Integral (log and exp) 

Question (Someone asked me for help on this integral and I couldn't figure it out myself.)

$$ \int_{-∞}^∞ log(1+ae^{-t^2})dt  $$

Even taking the Taylor series such that $log(1+ae^{-t^2})$ ~ $log(ae^{-t^2})$
Integral of $Log(a)$ + $Log(e^{-t^2})$
Integral of $Log(a)$ + $-t^2$ ...
still doesn't converge at infinity.
Any ideas?

If $|a| \lt 1$ and using the fact that
$$\int_{-\infty}^{\infty} dt \, e^{-k t^2} = \sqrt{\frac{\pi}{k}} $$
we, get, using the Taylor expansion of the log:
$$\sqrt{\pi} \sum_{k=1}^{\infty} \frac{(-a)^{k+1}}{k^{3/2}} = \sqrt{\pi} \operatorname{Li}_{3/2}(-a)$$
