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)?


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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