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# Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log,Gamma)

1周前 (05-04) 6次浏览

##### 0.1Bearbeiten
{displaystyle int _{0}^{1}log Gamma (x),dx=log {sqrt {2pi }}}
1. Beweis

{displaystyle 2int _{0}^{1}log Gamma (x),dx=int _{0}^{1}log Gamma (x),dx+int _{0}^{1}log Gamma (1-x),dx=int _{0}^{1}log {Big (}Gamma (x),Gamma (1-x){Big )},dx}

{displaystyle =int _{0}^{1}log left({frac {pi }{sin pi x}}right)dx=log pi -int _{0}^{1}log sin pi x,dx=log pi +log 2,Rightarrow ,int _{0}^{1}log Gamma (x),dx={frac {1}{2}}log(2pi )}

2. Beweis

Die Riemannsche Approximationssumme {displaystyle sum _{k=1}^{n-1}log Gamma !left({frac {k}{n}}right)cdot {frac {1}{n}}} vereinfacht sich zu

{displaystyle log left(prod _{k=1}^{n-1}Gamma !left({frac {k}{n}}right)right)cdot {frac {1}{n}}=log left({frac {{sqrt {2pi }}^{,n-1}}{sqrt {n}}}right)cdot {frac {1}{n}}={frac {(n-1)log {sqrt {2pi }}-log {sqrt {n}}}{n}}},

und konvergiert daher gegen {displaystyle log {sqrt {2pi }}} für {displaystyle nto infty ,}.

##### 0.2Bearbeiten
{displaystyle int _{1/4}^{3/4}log Gamma (x),dx={frac {1}{2}}left(log {sqrt {2pi }}-{frac {G}{pi }}right)}
1. Beweis

{displaystyle I:=int _{1/4}^{3/4}log Gamma (x),dx=int _{1/4}^{3/4}log Gamma (1-x),dx}

{displaystyle Rightarrow ,2I=int _{1/4}^{3/4}log {Big (}Gamma (x)Gamma (1-x){Big )},dx=int _{1/4}^{3/4}log left({frac {pi }{sin pi x}}right)dx={frac {1}{2}}log pi -int _{1/4}^{3/4}log(sin pi x),dx},

wobei {displaystyle int _{1/4}^{3/4}log(sin pi x),dx=int _{-1/4}^{1/4}log(cos pi x),dx=2int _{0}^{1/4}log(cos pi x),dx} {displaystyle ={frac {1}{pi }}int _{0}^{frac {pi }{2}}log left(cos {frac {x}{2}}right)dx={frac {G}{pi }}-{frac {1}{2}}log 2} ist.

Also ist {displaystyle 2I={frac {1}{2}}log(2pi )-{frac {G}{pi }}}.

2. Beweis

Für {displaystyle 0leq xleq 1} betrachte folgende Rechteck-Impuls-Funktion:

{displaystyle f(x)={frac {4}{pi }}sum _{k=0}^{infty }{frac {(-1)^{k}}{2k+1}},cos {Big (}(2k+1),2pi x{Big )}=left{{begin{matrix}+1&&0leq x<{frac {1}{4}},vee ,{frac {3}{4}}<xleq 1\\0&&x={frac {1}{4}},vee ,x={frac {3}{4}}\\-1&&{frac {1}{4}}<x<{frac {3}{4}}end{matrix}}right.}

{displaystyle int _{0}^{1}log Gamma (x),f(x),dx={frac {4}{pi }}sum _{k=0}^{infty }{frac {(-1)^{k}}{2k+1}}int _{0}^{1}log Gamma (x),cos {Big (}(2k+1),2pi x{Big )}dx={frac {4}{pi }}sum _{k=0}^{infty }{frac {(-1)^{k}}{2k+1}},{frac {1}{4,(2k+1)}}={frac {G}{pi }}}

Aus den Gleichungen

{displaystyle {text{I.}},quad int _{0}^{1/4}log Gamma (x),dx+int _{1/4}^{3/4}log Gamma (x),dx+int _{3/4}^{1}log Gamma (x),dx=log {sqrt {2pi }}}

{displaystyle {text{II.}}quad int _{0}^{1/4}log Gamma (x),dx-int _{1/4}^{3/4}log Gamma (x),dx+int _{3/4}^{1}log Gamma (x),dx={frac {G}{pi }}}

folgt     {displaystyle 2int _{1/4}^{3/4}log Gamma (x),dx=log {sqrt {2pi }}-{frac {G}{pi }}}.

##### 1.1Bearbeiten
{displaystyle int _{u}^{u+1}log Gamma (x),dx=u,{Big (}log(u)-1{Big )}+log {sqrt {2pi }}qquad u>0}
Beweis (Raabesche Formel)

{displaystyle int _{u}^{u+1}log Gamma (x),dx=int _{0}^{u+1}log Gamma (x),dx-int _{0}^{u}log Gamma (x),dx}

{displaystyle =int _{0}^{1}log Gamma (x),dx+int _{1}^{u+1}log Gamma (x),dx-int _{0}^{u}log Gamma (x),dx}

{displaystyle =log {sqrt {2pi }}+int _{0}^{u}log Gamma (x+1),dx-int _{0}^{u}log Gamma (x),dx}.

Wegen {displaystyle log Gamma (x+1)-log Gamma (x)=log x,} ist

{displaystyle int _{0}^{u}log Gamma (x+1)dx-int _{0}^{u}log Gamma (x)dx=u,{Big (}log(u)-1{Big )}}.