}}dx} \]
\[{\text{Let }}a,b,c \in \mathbb{R},{\text{ }}0 < a < b.{\text{ Show that}}\] $$\int\limits_0^1 {{{\left( {{x^a} - {x^b}} \right)}^c}dx} = \frac{{\Gamma \left( {\frac{{ac + 1}}{{b - a}}} \right) \cdot \Gamma \left( {c + 1} \right)}}{{\Gamma \left( {\frac{{bc + 1}}{{b - a}}} \right) \cdot \left( {bc + 1} \right)}}.$$
\[\int\limits_0^1 {\ln \Gamma \left( x \right)dx} \]
$$\prod\limits_{k = 1}^{ + \infty } {\frac{{2k\left( {4k + 3} \right)}}{{\left( {2k + 1} \right)\left( {4k + 1} \right)}}} = \frac{1}{6}{\rm B}\left( {\frac{1}{2},\frac{1}{4}} \right)$$