If $p=\frac{7}{8}$ then what should be the value of $\displaystyle\int\limits_1^n \frac{g(x)}{x^{p+1}} \mathrm dx $ when $$g(x) = x \log x \quad \text{or} \quad g(x) = \frac{x}{\log x}? $$

Wondering which way to proceed?

  1. an algebraic substitution,
  2. partial fractions,
  3. integration by parts, or
  4. reduction formulae.

Please don't suggest something like ("Learn basic Calculus first" etc).
Kindly help by solving if possible because I'm out of touch with calculus for nearly 15 yrs.

  • $\begingroup$ Yes U r correct...I meant $\int_1^ng(x)/x^{p+1}dx$ $\endgroup$ – Arnab Dutta Jun 3 '13 at 12:43
  • $\begingroup$ For the second one, $g(x)=x/\log x$, the integrand is not integrable due to the behaviour at the bound $1$. $\endgroup$ – Julien Jun 3 '13 at 12:58
  • 1
    $\begingroup$ @Sigur: you have edited it wrongly..please revert it back $\endgroup$ – Arnab Dutta Jun 3 '13 at 13:09

$$\frac{g(x)}{x^{p+1}}=\frac{x\log x}{x^{p+1}}=\frac{\log x}{x^p}$$

By parts:

$$u=\frac1{x^p}\;,\;\;u'=-\frac p{x^{p+1}}\\v'=\log x\;,\;\;v=x\log x-x$$


$$\int\limits_1^n\frac{\log x}{x^p}dx=\left.\left(\frac{\log x}{x^{p-1}}-\frac1{x^{p-1}}\right)\right|_1^n+p\int\limits_1^n\frac{\log x}{x^p}dx-p\int\limits_1^n\frac1{x^p}dx\ldots\ldots$$

  • $\begingroup$ Thanks @DonA but Whats the ultimate value? $\endgroup$ – Arnab Dutta Jun 3 '13 at 12:48
  • $\begingroup$ It would be more direct to differentiate $\log x$ in the integration by parts. When $p\neq 1$. $\endgroup$ – Julien Jun 3 '13 at 12:52
  • $\begingroup$ You calculate it, @ArnabDutta . Do some effort, pass from side to side...if you get stuck somewhere write back. $\endgroup$ – DonAntonio Jun 3 '13 at 12:52
  • $\begingroup$ Julien is right: it'd be better (well, or simpler, easier...) to put $\,u=\log x\ldots$ etc. $\endgroup$ – DonAntonio Jun 3 '13 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.