Consider the function defined for s∈(1,∞) by
f(s)=∫0∞ex−1xs−1dx.
Show that f has an analytic continuation to {s∈C∣Res>0,s=1} with a simple pole at s=1. Compute the residue at s=1.
Solution.
We have as−1=e(s−1)loga for a>0. We split f up into
f(s)=∫01ex−1xs−1dx+∫1∞ex−1xs−1dx=:I1(s)+I2(s).
Notice that
I2(s)=(∫1∞n=0∑∞ex−1(logx)ndx)(s−1)n=n=0∑∞(∫1∞ex−1(logx)ndx)(s−1)n,
by the monotone convergence theorem, since each term in the sum is non-negative. Since I2 has no singularities, this series expansion is valid on the entire complex plane, so I2 is entire.
It remains to show that I1 is analytic on the right-half plane except at s=1. Observe that
g(s):=I1(s)−s−11=∫01xs−1(ex−11−x1)dx.
Then by L'Hôpital's,
x→0lim(ex−11−x1)=x→0limx(ex−1)x−ex+1=x→0limex−1+xex1−ex=x→0limex+ex+xex−ex=−21,
so the integrand is continuous on [0,1], and so g is well-defined. Hence, the integral converges whenever s−1>1⟹s>0 and so g is analytic on Res>0 by differentiating under the integral sign. We then have the representation
I1(s)=g(s)+s−11,
so I1(s) is analytic on Res>0 except at s=1, and it has a simple pole at s=1, and so f(s) satisfies the same properties. To calculate the residue at s=1,
s→1lim(s−1)f(s)=s→1lim((s−1)g(s)+1+(s−1)I2(s))=1,
and this completes the proof.