It does converge. (Not absolutely, of course.)
Hello,
I'll try the proof I read a time ago...
Let
Integrate by parts with :
and . So we have and , and we'll take c=1. So
But we know by Taylor series that when x is near from 0.
So we can make a=0 since and have finite limits when
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See
By extension, the integrand is continuous for all x in [0,1] and finite. Thus the integral exists and it is finite value. So it doesn't intervene in the convergence or divergence of
(actually, you have to check this explanation...it's the sort of parts in which I am always wrong)
See
This obviously converges.
See
This converges by comparison with the Riemann integral, since :
.