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

0.1Bearbeiten

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

Beweis

Verwende die Fourierreihe {displaystyle -log left(2sin {frac {x}{2}} ight)=sum _{n=1}^{infty }{frac {cos nx}{n}}}.

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

 

0.2Bearbeiten

{displaystyle int _{0}^{frac {pi }{3}}log ^{2}left(2sin {frac {x}{2}} ight)\,dx={frac {7pi ^{3}}{108}}}

Beweis

Es sei {displaystyle mathbb {H} =mathbb {C} setminus {zin mathbb {R} \,|\,zleq 0}} und {displaystyle ngeq 2\,} eine natürliche Zahl.

Die Funktion {displaystyle f_{n}:mathbb {H} o mathbb {C} \,,\,zmapsto {frac {(-log z)^{n-1}}{1-z}}} ist auf ganz {displaystyle mathbb {H} \,} holomorph,

wenn man sie an ihrer hebbaren Definitionslücke {displaystyle z=1\,} stetig fortsetzt.


{displaystyle F_{n}:\,\,]-pi ,pi [ o mathbb {C} \,,\, heta mapsto int _{1}^{e^{-i heta }}f_{n}(z)\,dz} ist nach der Substitution {displaystyle z o {frac {1}{z}}}

gleich {displaystyle int _{1}^{e^{i heta }}{frac {(log z)^{n-1}}{1-{frac {1}{z}}}}\,{frac {-dz}{z^{2}}}=int _{1}^{e^{i heta }}{frac {(log z)^{n-1}}{z\,(1-z)}}\,dz}.

Und das ist nach der Partialbruchzerlegung {displaystyle {frac {1}{z\,(1-z)}}={frac {1}{z}}+{frac {1}{1-z}}}

gleich {displaystyle int _{1}^{e^{i heta }}{frac {(log z)^{n-1}}{z}}\,dz+int _{1}^{e^{i heta }}{frac {(log z)^{n-1}}{1-z}}\,dz=left[{frac {1}{n}}\,(log z)^{n} ight]_{1}^{e^{i heta }}+(-1)^{n-1}int _{1}^{e^{i heta }}f_{n}(z)\,dz}.

Also ist {displaystyle F_{n}( heta )={frac {(i heta )^{n}}{n}}+{overline {F_{n}( heta )}}qquad (1)}


{displaystyle G_{n}:\,\,]0,2pi [ o mathbb {C} \,,\, heta mapsto int _{0}^{1-e^{i heta }}f_{n}(z)\,dz} ist nach der Substitution {displaystyle z o 1-z\,} gleich

{displaystyle -int _{1}^{e^{i heta }}{frac {left[-log(1-z) ight]^{n-1}}{z}}\,dz}. Und das ist nach der Substitution {displaystyle z o e^{ix}\,} gleich

{displaystyle -iint _{0}^{ heta }left[-log(1-e^{ix}) ight]^{n-1}\,dx}, wobei {displaystyle -log(1-e^{ix})=-log left(2sin {frac {x}{2}} ight)+i\,{frac {pi -x}{2}}} ist.

Also ist {displaystyle G_{n}( heta )=-iint _{0}^{ heta }left[-log left(2sin {frac {x}{2}} ight)+i\,{frac {pi -x}{2}} ight]^{n-1}\,dxqquad (2)}


{displaystyle G_{n}( heta )=int _{0}^{1-e^{i heta }}f_{n}(z)\,dz} lässt sich aufspalten in {displaystyle int _{0}^{1}f_{n}(z)\,dz+int _{1}^{1-e^{i heta }}f_{n}(z)\,dz},

wobei {displaystyle int _{0}^{1}f_{n}(z)\,dz=Gamma (n)\,zeta (n)} ist. Setzt man {displaystyle heta ={frac {pi }{3}}\,}, so ist {displaystyle 1-e^{i heta }=e^{-i heta }\,}.

Daher gilt {displaystyle G_{n}left({frac {pi }{3}} ight)=Gamma (n)zeta (n)+F_{n}left({frac {pi }{3}} ight)qquad (3)}



Betrachte nun den Fall {displaystyle heta ={frac {pi }{3}}} und {displaystyle n=3\,:}

Aus {displaystyle (1)\,\,\,F_{3}left({frac {pi }{3}} ight)={frac {left(i{frac {pi }{3}} ight)^{3}}{3}}+{overline {F_{3}left({frac {pi }{3}} ight)}}}

folgt {displaystyle { ext{Im}}left[F_{3}left({frac {pi }{3}} ight) ight]={frac {1}{2i}}\,left(F_{3}left({frac {pi }{3}} ight)-{overline {F_{3}left({frac {pi }{3}} ight)}} ight)={frac {1}{2i}}\,{frac {i^{3}pi ^{3}}{3^{4}}}=-{frac {pi ^{3}}{162}}}.

Aus {displaystyle (2)\,\,\,G_{3}left({frac {pi }{3}} ight)=-iint _{0}^{frac {pi }{3}}left[-log left(2sin {frac {x}{2}} ight)+i\,{frac {pi -x}{2}} ight]^{2}\,dx}

{displaystyle =int _{0}^{frac {pi }{3}}(pi -x)\,log left(2sin {frac {x}{2}} ight)\,dx-iint _{0}^{frac {pi }{3}}left[log ^{2}left(2sin {frac {x}{2}} ight)-left({frac {pi -x}{2}} ight)^{2} ight]dx}

folgt {displaystyle { ext{Im}}left[G_{3}left({frac {pi }{3}} ight) ight]=int _{0}^{frac {pi }{3}}left({frac {pi -x}{2}} ight)^{2}\,dx-int _{0}^{frac {pi }{3}}log ^{2}left(2sin {frac {x}{2}} ight)\,dx}.

