Asymptotic Expansion of integral with logarithm

Question

I want to show that as $\epsilon \to 0$ : $$ \int_0 ^1 \frac{\ln(x)}{x+\epsilon}= -\frac{1}{2} \ln ^2 \left( \frac{1}{\epsilon} \right) -\frac{\pi^2}{6} + \epsilon \left( 1-\frac{\epsilon}{4} + \frac{\epsilon^2}{9} - \frac{\epsilon^3}{16} + ... \right). $$

I started by splitting the range of integration to: $\overbrace{\int_0 ^\delta}^{:=I_1} + \overbrace{\int_\delta ^ 1}^{:=I_2}$ where $\epsilon \ll \delta \ll 1 $.

To calculate $I_2$, I used a taylor expansion around $\epsilon=0$ to obtain $$ I_2= -\frac{1}{2} \ln ^2 \delta - \epsilon \left( \frac{\ln \delta +1 - \delta}{\delta} \right) . $$

As for $I_1$, I cannot use taylor expansion of $\ln (x)$, because it doesn't have one, so what can I do to approximate it?

Answer 1

This is too long for a comment.

$$\int \frac{\log(x)}{x+\epsilon}=\text{Li}_2\left(-\frac{x}{\epsilon }\right)+\log (x) \log \left(1+\frac{x}{\epsilon }\right)$$ where appears the polylogarithm function. $$\int_0 ^1 \frac{\log(x)}{x+\epsilon}=\text{Li}_2\left(-\frac{1}{\epsilon }\right)\qquad \text{if} \qquad \Re(\epsilon )>0\lor \Re(\epsilon )\leq -1\lor \epsilon \notin \mathbb{R}$$ the development of which being exactly your rhs.

Edit

Looking at DLMF, equation $25.12.4$ gives $$\text{Li}_2\left(\frac{1}{z}\right)+\text{Li}_2(z)=-\frac{\pi^2}6-\frac 12\log^2(-z)$$ and, by definition (equation $25.12.1$) $$\text{Li}_2(z)=\sum_{n=1}^\infty \frac{z^n}{n^2}$$ making $$\text{Li}_2\left(-\frac{1}{z}\right)=-\frac{\pi^2}6-\frac 12\log^2(z)-\sum_{n=1}^\infty (-1)^n\frac{z^n}{n^2}$$ which is your expansion.

You also could find here other interesting relations between $\text{Li}_2(z)$ and $\text{Li}_2\left(\frac{1}{z}\right)$.