Does the Integral from e to Infinity of 67/(x(ln(x))^3) Converge?

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Homework Statement



integral [from e to infinity of ] 67 / (x(ln(x))^3)

read as the integral from e to infinity of

67 divided by x times cubed lnx.


Homework Equations





The Attempt at a Solution





i know it converges,

but i got the value

67/2
 
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tnutty said:

Homework Statement



integral [from e to infinity of ] 67 / (x(ln(x))^3)

read as the integral from e to infinity of

67 divided by x times cubed lnx.

The Attempt at a Solution



i know it converges,

but i got the value

67/2
And I would agree with you...

Using the substitution u=\ln{x}, the integral

\int_e^\infty \frac{1}{x(\ln{x})^3} \, dx

becomes

-\frac{1}{2} \int_1^\infty \frac{1}{u^2} \, du.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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