Und aus {displaystyle (3)\,\,\,G_{3}left({frac {pi }{3}} ight)=2zeta (3)+F_{3}left({frac {pi }{3}} ight)} folgt {displaystyle { ext{Im}}left[G_{3}left({frac {pi }{3}} ight) ight]={ ext{Im}}left[F_{3}left({frac {pi }{3}} ight) ight]}.

Also ist {displaystyle int _{0}^{frac {pi }{3}}log ^{2}left(2sin {frac {x}{2}} ight)\,dx-int _{0}^{frac {pi }{3}}{frac {(pi -x)^{2}}{4}}\,dx={frac {pi ^{3}}{162}}}.

Und somit ist {displaystyle int _{0}^{frac {pi }{3}}log ^{2}left(2sin {frac {x}{2}} ight)\,dx={frac {7pi ^{3}}{108}}}.

 

0.3Bearbeiten

{displaystyle int _{0}^{pi }log ^{2}left(2sin {frac {x}{2}} ight)\,dx={frac {pi ^{3}}{12}}}

ohne Beweis

 

0.4Bearbeiten

{displaystyle int _{0}^{frac {pi }{3}}xlog ^{2}left(2sin {frac {x}{2}} ight)\,dx={frac {17pi ^{4}}{6480}}}

Beweis

Betrachte nun den Fall {displaystyle heta ={frac {pi }{3}}} und {displaystyle n=4\,:}

Aus {displaystyle (1)\,\,\,F_{4}left({frac {pi }{3}} ight)={frac {left(i{frac {pi }{3}} ight)^{4}}{4}}-{overline {F_{4}left({frac {pi }{3}} ight)}}}

folgt {displaystyle { ext{Re}}left[F_{4}left({frac {pi }{3}} ight) ight]={frac {1}{2}}\,left(F_{4}left({frac {pi }{3}} ight)+{overline {F_{4}left({frac {pi }{3}} ight)}} ight)={frac {1}{2}}\,{frac {pi ^{4}}{4cdot 3^{4}}}={frac {pi ^{4}}{648}}}.

Aus {displaystyle (2)\,\,\,G_{4}left({frac {pi }{3}} ight)=-iint _{0}^{frac {pi }{3}}left[-log left(2sin {frac {x}{2}} ight)+i\,{frac {pi -x}{2}} ight]^{3}\,dx}

{displaystyle =int _{0}^{frac {pi }{3}}left[3log ^{2}left(2sin {frac {x}{2}} ight){frac {pi -x}{2}}-left({frac {pi -x}{2}} ight)^{3} ight]\,dx+iint _{0}^{frac {pi }{3}}left[log ^{3}left(2sin {frac {x}{2}} ight)-3log left(2sin {frac {x}{2}} ight)\,left({frac {pi -x}{2}} ight)^{2} ight]\,dx}

folgt {displaystyle { ext{Re}}left[G_{4}left({frac {pi }{3}} ight) ight]={frac {3pi }{2}}int _{0}^{frac {pi }{3}}log ^{2}left(2sin {frac {x}{2}} ight)\,dx-{frac {3}{2}}int _{0}^{frac {pi }{3}}xlog ^{2}left(2sin {frac {x}{2}} ight)\,dx-int _{0}^{frac {pi }{3}}{frac {(pi -x)^{3}}{8}}dx}.

Aus dem Fall {displaystyle n=3\,} ist bereits bekannt, dass {displaystyle int _{0}^{frac {pi }{3}}log ^{2}left(2sin {frac {x}{2}} ight)\,dx={frac {7pi ^{3}}{108}}} ist.

Also ist {displaystyle { ext{Re}}left[G_{4}left({frac {pi }{3}} ight) ight]=-{frac {3}{2}}int _{0}^{frac {pi }{3}}xlog ^{2}left(2sin {frac {x}{2}} ight)\,dx+{frac {187pi ^{4}}{2592}}}.

Und aus {displaystyle (3)\,\,\,G_{4}left({frac {pi }{3}} ight)=Gamma (4)zeta (4)+F_{4}left({frac {pi }{3}} ight)} folgt {displaystyle { ext{Re}}left[G_{4}left({frac {pi }{3}} ight) ight]-6cdot {frac {pi ^{4}}{90}}={ ext{Re}}left[F_{4}left({frac {pi }{3}} ight) ight]}.

Also ist {displaystyle -{frac {3}{2}}int _{0}^{frac {pi }{3}}xlog ^{2}left(2sin {frac {x}{2}} ight)\,dx+{frac {187pi ^{4}}{2592}}-6cdot {frac {pi ^{4}}{90}}={frac {pi ^{4}}{648}}}.

Und somit ist {displaystyle int _{0}^{frac {pi }{3}}xlog ^{2}left(2sin {frac {x}{2}} ight)\,dx={frac {17pi ^{4}}{6480}}}.