Integrate dx/(xlogx): Limits 1 to n - Result

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SUMMARY

The integral \(\int \frac{dx}{x \log x}\) from 1 to \(n\) does not converge. The substitution \(u = \ln x\) simplifies the integrand to \(\frac{du}{u}\), but the limits change to 0 to \(\ln n\). Evaluating the integral reveals that it diverges due to the behavior of the function \(\frac{1}{x \log x}\) approaching infinity as \(x\) approaches 1. Thus, the area under the curve is infinite, confirming the non-convergence of the integral.

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prasoonsaurav
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Is there any way to integrate
[tex]\int dx/(xlogx)[/tex]

within the limits 1 and n?

If yes what is the result?
 
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Assuming that log x means ln x (if not a constant is just introduced), if you make the substitution u= ln x the integrand simplifies to du/u. However the limits are 0 to ln n, and so ln u would have to be evaluated at 0, which means the integral does not converge.
 
Yes the integral doesn not converge . This can be seen without integrating - the function 1 / (x * ln x) goes to infinity at x = 1 , and also func is continuous in the interval .. so the area under the curve is infinite
 

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