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Consider the function $$f_\epsilon(x):=\int_0^1 \frac{(1-z)^x-1}{\log(1-\epsilon z)}\, dz,$$ defined for a parameter $\epsilon \in (0,1]$ and $x \geq 0$. When $\epsilon=1$, this is $\log(1+x)$. One way to see this is to take the derivative of the function with respect to $x$ and see that the resulting integral evaluates to $1/(1+x)$, and moreover notice that $f_1(0)=0$.

In general, one can express the function as $$f_\epsilon(x)=\frac{1}{\epsilon}\log(1+x)+E_\epsilon(x).$$ It can be observed that, as $x$ tends to infinity, $E_\epsilon(x)$ converges to a constant depending on $\epsilon$. A plot of $C(\epsilon):=\lim_{x \to \infty} E_\epsilon(x)$, as a function of $\epsilon$ is given below: enter image description here And here is a plot of $\epsilon C(\epsilon)$, which seems to converge to the Euler–Mascheroni constant as $\epsilon\to 0$: enter image description here What is the value of $C(\epsilon)$? Can it be expressed in terms of common special functions (such as logarithmic integral)?

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Let us calculate $f_{\epsilon}(x) - \frac{1}{\epsilon}\log(1 + x)$ using your formula for $\log(1 + x)$ as $f_1(x)$(I didn't checked it but believe that it is correct):

$f_\epsilon(x) - \frac{1}{\epsilon}\log(1 + x) = \int_0^1 ((1 - z)^x - 1) \frac{-\log(1 - z) + \frac{1}{\epsilon}\log(1 - \epsilon z)}{\log(1 - z)\log(1 - \epsilon z)}dz$

It is easy to see using Taylor expansion that $K(\epsilon, z) = \frac{-\log(1 - z) + \frac{1}{\epsilon}\log(1 - \epsilon z)}{\log(1 - z)\log(1 - \epsilon z)}$ is $O(1)$ near $0$ for fixed $\epsilon$ and that it is integraible from $0$ to $1$. From these two facts it is easy to see that

$\int_0^1 (1 - z)^x \frac{-\log(1 - z) + \frac{1}{\epsilon}\log(1 - \epsilon z)}{\log(1 - z)\log(1 - \epsilon z)}dz$

goes to zero as $x$ goes to infinity(near $0$ we estimate $(1 - z)^x$ as $1$ and outside of neibourhood of $0$ it tends to $0$). So $C_\epsilon$ is equal to

$\int_0^1 K(\epsilon, z)dz = \lim\limits_{\delta \to \infty}-\int_\delta^1 \frac{1}{\log(1 - \epsilon z)}dz + \frac{1}{\epsilon}\int_\delta^1 \frac{1}{\log(1 - z)}dz = \frac{1}{\epsilon}\lim\limits_{\delta \to \infty} \int_\delta^1 \frac{1}{\log(1 - z)}dz - \int_{\epsilon \delta}^\epsilon \frac{1}{1 - \log(1 - z)}dz = \lim\limits_{\delta \to 0}\frac{1}{\epsilon}\int_{\epsilon}^1 \frac{1}{\log(1 - z)}dz - \frac{1}{\epsilon}\int_{\epsilon \delta}^\delta \frac{1}{\log(1 - z)}dz$

and again from the Taylor expansion of $\log$ it is easy to see that this limit is equal to

$\frac{1}{\epsilon}(li(1 - \epsilon) - \log(\epsilon))$

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