I am trying to find an approximation to

$$
I = \int\_a^b \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-(x-\mu)^2/2 \sigma^2}\log(1+e^{-x}) \ \ dx.
$$
My attempt is as follows:

$$
\begin{align}
I &= \int\_a^b \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-(x-\mu)^2/2 \sigma^2} \left( \sum\_{i=1}^\infty \frac{e^{-ix}}{i} (-1)^{(i+1)} \right)\ dx\\
&= \sum\_{i=1}^\infty \frac{(-1)^{(i+1)}}{i}\int\_a^b \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-(x-\mu)^2/2 \sigma^2} e^{-ix} \ \ dx\\
&= \sum\_{i=1}^\infty \frac{(-1)^{(i+1)}}{i} k\_i \int\_a^b \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-(x-(\mu-i \sigma^2))^2/2 \sigma^2} \ \ dx,\\
\end{align}
$$
where 

$$
k\_i = e^{(\mu -i \sigma^2)^2-\mu^2}.
$$
The $k\_i$ increases exponentially with increasing $i$ and thus makes the sum divergent. I don't understand why this is happening although this sum should be finite (because I don't see any problem with the original integral).

P.S. I used MacLauren series in approximating natural logarithm.