Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,tan)

0.1Bearbeiten

{displaystyle int _{0}^{pi }x\, an x\,dx=-pi \,log 2}

Beweis

Setzt man {displaystyle f(z)=z\, an z}, so ist

{displaystyle int _{C_{varepsilon }}f\,dz+int _{K_{varepsilon }}f\,dz+int _{D_{varepsilon }}f\,dz+int _{gamma _{N}}f\,dz+int _{sigma _{N}}f\,dz+int _{delta _{N}}f\,dz=oint f\,dz=0}.

Nun ist {displaystyle lim _{varepsilon o 0+}int _{C_{varepsilon }}f\,dz+int _{D_{varepsilon }}f\,dz={ ext{p.V.}}int _{0}^{pi }x\, an x\,dx}

und {displaystyle lim _{varepsilon o 0+}int _{K_{varepsilon }}f\,dz=-ipi \,{ ext{res}}left(f,{frac {pi }{2}} ight)={frac {ipi ^{2}}{2}}}.

Und aus {displaystyle int _{gamma _{N}}f\,dz+int _{delta _{N}}f\,dz=int _{0}^{N}f(pi +iy)\,i\,dy-int _{0}^{N}f(iy)\,i\,dy=-pi int _{0}^{N} anh y\,dy=-pi log(cosh N)}

und {displaystyle int _{sigma _{N}}f\,dz=-int _{0}^{pi }f(x+iN)\,dx=pi log(2cosh N)-{frac {ipi ^{2}}{2}}}

folgt {displaystyle int _{gamma _{N}}f\,dz+int _{sigma _{N}}f\,dz+int _{delta _{N}}f\,dz=pi log 2-{frac {ipi ^{2}}{2}}}.

Also ist {displaystyle { ext{p.V.}}int _{0}^{pi }x\, an x\,dx+pi log 2=0}.

 

0.2Bearbeiten

{displaystyle int _{0}^{frac {pi }{4}}log(1+ an x)\,dx={frac {pi }{8}}log 2}

Beweis

Wegen {displaystyle 1+ an x={frac {cos x+sin x}{cos x}}={frac {{sqrt {2}}cos left({frac {pi }{4}}-x ight)}{cos x}}}

ist {displaystyle log(1+ an x)={frac {1}{2}}log 2+log cos left({frac {pi }{4}}-x ight)-log(cos x)}.

Da nach Substitution {displaystyle xmapsto {frac {pi }{4}}-x}

{displaystyle int _{0}^{frac {pi }{4}}log cos left({frac {pi }{4}}-x ight)\,dx=int _{0}^{frac {pi }{4}}log(cos x)\,dx} ist,

ist das gesuchte Integral {displaystyle int _{0}^{frac {pi }{4}}{frac {1}{2}}log 2\,dx={frac {pi }{8}}\,log 2} .

 

1.1Bearbeiten

{displaystyle int _{-infty }^{infty }{frac { an alpha x}{x}}\,dx=pi qquad alpha >0}

Beweis

Nach der Formel von Lobatschewski ist {displaystyle int _{-infty }^{infty }1cdot {frac { an x}{x}}\,dx=int _{0}^{pi }1\,dx=pi }.

Substituiert man {displaystyle x o alpha x\,}, so erhält man die behauptete Formel.

 

1.2Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}}{frac {1}{1+ an ^{alpha }x}}\,dx={frac {pi }{4}}qquad alpha in mathbb {C} setminus imathbb {R} ^{ imes }}

Beweis

Für {displaystyle alpha in mathbb {C} setminus imathbb {R} ^{ imes }} sei {displaystyle I(alpha )=int _{0}^{frac {pi }{2}}{frac {1}{1+ an ^{alpha }x}}\,dx}.

Nach Substitution {displaystyle xmapsto {frac {pi }{2}}-x} ist {displaystyle I(alpha )=int _{0}^{frac {pi }{2}}{frac {1}{1+cot ^{alpha }x}}\,dx=int _{0}^{frac {pi }{2}}{frac { an ^{alpha }x}{1+ an ^{alpha }x}}\,dx}.

Addiert man die verschiedenen Darstellungen von {displaystyle I(alpha )\,}, so ist {displaystyle 2I(alpha )=int _{0}^{frac {pi }{2}}{frac {1+ an ^{alpha }x}{1+ an ^{alpha }x}}\,dx={frac {pi }{2}}}.

Unabhängig von {displaystyle alpha \,} gilt also {displaystyle I(alpha )={frac {pi }{4}}}.

 

1.3Bearbeiten

{displaystyle int _{0}^{frac {pi }{2}} an ^{2alpha -1}\,x;dx={frac {pi }{2sin alpha pi }}qquad 0<{ ext{Re}}(alpha )<1}

Beweis

Verwende die Formel

{displaystyle 2int _{0}^{frac {pi }{2}}sin ^{2alpha -1}x\,cos ^{2eta -1}x\,dx={frac {Gamma (alpha )\,Gamma (eta )}{Gamma (alpha +eta )}}qquad { ext{Re}}(alpha ),{ ext{Re}}(eta )>0}.

Ist {displaystyle 0<{ ext{Re}}(alpha )<1\,} und setzt man {displaystyle eta =1-alpha \,}, so ist auch {displaystyle 0<{ ext{Re}}(eta )<1\,}.

Also ist {displaystyle 2int _{0}^{frac {pi }{2}} an ^{2alpha -1}x;dx=Gamma (alpha )\,Gamma (1-alpha )}.

Und das ist {displaystyle {frac {pi }{sin alpha pi }}} nach dem Eulerschen Ergänzungssatz.