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

1.1Bearbeiten

{displaystyle int _{-infty }^{infty }e^{-x^{2}}\,cos(2ax)\,dx={sqrt {pi }}cdot e^{-a^{2}}qquad ain mathbb {C} }

1. Beweis

Verwende die Reihenentwicklung {displaystyle cos(2ax)=sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k)!}}\,(2ax)^{2k}}.

{displaystyle I:=int _{-infty }^{infty }e^{-x^{2}}\,cos(2ax)\,dx=sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k)!}}\,(2a)^{2k}\,int _{-infty }^{infty }x^{2k}\,e^{-x^{2}}\,dx}

Dabei ist {displaystyle int _{-infty }^{infty }x^{2k}\,e^{-x^{2}}\,dx=2int _{0}^{infty }x^{2k}\,e^{-x^{2}}\,dx=2int _{0}^{infty }y^{k}\,e^{-y}\,{frac {dy}{2{sqrt {y}}}}=Gamma left(k+{frac {1}{2}} ight)={frac {sqrt {pi }}{2^{2k-1}}}\,{frac {Gamma (2k)}{Gamma (k)}}={frac {sqrt {pi }}{2^{2k}}}\,{frac {(2k)!}{k!}}}.

{displaystyle I=sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k)!}}\,(2a)^{2k}\,{frac {sqrt {pi }}{2^{2k}}}\,{frac {(2k)!}{k!}}={sqrt {pi }}cdot sum _{k=0}^{infty }{frac {(-1)^{k}}{k!}}\,a^{2k}={sqrt {pi }}cdot e^{-a^{2}}}

2. Beweis

Integriert man die holomorphe Funktion {displaystyle f(z)=e^{-z^{2}}} längs der geschlossenen Kurve in der Zeichnung, so ist {displaystyle oint f(z)\,dz=0}.

Die Integrale über den vertikalen Strecken verschwinden für {displaystyle R o infty \,}. Also ist {displaystyle {sqrt {pi }}=int _{-infty }^{infty }f(x)\,dx=int _{-infty }^{infty }f(x+ia)\,dx}

{displaystyle f(x+ia)=e^{-(x+ia)^{2}}=e^{-x^{2}-icdot 2ax+a^{2}}=e^{a^{2}}cdot e^{-x^{2}}\,{Big (}cos(2ax)-isin(2ax){Big )}}

{displaystyle Rightarrow \,int _{-infty }^{infty }f(x+ia)\,dx=e^{a^{2}}cdot {Bigg [}int _{-infty }^{infty }e^{-x^{2}}cos(2ax)\,dx-iunderbrace {int _{-infty }^{infty }e^{-x^{2}}sin(2ax)\,dx} _{=0}{Bigg ]}}
{displaystyle Rightarrow \,int _{-infty }^{infty }e^{-x^{2}}\,cos(2ax)\,dx={sqrt {pi }}cdot e^{-a^{2}}}

 

3.1Bearbeiten

{displaystyle int _{0}^{infty }e^{-ax}\,cos(bx)\,x^{s-1}\,dx={frac {Gamma (s)}{{sqrt {a^{2}+b^{2}}}^{s}}}\,cos left(s\,arctan {frac {b}{a}} ight)qquad a>0\,,\,bin mathbb {R} \,,\,{ ext{Re}}(s)>0}

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