Leading behavior of a logarithmic integral The integral
$$\int_0^1 du \frac{\ln u}{u^2+b^2}$$
is governed by the divergence at $u=0$. I'm interested in the leading behavior of the integral, and it seems reasonable to expand around $u=0$ (as it is typically done in Laplace's method for instance), but what should be done next?

Here's my attempt. First we start off by integrating by parts.
\begin{equation}
\int_0^1 \frac{\ln u}{u^2+b^2} du = \frac{1}{b} \arctan\left( \frac{u}{b}  \right) \ln u \bigg |_0^1 - \frac{1}{b} \int_0^1 \frac{\arctan\left(\frac{u}{b} \right)}{u} du.
\end{equation}
Assuming $b\neq0$, we can evaluate the limit
\begin{equation}
\lim_{u\rightarrow 1} \arctan\left( \frac{u}{b}  \right) \ln u   =0, \\
\lim_{u \rightarrow 0} \arctan\left( \frac{u}{b}  \right) \ln u  \text{ - undefined}.
\end{equation}
Applying L'Hôpital's rule to the second limit
\begin{equation}
\lim_{u \rightarrow 0} \frac{1}{b} \arctan\left( \frac{u}{b}  \right) \ln u = \lim_{u \rightarrow 0} \frac{1}{b}  \frac{\ln u}{\frac{1}{\arctan\left(\frac{u}{b} \right)}} =\frac{1}{b}  \lim_{u \rightarrow 0} \frac{ \frac{1}{u}}{-\left(\frac{1}{\arctan\left(\frac{u}{b} \right)} \right)^2  \frac{b}{u^2+b^2} }\\
= -\frac{1}{b^2} \lim_{u\rightarrow 0} \frac{\arctan \left( \frac{u}{b} \right)^2 (u^2+b^2)}{u}=0,
\end{equation}
since $\arctan(x) \sim x$, for small $x$.
We see that 
\begin{equation}
\int_0^1 \frac{\ln u}{u^2+b^2} du = - \frac{1}{b} \int_0^1 \frac{\arctan\left( \frac{u}{b}\right)}{u} du.
\end{equation}
Since you asked for a power series, taylor expanding gives us the following result:
\begin{equation}
-\frac{1}{b} \int_0^1  \frac{\arctan\left( \frac{u}{b} \right)}{u} = -\frac{1}{b}\int_0^1 \frac{1}{u}\left( \left(\frac{u}{b}\right) - \frac{1}{3}\left(\frac{u}{b}\right)^3 + \frac{1}{5} \left(\frac{u}{b}\right)^5 + O(u^7) \right) \\
=-\frac{1}{b^2}\int_0^1 \left(1 - \frac{1}{3b^2} u^2 + \frac{1}{5b^4} u^4 + O(u^6) \right)\\
= - \frac{1}{b^2} \left(1 - \frac{1}{3^2 b^2} + \frac{1}{5^2 b^4} + O\left(\frac{1}{b^6} \right) \right)
= - \frac{1}{b^2} \sum_{i=0}^{\infty} \frac{(-1)^{i}}{(2i+1)^{2i} b^{2i}}
\end{equation}
Note that the $b\rightarrow0$ divergence makes sense, since the initial integral is also divergent for $b=0$. I suspect that this could probably be reexpressed in terms of the zeta function for the case of $b=1$. 
I hope this helps.
