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Spring 2021 - Problem 10

analytic continuation

Define

f(z)=∫01tzet−1 dt,z∈C, Re⁡z>0.f\p{z} = \int_0^1 \frac{t^z}{e^t - 1} \,\diff{t}, \quad z \in \C,\ \Re{z} > 0.

Show that ff is an analytic function in {z∈C | Re⁡z>0}\set{z \in \C \st \Re{z} > 0} and that it admits a meromorphic continuation f^\hat{f} to the region {z∈C | Re⁡z>−1}\set{z \in \C \st \Re{z} > -1}. Compute the residue of f^\hat{f} at z=0z = 0.
Solution.

Notice that because 1+t≤et1 + t \leq e^t, we have t≤et−1t \leq e^t - 1, so

∣tzet−1∣≤tRe⁡zt=tRe⁡z−1∈L1([0,1]),\abs{\frac{t^z}{e^t - 1}} \leq \frac{t^{\Re{z}}}{t} = t^{\Re{z}-1} \in L^1\p{\br{0,1}},

since Re⁡z−1>−1\Re{z} - 1 > -1. Thus, ff is well-defined, and we may also apply dominated convergence to see that ff is continuous. Hence, given any closed γ⊆{Re⁡z>0}\gamma \subseteq \set{\Re{z} > 0}, we have a=inf⁡z∈γRe⁡z>0a = \inf_{z \in \gamma} \Re{z} > 0, and so

∫γ∫01∣tzet−1∣ dt dz≤∣γ∣∫01ta−1 dt<∞.\int_\gamma \int_0^1 \abs{\frac{t^z}{e^t - 1}} \,\diff{t} \,\diff{z} \leq \abs{\gamma} \int_0^1 t^{a - 1} \,\diff{t} < \infty.

Thus, we may apply Fubini's theorem to get

∫γf(z) dz=∫011et−1∫γtz dz dt=0,\int_\gamma f\p{z} \,\diff{z} = \int_0^1 \frac{1}{e^t - 1} \int_\gamma t^z \,\diff{z} \,\diff{t} = 0,

since the inner integral is 00 for every t∈[0,1]t \in \br{0, 1}. Thus, by Morera's theorem, ff is holomorphic on {Re⁡z>0}\set{\Re{z} > 0}. For the second claim, we integrate by parts to see

f(z)=∫01tz−1⋅tet−1 dt=tzz(et−1)∣01+1z∫01tz⋅et(t−1)+1(et−1)2 dt=1z(e−1)+1z∫01tz[et(t−1)+1](et−1)2 dt.\begin{aligned} f\p{z} &= \int_0^1 t^{z-1} \cdot \frac{t}{e^t - 1} \,\diff{t} \\ &= \left. \frac{t^z}{z\p{e^t - 1}} \right\rvert_0^1 + \frac{1}{z} \int_0^1 t^z \cdot \frac{e^t\p{t - 1} + 1}{\p{e^t - 1}^2} \,\diff{t} \\ &= \frac{1}{z\p{e - 1}} + \frac{1}{z} \int_0^1 \frac{t^z\br{e^t\p{t - 1} + 1}}{\p{e^t - 1}^2} \,\diff{t}. \end{aligned}

and the integral term is holomorphic on {Re⁡z>−1}\set{\Re{z} > -1} by the same argument as before. Indeed, notice

et(t−1)+1≤(1+t)(t−1)+1=t2.e^t\p{t - 1} + 1 \leq \p{1 + t}\p{t - 1} + 1 = t^2.

Thus,

∣tz[et(t−1)+1](et−1)2∣≤tRe⁡zt2t2=tRe⁡z\abs{\frac{t^z\br{e^t\p{t - 1} + 1}}{\p{e^t - 1}^2}} \leq \frac{t^{\Re{z}} t^2}{t^2} = t^{\Re{z}}

which is integrable on Re⁡z>−1\Re{z} > -1. Finally, the residue at z=0z = 0 is given by

Res⁡(f;0)=1e−1+∫01et(t−1)+1(et−1)2 dt=1e−1−tet−1∣01=1.\begin{aligned} \Res{f}{0} &= \frac{1}{e - 1} + \int_0^1 \frac{e^t\p{t - 1} + 1}{\p{e^t - 1}^2} \,\diff{t} \\ &= \frac{1}{e - 1} - \left. \frac{t}{e^t - 1} \right\rvert_0^1 \\ &= 1. \end{aligned}