Zurück zu Unendliche Reihen
Aus der Formel ∫ 0 ∞ e − s x x a 2 + x 2 d x = sin ( a s ) ( π 2 − Si ( a s ) ) − cos ( a s ) Ci ( a s ) {\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)-\cos(as)\,{\text{Ci}}(as)} folgt ∫ 0 ∞ e − n π x x 1 + x 2 d x = ( − 1 ) n − 1 Ci ( n π ) {\displaystyle \int _{0}^{\infty }e^{-n\pi x}\,{\frac {x}{1+x^{2}}}\,dx=(-1)^{n-1}\,{\text{Ci}}(n\pi )} . Also ist ∑ n = 1 ∞ Ci ( n π ) = ∫ 0 ∞ ∑ n = 1 ∞ ( − 1 ) n − 1 e − n π x x 1 + x 2 d x = ∫ 0 ∞ 1 e π x + 1 x 1 + x 2 d x {\displaystyle \sum _{n=1}^{\infty }{\text{Ci}}(n\pi )=\int _{0}^{\infty }\sum _{n=1}^{\infty }(-1)^{n-1}\,e^{-n\pi x}\,{\frac {x}{1+x^{2}}}\,dx=\int _{0}^{\infty }{\frac {1}{e^{\pi x}+1}}\,{\frac {x}{1+x^{2}}}\,dx} = ∫ 0 ∞ 1 e π x + 1 ∫ 0 ∞ sin x t e − t d t d x = ∫ 0 ∞ ∫ 0 ∞ sin x t e π x + 1 d x e − t d t = 1 2 ∫ 0 ∞ ( 1 t − 1 sinh t ) e − t d t {\displaystyle =\int _{0}^{\infty }{\frac {1}{e^{\pi x}+1}}\,\int _{0}^{\infty }\sin xt\,e^{-t}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }{\frac {\sin xt}{e^{\pi x}+1}}\,dx\,e^{-t}\,dt={\frac {1}{2}}\int _{0}^{\infty }\left({\frac {1}{t}}-{\frac {1}{\sinh t}}\right)\,e^{-t}\,dt} . ⇒ 2 ∑ n = 1 ∞ Ci ( n π ) = ∫ 0 ∞ ( 1 t − 2 e − 2 t 1 − e − 2 t ) d t = − ∫ 0 1 ( 1 log u + 2 u 1 − u 2 ) d u {\displaystyle \Rightarrow \,2\sum _{n=1}^{\infty }{\text{Ci}}(n\pi )=\int _{0}^{\infty }\left({\frac {1}{t}}-{\frac {2e^{-2t}}{1-e^{-2t}}}\right)dt=-\int _{0}^{1}\left({\frac {1}{\log u}}+{\frac {2u}{1-u^{2}}}\right)du} = − ∫ 0 1 ( 1 log u + 1 1 − u ) d u + ∫ 0 1 d u 1 + u = − γ + log 2 {\displaystyle =-\int _{0}^{1}\left({\frac {1}{\log u}}+{\frac {1}{1-u}}\right)du+\int _{0}^{1}{\frac {du}{1+u}}=-\gamma +\log 2}
Aus der Formel ∫ 0 ∞ e − s x a a 2 + x 2 d x = sin ( a s ) Ci ( a s ) − cos ( a s ) ( Si ( a s ) − π 2 ) {\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\sin(as)\,{\text{Ci}}(as)-\cos(as)\,\left({\text{Si}}(as)-{\frac {\pi }{2}}\right)} folgt ∫ 0 ∞ e − ( n + 1 2 ) π x 1 1 + x 2 d x = ( − 1 ) n Ci ( ( n + 1 2 ) π ) {\displaystyle \int _{0}^{\infty }e^{-\left(n+{\frac {1}{2}}\right)\pi x}\,{\frac {1}{1+x^{2}}}\,dx=(-1)^{n}\,{\text{Ci}}\left(\left(n+{\frac {1}{2}}\right)\pi \right)} . Also ist ∑ n = 0 ∞ Ci ( ( n + 1 2 ) π ) = ∫ 0 ∞ ∑ n = 0 ∞ ( − 1 ) n e − ( n + 1 2 ) π x 1 1 + x 2 d x {\displaystyle \sum _{n=0}^{\infty }{\text{Ci}}\left(\left(n+{\frac {1}{2}}\right)\pi \right)=\int _{0}^{\infty }\sum _{n=0}^{\infty }(-1)^{n}\,e^{-\left(n+{\frac {1}{2}}\right)\pi x}\,{\frac {1}{1+x^{2}}}\,dx} = ∫ 0 ∞ e − π x 2 1 + e − π x 1 1 + x 2 d x = 1 2 ∫ 0 ∞ 1 cosh π x 2 d x 1 + x 2 = 1 4 [ ψ ( 1 4 + 3 4 ) − ψ ( 1 4 + 1 4 ) ] = log 2 2 {\displaystyle =\int _{0}^{\infty }{\frac {e^{-{\frac {\pi x}{2}}}}{1+e^{-\pi x}}}\,{\frac {1}{1+x^{2}}}\,dx={\frac {1}{2}}\int _{0}^{\infty }{\frac {1}{\cosh {\frac {\pi x}{2}}}}\,{\frac {dx}{1+x^{2}}}={\frac {1}{4}}\left[\psi \left({\frac {1}{4}}+{\frac {3}{4}}\right)-\psi \left({\frac {1}{4}}+{\frac {1}{4}}\right)\right]={\frac {\log 2}{2}}} .
