0
$\begingroup$

This question already has an answer here:

Evaluation of $$\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx$$

Let $$I = \int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx = \int^{1}_{0}\frac{(1-x)+\ln x}{(1-x)\ln x}dx$$

Now How can i solve after that , Help required, Thanks

$\endgroup$

marked as duplicate by Ng Chung Tak, Community Nov 25 '16 at 17:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Is there any reason to expect a reasonably nice closed form? $\endgroup$ – Daniel Fischer Nov 25 '16 at 14:22
  • $\begingroup$ To Daniel Fischer i don,t have any idea,it may be. $\endgroup$ – juantheron Nov 25 '16 at 14:23
  • 6
    $\begingroup$ See the definition of Euler–Mascheroni constant. $\endgroup$ – Mc Cheng Nov 25 '16 at 14:27
2
$\begingroup$

It is actually know that this integral is equal to the Euler Mascheroni constant.

$$\gamma=\int _0^1\left ({1\over1-x}+{1\over\ln x}\right)dx\approx0.57721566490153286060651209008240243104215933593992$$

$\endgroup$

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