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Frage zur Substitution: Integral von 2 bis +inf: 1/(x(lnx^2)dx. Folgende Substitution: u=lnx --> mit dieser Substitution führt meine Uni Professorin an: Integral von ln2 bis +inf: 1/u^2. wieso ist 1/u^2 das selbe wie 1/x(lnx^2) mit dieser substitution?

 07.01.2017
 #1
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Frage zur Substitution: Integral von 2 bis +inf: 1/(x(lnx^2)dx. Folgende Substitution: u=lnx --> mit dieser Substitution führt meine Uni Professorin an: Integral von ln2 bis +inf: 1/u^2. wieso ist 1/u^2 das selbe wie 1/x(lnx^2) mit dieser substitution?

 

Please understand that I only speak English:)

 

21x(ln(x2))dx=212xlnxdx letu=lnxx=eudxdu=eudx=eudu =212xlnxdx=212euu×eudu=212udu=1221udu=12[lnu]2

 

This integral does not converge so the answer is infinite

 07.01.2017
 #5
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There is a tine error with the first one, the integral should change to   ln2 to infinity.

BUT it is still divergent. :)

Melody  08.01.2017
 #2
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You must mean somthing different - Maybe.....

 

21x(ln2(x))dx letu=lnxx=eudxdu=eudx=euduand when x=2u=ln2 when x=u=ln()= ln21x(ln2(x))dx=ln21euu2×eudu=ln21u2du=ln2u2du=[u11]ln2=[1u]ln2=01ln2=1ln2

 

  That is more like it :)

 07.01.2017
 #3
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+5

Thanks Melody. Super fast and super clean answere, cheers for that :) 

i did mean the first case you calculated, and now i noticed that my (sometimes) confused Professor just messed up the writing. She should have written: 1/xln^2x, she did write: 1/xln(x^2). Did just notice that 'cause you said "does not converge" :) 

Keep up the good work, Daniel 

 07.01.2017
 #4
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hi Daniel :)

 

Your teacher wrote

1/(x(lnx^2)

 

Which need a closing bracket to start with

1/(x(lnx^2))

 

but yes she should have written 

1/(x(ln^2x) 

 

OR better still in my view would be

1/(x(lnx)^2)

 

I knew my interpretation was wrong because I did not get the solution that your teach got.  :)

 08.01.2017

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