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In a previous question I asked about convergence of the integral, now I would like to see how to evaluate it. I believe the result is $\gamma$.

$$\int_0^1 \frac{1}{1-x} + \frac{1}{\log(x)} dx$$

I have tried the substitution $t = \log(x)$ which transforms the integral to

$$\int_{- \infty}^0 \left( \frac{1}{1-e^{t}} + \frac{1}{t} \right) e^{t} dt$$

I can then clean up the bounds a little

$$\int_0^{\infty} \frac{1}{e^t-1} - \frac{e^{-t}}{t} dt$$

at this point I am not sure what to do, I am aiming to expand something out to an infinite series, and perhaps cancel one of the two terms inside the integral with one of the parts.

Arctic Char
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  • The answer that introduces a $s$ parameter to solve the integral does prove (a generalization of) what i want to prove, but I find it a very ingenious and difficult argument. Can we approach this integral in a more basic way? (I am a relative beginner to integrals) –  Aug 17 '21 at 14:08
  • What about the accepted answer? It does not use special functions or difficult methods (except the justification of interchanging limit and integral perhaps, but that's it). – vitamin d Aug 17 '21 at 14:12
  • @vitamind. Yoou are right and I am stupid ! I delete my comment. Cheers :-( – Claude Leibovici Aug 17 '21 at 14:18
  • If you use a series expansion around $x=1$ up to $O\left((x-1)^{101}\right)$ you will only obtain $0.576988$ that is to say a relative arror of $0.04$%. Up to $O\left((x-1)^{1001}\right)$ would give a relative arror of $0.002$% – Claude Leibovici Aug 17 '21 at 14:28
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    You can use Approach0 to search for mathematical expressions in MSE. You can note this for future use. The syntax is pretty much the same, even easier. – Sarvesh Ravichandran Iyer Aug 19 '21 at 22:29
  • @TeresaLisbon that is great! Thanks!\ –  Aug 20 '21 at 08:04
  • @river Good to be of help. – Sarvesh Ravichandran Iyer Aug 20 '21 at 10:58

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