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

1周前 (05-04) 6次浏览

##### 1.1Bearbeiten
{displaystyle int _{0}^{infty }left({frac {sin x}{x}}right)^{2},e^{-2ax},dx=a,log left({frac {a}{sqrt {1+a^{2}}}}right)+operatorname {arccot} a}
ohne Beweis

##### 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} ,,,,,{text{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} ,,,,,{text{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^{beta x}}},dx={frac {1}{2alpha }}-{frac {pi }{2beta }};{text{coth}}left({frac {alpha pi }{beta }}right)}
Beweis

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

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

##### 2.4Bearbeiten
{displaystyle int _{0}^{infty }{frac {sin alpha x}{1+e^{beta x}}},dx={frac {1}{2alpha }}-{frac {pi }{2beta }};{text{csch}}left({frac {alpha pi }{beta }}right)}
Beweis

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

Und das ist {displaystyle -sum _{k=1}^{infty }(-1)^{k},{frac {alpha }{k^{2}beta ^{2}+alpha ^{2}}}={frac {1}{2alpha }}-{frac {1}{2}}sum _{k=-infty }^{infty }(-1)^{k},{frac {alpha }{k^{2}beta ^{2}+alpha ^{2}}}={frac {1}{2alpha }}-{frac {pi }{2beta }};{text{csch}}left({frac {alpha pi }{beta }}right)}.

##### 2.5Bearbeiten
{displaystyle int _{0}^{infty }e^{-ax},sin bx,dx={frac {b}{a^{2}+b^{2}}}}
ohne Beweis

##### 2.6Bearbeiten
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}}right)^{2k+1}=arctan left({frac {b}{a}}right)}.

##### 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}}right)qquad a>0,,,bin mathbb {R} ,,,{text{Re}}(s)>0}
ohne Beweis