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# Substitution

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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:)

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

This integral does not converge so the answer is infinite

07.01.2017
#5
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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.....

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

That is more like it :)

07.01.2017
#3
+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
+118617
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hi Daniel :)

1/(x(lnx^2)

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