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

0.1Bearbeiten

{displaystyle int _{0}^{pi }log left(cos {frac {x}{2}} ight)\,dx=-pi log 2}

Beweis

Aus der Fourierreihendarstellung {displaystyle log left(2cos {frac {x}{2}} ight)=sum _{k=1}^{infty }{frac {(-1)^{k-1}}{k}}\,cos kx}

folgt {displaystyle int _{0}^{pi }log left(2cos {frac {x}{2}} ight)dx=sum _{k=1}^{infty }{frac {(-1)^{k-1}}{k}}\,int _{0}^{pi }cos kx\,dx=0}.

 

 

0.2Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}log left(cos {frac {x}{2}} ight)\,dx=G-{frac {pi }{2}}log 2}

Beweis

Aus der Fourierreihendarstellung {displaystyle log left(2cos {frac {x}{2}} ight)=sum _{k=1}^{infty }{frac {(-1)^{k-1}}{k}}\,cos kx}

folgt {displaystyle int _{0}^{frac {pi }{2}}log left(2cos {frac {x}{2}} ight)dx=sum _{k=1}^{infty }{frac {(-1)^{k-1}}{k}}\,int _{0}^{frac {pi }{2}}cos kx\,dx=sum _{k=1}^{infty }{frac {(-1)^{k-1}\,sin {frac {kpi }{2}}}{k^{2}}}=sum _{n=0}^{infty }{frac {(-1)^{n}}{(2n+1)^{2}}}=G}.

 

 

0.3Bearbeiten

{displaystyle int _{0}^{pi }x^{2}\,log ^{2}left(2cos {frac {x}{2}} ight)\,dx={frac {11pi ^{5}}{180}}}

Beweis

Nach der Cauchyschen Cosinus-Integralformel ist {displaystyle f(y):={frac {1}{pi }}int _{0}^{pi }left(2cos {frac { heta }{2}} ight)^{x}cos y heta \,d heta ={frac {Gamma (x+1)}{Gamma left({frac {x}{2}}+y+1 ight)\,Gamma left({frac {x}{2}}-y+1 ight)}}}.

Durch logarithmisches Differenzieren ergibt sich:

{displaystyle log f(y)=log Gamma (x+1)-log Gamma left({frac {x}{2}}+y+1 ight)-log Gamma left({frac {x}{2}}-y+1 ight)}

{displaystyle Rightarrow \,{frac {f'(y)}{f(y)}}=-psi left({frac {x}{2}}+y+1 ight)+psi left({frac {x}{2}}-y+1 ight)\,Rightarrow \,f'(y)=f(y)cdot left(psi left({frac {x}{2}}-y+1 ight)-psi left({frac {x}{2}}+y+1 ight) ight)\,Rightarrow \,f'(0)=0}

Nach der Produktregel ist dann

{displaystyle f''(y)=f'(y)cdot left(psi left({frac {x}{2}}-y+1 ight)-psi left({frac {x}{2}}+y+1 ight) ight)+f(y)cdot left(-psi 'left({frac {x}{2}}-y+1 ight)-psi 'left({frac {x}{2}}+y+1 ight) ight)}

{displaystyle Rightarrow \,f''(0)=-2cdot f(0)cdot psi 'left({frac {x}{2}}+1 ight)=-2cdot {frac {Gamma (x+1)}{Gamma ^{2}left({frac {x}{2}}+1 ight)}}cdot psi 'left({frac {x}{2}}+1 ight)}.

Also ist {displaystyle g(x):={frac {1}{pi }}int _{0}^{pi }left(2cos {frac { heta }{2}} ight)^{x}\, heta ^{2}\,d heta =2cdot {frac {Gamma (x+1)}{Gamma ^{2}left({frac {x}{2}}+1 ight)}}cdot psi 'left({frac {x}{2}}+1 ight)}.

