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

1.1Bearbeiten

{displaystyle int _{0}^{infty }left({frac {sin x}{x}} ight)^{2}\,e^{-2ax}\,dx=a\,log left({frac {a}{sqrt {1+a^{2}}}} ight)+operatorname {arccot} a}

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2.1Bearbeiten

{displaystyle int _{0}^{infty }e^{-alpha x}\,sin ^{2n}x\,dx={frac {(2n)!}{alpha \,(alpha ^{2}+2^{2})(alpha ^{2}+4^{2})cdots (alpha ^{2}+(2n)^{2})}}qquad nin mathbb {N} \,\,,\,\,{ ext{Re}}(alpha )>0}

Beweis

Es sei {displaystyle I_{n}=int _{0}^{infty }e^{-alpha x}\,sin ^{n}x\,dx}.

Durch zweimalige partielle Integration erhält man die Rekursion {displaystyle I_{n}=I_{n-2}\,{frac {(n-1)\,n}{alpha ^{2}+n^{2}}}}.

Also ist

{displaystyle I_{2n}=I_{0}cdot {frac {1cdot 2}{alpha ^{2}+2^{2}}}cdot {frac {3cdot 4}{alpha ^{2}+4^{2}}}cdots {frac {(2n-1)\,2n}{alpha ^{2}+(2n)^{2}}}={frac {(2n)!}{alpha \,(alpha ^{2}+2^{2})(alpha ^{2}+4^{2})cdots (alpha ^{2}+(2n)^{2})}}}.

 

2.2Bearbeiten

{displaystyle int _{0}^{infty }e^{-alpha x}\,sin ^{2n+1}x\,dx={frac {(2n+1)!}{(alpha ^{2}+1)(alpha ^{2}+3^{2})cdots (alpha ^{2}+(2n+1)^{2})}}qquad nin mathbb {N} \,\,,\,\,{ ext{Re}}(alpha )>0}

Beweis

Es sei {displaystyle I_{n}=int _{0}^{infty }e^{-alpha x}\,sin ^{n}x\,dx}.

Durch zweimalige partielle Integration erhält man die Rekursion {displaystyle I_{n}=I_{n-2}\,{frac {(n-1)\,n}{alpha ^{2}+n^{2}}}}.

Also ist

{displaystyle I_{2n+1}=I_{1}cdot {frac {2cdot 3}{alpha ^{2}+3^{2}}}cdot {frac {4cdot 5}{alpha ^{2}+5^{2}}}cdots {frac {n\,(2n+1)}{alpha ^{2}+(2n+1)^{2}}}={frac {(2n+1)!}{(alpha ^{2}+1)(alpha ^{2}+3^{2})cdots (alpha ^{2}+(2n+1)^{2})}}}.

 

2.3Bearbeiten

{displaystyle int _{0}^{infty }{frac {sin alpha x}{1-e^{eta x}}}\,dx={frac {1}{2alpha }}-{frac {pi }{2eta }};{ ext{coth}}left({frac {alpha pi }{eta }} ight)}

Beweis

Aus {displaystyle {frac {sin alpha x}{1-e^{eta x}}}=-sum _{k=1}^{infty }e^{-keta x}\,sin alpha x} folgt {displaystyle int _{0}^{infty }{frac {sin alpha x}{1-e^{eta x}}}\,dx=-sum _{k=1}^{infty }int _{0}^{infty }e^{-keta x}\,sin alpha x\,dx}.

Und das ist {displaystyle -sum _{k=1}^{infty }{frac {alpha }{k^{2}eta ^{2}+alpha ^{2}}}={frac {1}{2alpha }}-{frac {1}{2}}sum _{k=-infty }^{infty }{frac {alpha }{k^{2}eta ^{2}+alpha ^{2}}}={frac {1}{2alpha }}-{frac {pi }{2eta }};{ ext{coth}}left({frac {alpha pi }{eta }} ight)}.

 

2.4Bearbeiten

{displaystyle int _{0}^{infty }{frac {sin alpha x}{1+e^{eta x}}}\,dx={frac {1}{2alpha }}-{frac {pi }{2eta }};{ ext{csch}}left({frac {alpha pi }{eta }} ight)}

Beweis

Aus {displaystyle {frac {sin alpha x}{1+e^{eta x}}}=-sum _{k=1}^{infty }(-1)^{k}\,e^{-keta x}\,sin alpha x} folgt {displaystyle int _{0}^{infty }{frac {sin alpha x}{1+e^{eta x}}}\,dx=-sum _{k=1}^{infty }(-1)^{k}int _{0}^{infty }e^{-keta x}\,sin alpha x\,dx}.

Und das ist {displaystyle -sum _{k=1}^{infty }(-1)^{k}\,{frac {alpha }{k^{2}eta ^{2}+alpha ^{2}}}={frac {1}{2alpha }}-{frac {1}{2}}sum _{k=-infty }^{infty }(-1)^{k}\,{frac {alpha }{k^{2}eta ^{2}+alpha ^{2}}}={frac {1}{2alpha }}-{frac {pi }{2eta }};{ ext{csch}}left({frac {alpha pi }{eta }} ight)}.

 

2.5Bearbeiten

{displaystyle int _{0}^{infty }e^{-ax}\,sin bx\,dx={frac {b}{a^{2}+b^{2}}}}

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2.6Bearbeiten

{displaystyle int _{0}^{infty }e^{-ax}\,{frac {sin bx}{x}}\,dx=arctan left({frac {b}{a}} ight)qquad { ext{Re}}(a)geq |{ ext{Im}}(b)|quad ,quad {frac {b}{a}} eq pm i}

Beweis für a>b>0

Aus der Reihenentwicklung {displaystyle sin bx=sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k+1)!}}\,(bx)^{2k+1}}

folgt {displaystyle e^{-ax}\,{frac {sin bx}{x}}=sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,x^{2k}\,e^{-ax}}.

Also ist {displaystyle int _{0}^{infty }e^{-ax}\,{frac {sin bx}{x}}\,dx=sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,int _{0}^{infty }x^{2k}\,e^{-ax}\,dx}

{displaystyle =sum _{k=0}^{infty }{frac {(-1)^{k}}{(2k+1)!}}\,b^{2k+1}\,{frac {(2k)!}{a^{2k+1}}}=sum _{k=0}^{infty }{frac {(-1)^{k}}{2k+1}}\,left({frac {b}{a}} ight)^{2k+1}=arctan left({frac {b}{a}} ight)}.

 

3.1Bearbeiten

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

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