I need to evaluate the following integral in the limit $\kappa \ll 1$
$$\int_0^\infty exp(-\kappa t) f(t)\, dt,$$
where
$$f(x) = (1+x)(1-2x)\frac{u(x) \ln(u(x))}{u(x)^2 - 1},$$
$$u(x) = \frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}+1}.$$
The straight forward way to do a Taylor expansion fails, since the occurring integrals are divergent. I don't know any method to deal with this type of integrals. I know that the occurring series should look something like
$$ a + (b_1 + b_2\cdot \ln(\kappa)) \cdot \kappa + (c_1 + c_2 \cdot \ln(\kappa) + c_3 \cdot \ln(\kappa)^2) \cdot \kappa + \mathcal{O}(\kappa^3) .$$
I was able to obtain the 'first order' term by expanding $f(x)$ wrt. $1/x$ around $0$. However, I am not sure why this worked.
$$f(x) \sim \left(-\ln(2)-\frac{1}{2}\cdot \ln(x)\right)\cdot x-\frac{1}{4}, \quad x \to \infty$$
Nonetheless, the terms after have the form
$$\frac{\ln(x)}{x}, \qquad \frac{1}{x}, \qquad \ldots$$
and, thus, yield $Ei(y)$ for $y \downarrow 0$, which is divergent.
The first terms in the series should be $$-\dfrac{1}{4\cdot \kappa}+\frac{1}{2}\cdot\dfrac{-1+\gamma_E+\ln\left(\dfrac{\kappa}{4}\right)}{\kappa^2} + \ldots$$ and I am interested in the higher terms, i.e. $\kappa^0$ and $\ln$'s.
Edit: I have followed user619894's comment and done the expansion as suggested. For the $\int_0^L$ integration I get $$\frac{1}{8} \left(2 \left(\sqrt{L (L+1)}-2 \sqrt{L^3 (L+1)}\right) \sinh ^{-1}\left(\sqrt{L}\right)+(L-1) L+3 \coth ^{-1}\left(\sqrt{\frac{1}{L}+1}\right)^2\right)$$ While for $\int_\Lambda^\infty$ I get $$(1/(32 k^2))e^{-k L} (-16 - 8 k + 16 e^{k L} Ei[-k L] + 3 e^{k L} k^2 \Gamma[0, k L] - 16 Log[4] - 16 k L Log[4] + 3 e^{k L} k^2 \Gamma[0, k L] Log[16] - 16 Log[L] - 16 k L Log[L] + 6 e^{k L} k^2 \Gamma[0, k L] Log[L] + 6 e^{k L} k^2 \text{MeijerG}[{{}, {1, 1}}, {{0, 0, 0}, {}}, k L])$$ Afterward, I expanded integral 1 wrt. $L\gg1$ and integral 2 wrt. $\kappa L \ll 1$. Adding then yields $$\frac{\log (k)+\gamma -1-2 \log (2)}{2 k^2}-\frac{1}{4 k}+\frac{1}{64} \left(6 \log ^2(k)-24 \log (2) \log (k)+12 \gamma \log (k)-6 \log (k)+6 \gamma ^2+\pi ^2-6 \gamma +3+24 \log ^2(2)-24 \gamma \log (2)+12 \log (2)\right) + \mathcal{O}(k) + o(L^{-1})$$