Question 363011
a. Let {{{w = sqrt(x)-pi}}}.  Then {{{x = (w+pi)^2}}}, and {{{dx = 2sqrt(x)dw}}}.  Then{{{ (root(3,x)/(sqrt(x)-pi))dx  = ((w+pi)^(5/3)/w)dw}}}.  Now 
{{{(w+pi)^(5/3)/w > = w^(5/3)/w  =  w^(2/3)}}}.  It is enough to check the integral for {{{x>=pi^2}}} for divergence/convergence (Why?).  We have then  {{{ int (root(3,x)/(sqrt(x)-pi), dx, pi^2, infinity ) = int((w+pi)^(5/3)/w, dw, 0, infinity) >= int(w^(2/3), dw, 0,infinity) }}}
But the last integral has infinite value, and thus, the given integral itself has infinite value.  The integral is divergent.

b.{{{ int ( e^(-x^2), dx, 1, infinity )<= int(e^(-x),dx, 1, infinity) = 1/e }}}.  Therefore the given integral is convergent.