Consider, for example, the Beta function:
$$ B(\alpha,1+\alpha) = \frac{\Gamma(\alpha)\Gamma(1+\alpha)}{\Gamma(1+2\alpha)} = \int_0^1 dt\ t^\alpha (1-t)^{\alpha-1}. $$
Expanding the integrated result in small $\alpha$, there is a leading term $1/\alpha$ that is divergent. However, if one tries to expand the integrand and then perform the integral, this term is "lost" and instead the leading contribution is a divergent integral:
$$ \int_0^1 dt\ \frac{1}{1-t}.$$
Now, is there a smart way to recover the leading $1/\alpha$ term by expanding before integration?