Setzt man {displaystyle h(x):={frac {Gamma (x+1)}{Gamma ^{2}left({frac {x}{2}}+1 ight)}}}, so ist nach logarithmischer Differenzation {displaystyle log h(x)=log Gamma (x+1)-2log Gamma left({frac {x}{2}}+1 ight)}

{displaystyle Rightarrow \,{frac {h'(x)}{h(x)}}=psi (x+1)-2cdot psi left({frac {x}{2}}+1 ight)cdot {frac {1}{2}}\,Rightarrow \,h'(x)=h(x)cdot left(psi (x+1)-psi left({frac {x}{2}}+1 ight) ight)\,Rightarrow \,h'(0)=0}.

Nach der Produktregel ist {displaystyle h''(x)=h'(x)cdot left(psi (x+1)-psi left({frac {x}{2}}+1 ight) ight)+h(x)cdot left(psi '(x+1)-psi 'left({frac {x}{2}}+1 ight)cdot {frac {1}{2}} ight)}

{displaystyle Rightarrow \,h''(0)=psi '(1)-{frac {1}{2}}psi '(1)={frac {1}{2}}psi '(1)={frac {pi ^{2}}{12}}}.

Die Taylorreihenentwicklung von {displaystyle h\,} beginnt daher wie folgt {displaystyle h(x)=1+{frac {pi ^{2}}{24}}cdot x^{2}+...}

Und wegen {displaystyle psi (x+1)=-gamma +zeta (2)cdot x-zeta (3)cdot x^{2}+zeta (4)cdot x^{3}-...} ist {displaystyle psi '(x+1)=zeta (2)-2zeta (3)cdot x+3zeta (4)cdot x^{2}-...}

und somit ist {displaystyle psi 'left({frac {x}{2}}+1 ight)=zeta (2)-zeta (3)cdot x+{frac {3}{4}}zeta (4)cdot x^{2}-...}

{displaystyle h(x)cdot psi 'left({frac {x}{2}}+1 ight)=left(1+{frac {pi ^{2}}{24}}cdot x^{2}+... ight)cdot left(zeta (2)-zeta (3)cdot x+{frac {3}{4}}zeta (4)cdot x^{2}-... ight)=zeta (2)-zeta (3)cdot x+left({frac {pi ^{2}}{24}}zeta (2)+{frac {3}{4}}zeta (4) ight)cdot x^{2}+...}

{displaystyle g''(0)={frac {1}{pi }}int _{0}^{pi }log ^{2}left(2cos {frac { heta }{2}} ight)cdot heta ^{2}\,d heta =2cdot 2cdot left({frac {pi ^{2}}{24}}zeta (2)+{frac {3}{4}}zeta (4) ight)={frac {11\,pi ^{4}}{180}}}

 

 

0.4Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}{frac {x^{2}}{x^{2}+log ^{2}(2cos x)}}\,dx={frac {pi }{8}}left(1-gamma +log 2pi ight)}

ohne Beweis

 

 

 

1.1Bearbeiten

{displaystyle int _{0}^{pi }log left(1-2alpha cos x+alpha ^{2} ight)dx=left{{egin{matrix}0&|alpha |leq 1\\2pi log |alpha |&|alpha |>1end{matrix}} ight.qquad ,qquad alpha in mathbb {R} }

Beweis

Für {displaystyle |alpha |leq 1} und {displaystyle 0<x<pi } ist

{displaystyle -{frac {1}{2}}log left(1-2alpha cos x+alpha ^{2} ight)=sum _{k=1}^{infty }{frac {alpha ^{k}\,cos kx}{k}}}.

{displaystyle Rightarrow \,I(alpha ):=int _{0}^{pi }log left(1-2alpha cos x+alpha ^{2} ight)dx=-2sum _{k=1}^{infty }{frac {alpha ^{k}}{k}}\,int _{0}^{pi }cos kx\,dx=0}

Und für {displaystyle |alpha |>1} ist

{displaystyle I(alpha )=int _{0}^{pi }left[log(alpha ^{2})+log left({frac {1}{alpha ^{2}}}+{frac {2}{alpha }}cos x+1 ight) ight]dx=2pi log |alpha |+underbrace {Ileft({frac {1}{alpha }} ight)} _{=0}}.