+0  
 
0
399
5
avatar

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
avatar+102763 
0

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:)

 

\(\quad \displaystyle\int_2^\infty \frac{1}{x(ln(x^2))}\:dx\\ =\displaystyle\int_2^\infty \frac{1}{2xlnx}\:dx\\~\\ \qquad\qquad let \\ \qquad\qquad u=lnx\\ \qquad\qquad x=e^u\\ \qquad\qquad \frac{dx}{du}=e^u\\ \qquad\qquad dx=e^u du\\~\\ =\displaystyle\int_2^\infty \frac{1}{2xlnx}\:dx\\ =\displaystyle\int_2^\infty \frac{1}{2e^uu}\times e^u\:du\\ =\displaystyle\int_2^\infty \frac{1}{2u}\:du\\ =\dfrac{1}{2}\displaystyle\int_2^\infty \frac{1}{u}\:du\\ =\dfrac{1}{2}\left[ ln\:u \right]_2^\infty\)

 

This integral does not converge so the answer is infinite

 07.01.2017
 #5
avatar+102763 
0

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
avatar+102763 
0

 

 

You must mean somthing different - Maybe.....

 

\(\quad \displaystyle\int_2^\infty \frac{1}{x(ln^2(x))}\:dx\\~\\ \qquad\qquad let \\ \qquad\qquad u=lnx\\ \qquad\qquad x=e^u\\ \qquad\qquad \frac{dx}{du}=e^u\\ \qquad\qquad dx=e^u du\\ \qquad\qquad and\\ \qquad\qquad \text{ when } x=2 \quad u=ln2\\ \qquad\qquad \text{ when } x=\infty \quad u=ln(\infty) =\infty\\~\\ \quad \displaystyle\int_{ln2}^\infty \frac{1}{x(ln^2(x))}\:dx\\ =\displaystyle\int_{ln2}^\infty \frac{1}{e^uu^2}\times e^u\:du\\ =\displaystyle\int_{ln2}^\infty \frac{1}{u^2}\:du\\ =\displaystyle\int_{ln2}^\infty u^{-2}\:du\\ =\left[ \frac{u^{-1}}{-1} \right]_{ln2}^\infty\\ =\left[ \frac{-1}{u} \right]_{ln2}^\infty\\ =0--\dfrac{1}{ln2}\\ =\dfrac{1}{ln2} \)

 

  That is more like it :)

 07.01.2017
 #3
avatar
+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
avatar+102763 
0

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

18 Benutzer online

avatar
avatar