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
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1$\begingroup$ Is there any reason to expect a reasonably nice closed form? $\endgroup$ – Daniel Fischer♦ Nov 25 '16 at 14:22
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$\begingroup$ To Daniel Fischer i don,t have any idea,it may be. $\endgroup$ – juantheron Nov 25 '16 at 14:23
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6$\begingroup$ See the definition of Euler–Mascheroni constant. $\endgroup$ – Mc Cheng Nov 25 '16 at 14:27
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It is actually know that this integral is equal to the Euler Mascheroni constant.
$$\gamma=\int _0^1\left ({1\over1-x}+{1\over\ln x}\right)dx\approx0.57721566490153286060651209008240243104215933593992$$
answered Nov 25 '16 at 17:08
