Sorry, one more precision:
-\infty < x < \infty
So that,
\sum_{j=0}^{k-i}(...)\int e^{-ax^2}x^{2i+j} dx = \sum_{p=0}^{k-i}(...)\{\int_{-\infty}^{\infty} e^{-ax^2}x^{2(i+p)} dx + \int_{-\infty}^{\infty} e^{-ax^2}x^{2(i+p)+1} dx\}
given that j is either odd or even,i.e., j=2p or...