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Spring 2014 - Problem 11

calculation, residue theorem

For each p∈(−1,1)p \in \p{-1, 1}, compute the improper Riemann integral

∫0∞xpx2+1 dx.\int_0^\infty \frac{x^p}{x^2 + 1} \,\diff{x}.
Solution.

Let f(z)=zpz2+1f\p{z} = \frac{z^p}{z^2 + 1} and for 0<ε<1<R0 < \epsilon < 1 < R, consider the contour γε,R\gamma_{\epsilon,R} given by

γε,R=[−R,−ε]∪{εeiθ∣0≤θ≤π}∪[ε,R]∪{Reiθ∣0≤θ≤R}.\gamma_{\epsilon,R} = \br{-R, -\epsilon} \cup \set{\epsilon e^{i\theta} \mid 0 \leq \theta \leq \pi} \cup \br{\epsilon, R} \cup \set{Re^{i\theta} \mid 0 \leq \theta \leq R}.

On the larger semicircle CRC_R,

∣∫CRf(z) dz∣≤∫0π∣iRp+1eipθeiθR2ei2θ+1∣ dθ≤Rp+1∫0π1R2−1 dθ=2πRp+1R2−1→R→∞0,\begin{aligned} \abs{\int_{C_R} f\p{z} \,\diff{z}} &\leq \int_0^\pi \abs{\frac{iR^{p+1}e^{ip\theta}e^{i\theta}}{R^2e^{i2\theta} + 1}} \,\diff\theta \\ &\leq R^{p+1} \int_0^\pi \frac{1}{R^2 - 1} \,\diff\theta \\ &= \frac{2\pi R^{p+1}}{R^2 - 1} \xrightarrow{R\to\infty} 0, \end{aligned}

since ∣p+1∣<2\abs{p + 1} < 2. On the smaller semicircle CεC_\epsilon,

∣∫Cεf(z) dz∣≤∫0π∣iεp+1eipθeiθε2ei2θ+1∣ dθ≤εp+1∫0π1ε2−1 dθ=2πεp+1ε2−1→ε→∞0,\begin{aligned} \abs{\int_{C_\epsilon} f\p{z} \,\diff{z}} &\leq \int_0^\pi \abs{\frac{i\epsilon^{p+1}e^{ip\theta}e^{i\theta}}{\epsilon^2e^{i2\theta} + 1}} \,\diff\theta \\ &\leq \epsilon^{p+1} \int_0^\pi \frac{1}{\epsilon^2 - 1} \,\diff\theta \\ &= \frac{2\pi \epsilon^{p+1}}{\epsilon^2 - 1} \xrightarrow{\epsilon\to\infty} 0, \end{aligned}

since ∣ε∣<1\abs{\epsilon} < 1, so the denominator is non-zero. On [−R,−ε]\br{-R, -\epsilon}, taking arguments in (−π2,3π2)\p{-\frac{\pi}{2}, \frac{3\pi}{2}}, we have

∫−R−εzpz2+1 dz=∫−R−εeplog⁡zz2+1 dz=∫−R−εep(log⁡x+iπ)x2+1 dz=eipπ∫−R−εxpx2+1 dx.\begin{aligned} \int_{-R}^{-\epsilon} \frac{z^p}{z^2 + 1} \,\diff{z} &= \int_{-R}^{-\epsilon} \frac{e^{p\log{z}}}{z^2 + 1} \,\diff{z} \\ &= \int_{-R}^{-\epsilon} \frac{e^{p\p{\log{x} + i\pi}}}{x^2 + 1} \,\diff{z} \\ &= e^{ip\pi} \int_{-R}^{-\epsilon} \frac{x^p}{x^2 + 1} \,\diff{x}. \end{aligned}

Next, by the residue theorem,

∫γε,Rf(z) dz=2πiRes⁡(f;i)=2πilim⁡z→i (z−i)zpz2+1=2πilim⁡z→izpz+i=πeipπ/2.\begin{aligned} \int_{\gamma_{\epsilon,R}} f\p{z} \,\diff{z} &= 2\pi i \Res{f}{i} \\ &= 2\pi i \lim_{z \to i}\,\p{z - i} \frac{z^p}{z^2 + 1} \\ &= 2\pi i \lim_{z \to i} \frac{z^p}{z + i} \\ &= \pi e^{ip\pi/2}. \end{aligned}

Thus, in the limit as ε→0\epsilon \to 0 and R→∞R \to \infty,

(1+eipπ)∫0∞xpx2+1 dx=πeipπ/2.\p{1 + e^{ip\pi}}\int_0^\infty \frac{x^p}{x^2 + 1} \,\diff{x} = \pi e^{ip\pi/2}.

To complete the computation, we compute

eipπ/21+eipπ=1e−ipπ/2+eipπ/2=12cos⁡pπ2  ⟹  ∫0∞xpx2+1 dx=π2sec⁡pπ2.\begin{aligned} \frac{e^{ip\pi/2}}{1 + e^{ip\pi}} &= \frac{1}{e^{-ip\pi/2} + e^{ip\pi/2}} \\ &= \frac{1}{2\cos\frac{p\pi}{2}} \\ \implies \int_0^\infty \frac{x^p}{x^2 + 1} \,\diff{x} &= \frac{\pi}{2}\sec\frac{p\pi}{2}. \end{aligned}