Aus der Formel ∫ 0 ∞ e − s x x a 2 + x 2 d x = sin ( a s ) ( π 2 − Si ( a s ) ) − cos ( a s ) Ci ( a s ) {\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)-\cos(as)\,{\text{Ci}}(as)} folgt ∫ 0 ∞ e − ( 2 n + 1 ) π x x 1 + x 2 d x = Ci ( ( 2 n + 1 ) π ) {\displaystyle \int _{0}^{\infty }e^{-(2n+1)\pi x}\,{\frac {x}{1+x^{2}}}\,dx={\text{Ci}}((2n+1)\pi )} . Also ist ∑ n = 0 ∞ Ci ( ( 2 n + 1 ) π ) = ∫ 0 ∞ ∑ n = 0 ∞ e − ( 2 n + 1 ) π x x 1 + x 2 d x = ∫ 0 ∞ e − π x 1 − e − 2 π x x 1 + x 2 d x {\displaystyle \sum _{n=0}^{\infty }{\text{Ci}}((2n+1)\pi )=\int _{0}^{\infty }\sum _{n=0}^{\infty }e^{-(2n+1)\pi x}\,{\frac {x}{1+x^{2}}}\,dx=\int _{0}^{\infty }{\frac {e^{-\pi x}}{1-e^{-2\pi x}}}\,{\frac {x}{1+x^{2}}}\,dx} = 1 2 ∫ 0 ∞ 1 sinh π x x x 2 + 1 d x = 1 2 ( log 2 − 1 2 ) {\displaystyle ={\frac {1}{2}}\int _{0}^{\infty }{\frac {1}{\sinh \pi x}}\,{\frac {x}{x^{2}+1}}\,dx={\frac {1}{2}}\,\left(\log 2-{\frac {1}{2}}\right)} (Abels Integral)
Setze a k = 1 k π ∫ 2 π k x ∞ sin t d t t = ∫ x ∞ sin 2 π k t k π d t t {\displaystyle a_{k}={\frac {1}{k\pi }}\int _{2\pi kx}^{\infty }\sin t\,\,{\frac {dt}{t}}=\int _{x}^{\infty }{\frac {\sin 2\pi kt}{k\pi }}\,{\frac {dt}{t}}} . Wir wollen uns folgende Formel zu nutze machen: ∑ k = 1 ∞ sin 2 π k t k π = ⌊ x ⌋ + n + 1 2 − t {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin 2\pi kt}{k\pi }}=\lfloor x\rfloor +n+{\frac {1}{2}}-t} für ⌊ x ⌋ + n < t < ⌊ x ⌋ + n + 1 {\displaystyle \lfloor x\rfloor +n<t<\lfloor x\rfloor +n+1} Hierzu führen wir folgende Zerlegung durch: ∫ x ∞ = ∫ x ⌊ x ⌋ + 1 + ∑ n = 1 ∞ ∫ ⌊ x ⌋ + n ⌊ x ⌋ + n + 1 {\displaystyle \int _{x}^{\infty }=\int _{x}^{\lfloor x\rfloor +1}+\sum _{n=1}^{\infty }\int _{\lfloor x\rfloor +n}^{\lfloor x\rfloor +n+1}} Nun ist ∑ k = 1 m − 1 a k = ∫ x ⌊ x ⌋ + 1 ( ⌊ x ⌋ + 1 2 − t ) d t t + ∑ n = 1 m − 1 ∫ ⌊ x ⌋ + n ⌊ x ⌋ + n + 1 ( ⌊ x ⌋ + n + 1 2 − t ) d t t {\displaystyle \sum _{k=1}^{m-1}a_{k}=\int _{x}^{\lfloor x\rfloor +1}\left(\lfloor x\rfloor +{\frac {1}{2}}-t\right){\frac {dt}{t}}+\sum _{n=1}^{m-1}\int _{\lfloor x\rfloor +n}^{\lfloor x\rfloor +n+1}\left(\lfloor x\rfloor +n+{\frac {1}{2}}-t\right){\frac {dt}{t}}} . Das erste Integral ist ( ⌊ x ⌋ + 1 2 ) log ( ⌊ x ⌋ + 1 ) − ( ⌊ x ⌋ + 1 2 ) log x − ( ⌊ x ⌋ + 1 − x ) {\displaystyle \left(\lfloor x\rfloor +{\frac {1}{2}}\right)\log \left(\lfloor x\rfloor +1\right)-\left(\lfloor x\rfloor +{\frac {1}{2}}\right)\log x-\left(\lfloor x\rfloor +1-x\right)} und ∫ ⌊ x ⌋ + n ⌊ x ⌋ + n + 1 ( ⌊ x ⌋ + n + 1 2 − t ) d t t = ( ⌊ x ⌋ + n + 1 2 ) log ( ⌊ x ⌋ + n + 1 ) − ( ⌊ x ⌋ + n − 1 2 ) log ( ⌊ x ⌋ + n ) − log ( ⌊ x ⌋ + n ) − 1 {\displaystyle \int _{\lfloor x\rfloor +n}^{\lfloor x\rfloor +n+1}\left(\lfloor x\rfloor +n+{\frac {1}{2}}-t\right){\frac {dt}{t}}=\left(\lfloor x\rfloor +n+{\frac {1}{2}}\right)\log \left(\lfloor x\rfloor +n+1\right)-\left(\lfloor x\rfloor +n-{\frac {1}{2}}\right)\log \left(\lfloor x\rfloor +n\right)-\log \left(\lfloor x\rfloor +n\right)-1} . Somit ist die letzte Reihe ( ⌊ x ⌋ + m − 1 2 ) log ( ⌊ x ⌋ + m ) − ( ⌊ x ⌋ + 1 2 ) log ( ⌊ x ⌋ + 1 ) − log ( ⌊ x ⌋ + m − 1 ) ! ⌊ x ⌋ ! − ( m − 1 ) {\displaystyle \left(\lfloor x\rfloor +m-{\frac {1}{2}}\right)\log(\lfloor x\rfloor +m)-\left(\lfloor x\rfloor +{\frac {1}{2}}\right)\log \left(\lfloor x\rfloor +1\right)-\log {\frac {(\lfloor x\rfloor +m-1)!}{\lfloor x\rfloor !}}-(m-1)} . ∑ k = 1 m − 1 a k = ( ⌊ x ⌋ + m − 1 2 ) log ( ⌊ x ⌋ + m ) − ( ⌊ x ⌋ + 1 2 ) log x − ( ⌊ x ⌋ − x ) − log ( ⌊ x ⌋ + m − 1 ) ! ⌊ x ⌋ ! − m {\displaystyle \sum _{k=1}^{m-1}a_{k}=\left(\lfloor x\rfloor +m-{\frac {1}{2}}\right)\log \left(\lfloor x\rfloor +m\right)-\left(\lfloor x\rfloor +{\frac {1}{2}}\right)\log x-\left(\lfloor x\rfloor -x\right)-\log {\frac {(\lfloor x\rfloor +m-1)!}{\lfloor x\rfloor !}}-m} → m → ∞ Stirling x − ( ⌊ x ⌋ + 1 2 ) log x + log ( ⌊ x ⌋ ! ) − log 2 π {\displaystyle {\xrightarrow[{m\to \infty }]{\text{Stirling}}}\;\,x-\left(\lfloor x\rfloor +{\frac {1}{2}}\right)\log x+\log(\lfloor x\rfloor !)-\log {\sqrt {2\pi }}}
4 π ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 cos ( ( 2 k + 1 ) π t ) = ( − 1 ) ⌊ t + 1 2 ⌋ {\displaystyle {\frac {4}{\pi }}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\cos {\Big (}(2k+1)\pi t{\Big )}=(-1)^{\lfloor t+{\frac {1}{2}}\rfloor }} für alle reellen t ∉ Z + 1 2 {\displaystyle t\notin \mathbb {Z} +{\frac {1}{2}}} R := − 4 π ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 Ci ( ( 2 k + 1 ) π x ) = ∫ x ∞ ( − 1 ) ⌊ t + 1 2 ⌋ d t t {\displaystyle R:=-{\frac {4}{\pi }}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,{\text{Ci}}{\Big (}(2k+1)\pi x{\Big )}=\int _{x}^{\infty }(-1)^{\lfloor t+{\frac {1}{2}}\rfloor }\,{\frac {dt}{t}}} = ∫ x ⌊ x ⌋ + 1 2 ( − 1 ) ⌊ t + 1 2 ⌋ d t t + ∑ n = ⌊ x ⌋ + 1 ∞ ∫ n − 1 2 n + 1 2 ( − 1 ) ⌊ t + 1 2 ⌋ d t t = ( − 1 ) ⌊ x + 1 2 ⌋ log ⌊ x ⌋ + 1 2 x + ∑ n = ⌊ x ⌋ + 1 ∞ ( − 1 ) n log n + 1 2 n − 1 2 ⏟ S {\displaystyle =\int _{x}^{\lfloor x\rfloor +{\frac {1}{2}}}(-1)^{\lfloor t+{\frac {1}{2}}\rfloor }\,{\frac {dt}{t}}+\sum _{n=\lfloor x\rfloor +1}^{\infty }\int _{n-{\frac {1}{2}}}^{n+{\frac {1}{2}}}(-1)^{\lfloor t+{\frac {1}{2}}\rfloor }\,{\frac {dt}{t}}=(-1)^{\lfloor x+{\frac {1}{2}}\rfloor }\,\log {\frac {\lfloor x\rfloor +{\frac {1}{2}}}{x}}+\underbrace {\sum _{n=\lfloor x\rfloor +1}^{\infty }(-1)^{n}\,\log {\frac {n+{\frac {1}{2}}}{n-{\frac {1}{2}}}}} _{S}} , dabei ist die letzte Reihe S = 2 log ( 2 ⋅ Γ ( 3 4 ) Γ ( 1 4 ) ) − ∑ n = 1 ⌊ x ⌋ ( − 1 ) n log n + 1 2 n − 1 2 {\displaystyle S=2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)-\sum _{n=1}^{\lfloor x\rfloor }(-1)^{n}\,\log {\frac {n+{\frac {1}{2}}}{n-{\frac {1}{2}}}}} . Hierbei wiederrum ist − ∑ n = 1 ⌊ x ⌋ ( − 1 ) n log n + 1 2 n − 1 2 = ∑ n = 1 ⌊ x ⌋ ( − 1 ) n log ( n − 1 2 ) − ∑ n = 1 ⌊ x ⌋ ( − 1 ) n log ( n + 1 2 ) {\displaystyle -\sum _{n=1}^{\lfloor x\rfloor }(-1)^{n}\,\log {\frac {n+{\frac {1}{2}}}{n-{\frac {1}{2}}}}=\sum _{n=1}^{\lfloor x\rfloor }(-1)^{n}\,\log \left(n-{\frac {1}{2}}\right)-\sum _{n=1}^{\lfloor x\rfloor }(-1)^{n}\,\log \left(n+{\frac {1}{2}}\right)} = − ∑ n = 0 ⌊ x ⌋ − 1 ( − 1 ) n log ( n + 1 2 ) + log ( 1 2 ) − ∑ n = 0 ⌊ x ⌋ − 1 ( − 1 ) n log ( n + 1 2 ) − ( − 1 ) ⌊ x ⌋ log ( ⌊ x ⌋ + 1 2 ) {\displaystyle =-\sum _{n=0}^{\lfloor x\rfloor -1}(-1)^{n}\,\log \left(n+{\frac {1}{2}}\right)+\log \left({\frac {1}{2}}\right)-\sum _{n=0}^{\lfloor x\rfloor -1}(-1)^{n}\,\log \left(n+{\frac {1}{2}}\right)-(-1)^{\lfloor x\rfloor }\log \left(\lfloor x\rfloor +{\frac {1}{2}}\right)} . S = 2 log ( 2 Γ ( 3 4 ) Γ ( 1 4 ) ) − 2 ∑ n = 0 ⌊ x ⌋ − 1 ( − 1 ) n log ( n + 1 2 ) − ( − 1 ) ⌊ x ⌋ log ( ⌊ x ⌋ + 1 2 ) {\displaystyle S=2\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)-2\sum _{n=0}^{\lfloor x\rfloor -1}(-1)^{n}\log \left(n+{\frac {1}{2}}\right)-(-1)^{\lfloor x\rfloor }\log \left(\lfloor x\rfloor +{\frac {1}{2}}\right)} R = − ( − 1 ) ⌊ x + 1 2 ⌋ log x + 2 log ( 2 Γ ( 3 4 ) Γ ( 1 4 ) ) − 2 ∑ n = 0 ⌊ x ⌋ − 1 ( − 1 ) n log ( n + 1 2 ) + ( ( − 1 ) ⌊ x + 1 2 ⌋ − ( − 1 ) ⌊ x ⌋ ) log ( ⌊ x ⌋ + 1 2 ) ⏟ T {\displaystyle R=-(-1)^{\lfloor x+{\frac {1}{2}}\rfloor }\log x+2\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)-2\sum _{n=0}^{\lfloor x\rfloor -1}(-1)^{n}\log \left(n+{\frac {1}{2}}\right)+\underbrace {\left((-1)^{\lfloor x+{\frac {1}{2}}\rfloor }-(-1)^{\lfloor x\rfloor }\right)\log \left(\lfloor x\rfloor +{\frac {1}{2}}\right)} _{T}} , wobei T = ( ( − 1 ) ⌊ x + 1 2 ⌋ − ⌊ x ⌋ − 1 ) ( − 1 ) ⌊ x ⌋ log ( ⌊ x ⌋ + 1 2 ) {\displaystyle T=\left((-1)^{\lfloor x+{\frac {1}{2}}\rfloor -\lfloor x\rfloor }-1\right)(-1)^{\lfloor x\rfloor }\log \left(\lfloor x\rfloor +{\frac {1}{2}}\right)} genau dann einen zusätzlichen Summanden liefert, wenn ⌊ x − 1 2 ⌋ > ⌊ x ⌋ − 1 {\displaystyle \left\lfloor x-{\frac {1}{2}}\right\rfloor >\lfloor x\rfloor -1} ist. Also ist R = − ( − 1 ) ⌊ x + 1 2 ⌋ log x + 2 log ( 2 Γ ( 3 4 ) Γ ( 1 4 ) ) − 2 ∑ n = 0 ⌊ x − 1 2 ⌋ ( − 1 ) n log ( n + 1 2 ) {\displaystyle R=-(-1)^{\lfloor x+{\frac {1}{2}}\rfloor }\log x+2\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)-2\sum _{n=0}^{\lfloor x-{\frac {1}{2}}\rfloor }(-1)^{n}\log \left(n+{\frac {1}{2}}\right)} . 4 π ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 Ci ( ( 2 k + 1 ) π x ) = ( − 1 ) ⌊ x + 1 2 ⌋ log x − 2 log ( 2 Γ ( 3 4 ) Γ ( 1 4 ) ) + 2 ∑ n = 0 ⌊ x − 1 2 ⌋ ( − 1 ) n log ( n + 1 2 ) {\displaystyle {\frac {4}{\pi }}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{2k+1}}\,{\text{Ci}}{\Big (}(2k+1)\pi x{\Big )}=(-1)^{\lfloor x+{\frac {1}{2}}\rfloor }\log x-2\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)+2\sum _{n=0}^{\lfloor x-{\frac {1}{2}}\rfloor }(-1)^{n}\log \left(n+{\frac {1}{2}}\right)} = ( − 1 ) ⌊ x + 1 2 ⌋ log ( π x ) − 2 log ( 2 π Γ ( 3 4 ) Γ ( 1 4 ) ) + 2 ∑ n = 0 ⌊ x − 1 2 ⌋ ( − 1 ) n log ( ( 2 n + 1 ) π 2 ) {\displaystyle =(-1)^{\lfloor x+{\frac {1}{2}}\rfloor }\log(\pi x)-2\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)+2\sum _{n=0}^{\lfloor x-{\frac {1}{2}}\rfloor }(-1)^{n}\log \left((2n+1){\frac {\pi }{2}}\right)}
für alle reellen 
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genau dann einen zusätzlichen Summanden liefert, wenn 
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![{\displaystyle =\int _{0}^{\infty }{\frac {e^{-{\frac {\pi x}{2}}}}{1+e^{-\pi x}}}\,{\frac {1}{1+x^{2}}}\,dx={\frac {1}{2}}\int _{0}^{\infty }{\frac {1}{\cosh {\frac {\pi x}{2}}}}\,{\frac {dx}{1+x^{2}}}={\frac {1}{4}}\left[\psi \left({\frac {1}{4}}+{\frac {3}{4}}\right)-\psi \left({\frac {1}{4}}+{\frac {1}{4}}\right)\right]={\frac {\log 2}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc70f79762000ee59d615b01e38c21e990ba67b7)
![{\displaystyle {\xrightarrow[{m\to \infty }]{\text{Stirling}}}\;\,x-\left(\lfloor x\rfloor +{\frac {1}{2}}\right)\log x+\log(\lfloor x\rfloor !)-\log {\sqrt {2\pi }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/01e60d2b3432909a201bf9941443bb8265bee307)
