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I was reading the list of integrals from Wikipedia, and I've created a variation of the integral $$\int \log_a(x)dx.$$ I show my example as a definite integral. I would like to know if is it possible to find an approximation of the integal of such logarithm in my example

$$\int_0^1\frac{\log x}{\log(\gamma+\psi(x+1))}dx,\tag{1}$$ where $\gamma$ is the Euler-Mascheroni constant and $\psi(s)$ the digamma function.

Question. How can we justify an approximation of $(1)$? Preferably using analysis and some asymptotic property of functions in the integrand. If this integral was in the literature and you need answer it as a reference request feel free to do it (then I am going to search and try understand such literaure). Many thanks.

See the integration that I've encoded and my integrand with Wolfram Alpha online calculator

int log(x)/log(EulerGamma+Digamma(x+1))dx, from x=0 to 1

plot log(x)/log(EulerGamma+Digamma(x+1)), from x=0 to 1

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  • $\begingroup$ Why don't you take $\int_0^1 \frac{\log (x+1)}{\log (x+2)}\,dx$ ? Looking simple, isn't it ? $\endgroup$ – Claude Leibovici Jan 9 '18 at 10:25
  • $\begingroup$ Yes I think that your example is good. Maybe was my blindness asking a more artificious question. Many thanks for your comment @ClaudeLeibovici $\endgroup$ – user243301 Jan 9 '18 at 10:33
  • $\begingroup$ Imagination and curiosity are very good qualities ! Continue that way. But, if I may, life is already difficult with even looking simple stuff. Cheers. $\endgroup$ – Claude Leibovici Jan 9 '18 at 10:45
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This is not an answer to the question.

Since I wrote it in a comment, let us consider something as simple as $$I=\int_0^1 \frac{\log (x+1)}{\log (x+2)}\,dx$$ which, more than likely, does not show a closed form.

The numerical integration leads to $\approx \color{red} {0.402665}$.

To approximate the result, I used $[n,2]$ Padé approximants built around $x=0$ ; these are "easy" to obtain. For example, for $n=2$, we have $$\frac{\log (x+1)}{\log (x+2)}=\frac{\frac{x}{\log (2)}+\frac{x^2}{2 \log (2)} } {1+x \left(1+\frac{1}{2 \log (2)}\right)+x^2 \left(\frac{1}{6}+\frac{3}{8 \log (2)}\right) }$$ Plot the two functions to see how close they are.

Integration does not make any problem and the result evaluates $\approx \color{red} {0.402}705$ which is not too bad.

Doing the same with $n=3$ would lead to $\approx \color{red} {0.4026}52$.

Now, I give up !

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  • $\begingroup$ Many thanks for your example. I presume that you are saying the Padé approximants showed, for example, from Wikipedia. $\endgroup$ – user243301 Jan 9 '18 at 11:37
  • $\begingroup$ @user243301. I am a fantic user of Padé approximants (the Wikipedia page is good). They are so better than series ! I hope you liked this one. Cheers. $\endgroup$ – Claude Leibovici Jan 9 '18 at 11:58
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    $\begingroup$ @user243301. I bet that making the expansion around $x=\frac 12$ would give better. It does ! for $n=2$, I get $\color{red} {0.40266}3$ $\endgroup$ – Claude Leibovici Jan 9 '18 at 12:06
  • $\begingroup$ Many thanks I read (the second paragraph of Wikipedia), about this approximation by rational functions versus Taylor expansion. $\endgroup$ – user243301 Jan 9 '18 at 12:14

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