<Home>Qualifying Exams><Analysis>Fall 2019 - Problem 12

Fall 2019 - Problem 12

analytic continuation

Show that

F(z)=∫1∞tz1+t3 dtF\p{z} = \int_1^\infty \frac{t^z}{\sqrt{1 + t^3}} \,\diff{t}

is well-defined (by the integral) and analytic in {z∈C∣Re⁡z<12}\set{z \in \C \mid \Re{z} < \frac{1}{2}}, and admits a meromorphic continuation to the region {z∈C∣Re⁡z<32}\set{z \in \C \mid \Re{z} < \frac{3}{2}}.
Solution.

Let a<12a < \frac{1}{2}. Then if Re⁡z≤a\Re{z} \leq a,

∣tz1+t3∣=tRe⁡z1+t3≤ta1+t3≤Cta−32.\abs{\frac{t^z}{\sqrt{1 + t^3}}} = \frac{t^{\Re{z}}}{\sqrt{1 + t^3}} \leq \frac{t^a}{\sqrt{1 + t^3}} \leq Ct^{a-\frac{3}{2}}.

Since a−32<−1a - \frac{3}{2} < -1, this is integrable. Thus, the integral converges absolutely, so FF is well-defined. In particular, by dominated convergence, FF is continuous.

To show that it is analytic, let γ⊆{Re⁡z<12}\gamma \subseteq \set{\Re{z} < \frac{1}{2}} be a closed rectangle. By compactness, sup⁡z∈γRe⁡z≤12−ε\sup_{z \in \gamma} \Re{z} \leq \frac{1}{2} - \epsilon for some ε>0\epsilon > 0. Since the integral in FF converges absolutely on {Re⁡z≤12−ε}\set{\Re{z} \leq \frac{1}{2} - \epsilon} and RR is compact, we may use Fubini's theorem in the following:

∫γF(z) dz=∫1∞11+t3∫γtz dz dt=0,\int_\gamma F\p{z} \,\diff{z} = \int_1^\infty \frac{1}{\sqrt{1 + t^3}} \int_\gamma t^z \,\diff{z} \,\diff{t} = 0,

since the inner integral vanishes for all t≥1t \geq 1 by Cauchy's theorem. Thus, by Morera's theorem, FF is analytic in the half-plane {Re⁡z<12}\set{\Re{z} < \frac{1}{2}}. Finally, observe that by integration by parts,

(z−12)F(z)=∫1∞(z−12)tz1+t3 dt=∫1∞(z−12)tz−32⋅t3/21+t3 dt=tz−12t3/21+t3∣1∞−∫1∞3tz−12t1/22(1+t3)3/2 dt=tz+11+t3∣1∞−∫1∞3tz2(1+t3)3/2 dt=−12−32∫1∞tz(1+t3)3/2 dt.\begin{aligned} \p{z - \frac{1}{2}}F\p{z} &= \int_1^\infty \p{z - \frac{1}{2}} \frac{t^z}{\sqrt{1 + t^3}} \,\diff{t} \\ &= \int_1^\infty \p{z - \frac{1}{2}} t^{z-\frac{3}{2}} \cdot \frac{t^{3/2}}{\sqrt{1 + t^3}} \,\diff{t} \\ &= \left. \frac{t^{z-\frac{1}{2}}t^{3/2}}{\sqrt{1 + t^3}} \right\rvert_1^\infty - \int_1^\infty \frac{3t^{z-\frac{1}{2}}t^{1/2}}{2\p{1 + t^3}^{3/2}} \,\diff{t} \\ &= \left. \frac{t^{z+1}}{\sqrt{1 + t^3}} \right\rvert_1^\infty - \int_1^\infty \frac{3t^z}{2\p{1 + t^3}^{3/2}} \,\diff{t} \\ &= -\frac{1}{\sqrt{2}} - \frac{3}{2} \int_1^\infty \frac{t^z}{\p{1 + t^3}^{3/2}} \,\diff{t}. \end{aligned}

This is holomorphic for

Re⁡z−92<−1  ⟹  Re⁡z<72\Re{z} - \frac{9}{2} < -1 \implies \Re{z} < \frac{7}{2}

by the exact same argument as before, so FF extends holomorphically.