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

1周前 (05-04) 7次浏览

##### 0.1Bearbeiten
{displaystyle int _{0}^{pi }log left(cos {frac {x}{2}}right),dx=-pi log 2}
Beweis

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

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

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

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

Nach der Cauchyschen Cosinus-Integralformel ist {displaystyle f(y):={frac {1}{pi }}int _{0}^{pi }left(2cos {frac {theta }{2}}right)^{x}cos ytheta ,dtheta ={frac {Gamma (x+1)}{Gamma left({frac {x}{2}}+y+1right),Gamma left({frac {x}{2}}-y+1right)}}}.

Durch logarithmisches Differenzieren ergibt sich:

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

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

Nach der Produktregel ist dann

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

{displaystyle Rightarrow ,f”(0)=-2cdot f(0)cdot psi ‘left({frac {x}{2}}+1right)=-2cdot {frac {Gamma (x+1)}{Gamma ^{2}left({frac {x}{2}}+1right)}}cdot psi ‘left({frac {x}{2}}+1right)}.

Also ist {displaystyle g(x):={frac {1}{pi }}int _{0}^{pi }left(2cos {frac {theta }{2}}right)^{x},theta ^{2},dtheta =2cdot {frac {Gamma (x+1)}{Gamma ^{2}left({frac {x}{2}}+1right)}}cdot psi ‘left({frac {x}{2}}+1right)}.

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

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

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

{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}}+1right)=zeta (2)-zeta (3)cdot x+{frac {3}{4}}zeta (4)cdot x^{2}-…}

{displaystyle h(x)cdot psi ‘left({frac {x}{2}}+1right)=left(1+{frac {pi ^{2}}{24}}cdot x^{2}+…right)cdot left(zeta (2)-zeta (3)cdot x+{frac {3}{4}}zeta (4)cdot x^{2}-…right)=zeta (2)-zeta (3)cdot x+left({frac {pi ^{2}}{24}}zeta (2)+{frac {3}{4}}zeta (4)right)cdot x^{2}+…}

{displaystyle g”(0)={frac {1}{pi }}int _{0}^{pi }log ^{2}left(2cos {frac {theta }{2}}right)cdot theta ^{2},dtheta =2cdot 2cdot left({frac {pi ^{2}}{24}}zeta (2)+{frac {3}{4}}zeta (4)right)={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 right)}
ohne Beweis

##### 1.1Bearbeiten
{displaystyle int _{0}^{pi }log left(1-2alpha cos x+alpha ^{2}right)dx=left{{begin{matrix}0&|alpha |leq 1\\2pi log |alpha |&|alpha |>1end{matrix}}right.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}right)=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}right)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+1right)right]dx=2pi log |alpha |+underbrace {Ileft({frac {1}{alpha }}right)} _{=0}}.