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

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}} ight)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}} ight)cdot {frac {1}{n}}} vereinfacht sich zu

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

und konvergiert daher gegen {displaystyle log {sqrt {2pi }}} für {displaystyle n o infty \,}.

 

0.2Bearbeiten

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

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}} ight)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}} ight)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{{egin{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}} ight.}

{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 { ext{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 { ext{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 )}}.