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Spring 2016 - Problem 7

calculation, residue theorem

Determine

∫0∞xa−1x+z dx,\int_0^\infty \frac{x^{a-1}}{x + z} \,\diff{x},

for 0<a<10 < a < 1 and Re⁡z>0\Re{z} > 0. Justify all manipulations.
Solution.

Set f(w)=wa−1w+zf\p{w} = \frac{w^{a-1}}{w + z} and for 0<ε<∣z∣<R0 < \epsilon < \abs{z} < R and α∈(0,π)\alpha \in \p{0, \pi}, let γε,R,α\gamma_{\epsilon,R,\alpha} be the boundary of

{z∈C∣ε<∣z∣<R and α<Arg⁡z<2π−α},\set{z \in \C \mid \epsilon < \abs{z} < R \text{ and } \alpha < \Arg{z} < 2\pi - \alpha},

where we take the standard branch of Arg⁡z\Arg{z} with branch cut on the positive real axis. On the arc CRC_R of radius RR, we have

∣∫CRf(w) dw∣=∣∫α2π−αRa−1ei(a−1)θRei(a−1)θ+ziReiθ dθ∣≤∫02πRaR−∣z∣ dθ=2πRaR−∣z∣→R→∞0,\begin{aligned} \abs{\int_{C_R} f\p{w} \,\diff{w}} &= \abs{\int_\alpha^{2\pi-\alpha} \frac{R^{a-1}e^{i\p{a-1}\theta}}{Re^{i\p{a-1}\theta} + z} iRe^{i\theta} \,\diff\theta} \\ &\leq \int_0^{2\pi} \frac{R^a}{R - \abs{z}} \,\diff\theta \\ &= \frac{2\pi R^a}{R - \abs{z}} \xrightarrow{R\to\infty} 0, \end{aligned}

since a<1a < 1, so this contribution tends to 00 uniformly in α\alpha. Similarly, on the arc CεC_\epsilon of radius ε\epsilon,

∣∫Cεf(w) dw∣≤2πεa∣z∣−ε→ε→0,\abs{\int_{C_\epsilon} f\p{w} \,\diff{w}} \leq \frac{2\pi \epsilon^a}{\abs{z} - \epsilon} \xrightarrow{\epsilon\to0},

since a>0a > 0, uniformly in α\alpha as well. Finally, in the limit as α→0\alpha \to 0, the contribution from the segment [εei(2π−α),Rei(2π−α)]\br{\epsilon e^{i\p{2\pi-\alpha}}, Re^{i\p{2\pi-\alpha}}} becomes

∫εRta−1ei(a−1)(2π−α)tei(2π−α)+zei(2π−α) dt→α→0e2πi(a−1)∫εRta−1t+z dt.\int_\epsilon^R \frac{t^{a-1} e^{i\p{a-1}\p{2\pi-\alpha}}}{te^{i\p{2\pi-\alpha}} + z} e^{i\p{2\pi-\alpha}} \,\diff{t} \xrightarrow{\alpha\to0} e^{2\pi i\p{a-1}} \int_\epsilon^R \frac{t^{a-1}}{t + z} \,\diff{t}.

Indeed, the integrand is bounded and [ε,R]\br{\epsilon, R} is compact, so we may apply dominated convergence to send α→0\alpha \to 0. ff only has one residue at w=−zw = -z, which is

Res⁡(f;−z)=(−z)a−1,\Res{f}{-z} = \p{-z}^{a-1},

which is well-defined since Re⁡z>0\Re{z} > 0, so −z-z avoids the positive real axis. Thus, by the residue theorem and sending α→0\alpha \to 0, ε→0\epsilon \to 0, and R→∞R \to \infty, we get

(1+e2πi(a−1))∫0∞xa−1x+z dx=2πi(−z)a−1  ⟹  ∫0∞xa−1x+z dx=2πi(−z)a−11+e2πi(a−1).\p{1 + e^{2\pi i\p{a-1}}} \int_0^\infty \frac{x^{a-1}}{x + z} \,\diff{x} = 2\pi i \p{-z}^{a-1} \implies \int_0^\infty \frac{x^{a-1}}{x + z} \,\diff{x} = \frac{2\pi i \p{-z}^{a-1}}{1 + e^{2\pi i\p{a-1}}}.