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Fall 2019 - Problem 10

calculation, residue theorem

Evaluate

lim⁡x→∞∫0xsin⁡t2 dt.\lim_{x\to\infty} \int_0^x \sin{t^2} \,\diff{t}.

Justify all steps.
Solution.

Let f(z)=eiz2f\p{z} = e^{iz^2} so that if t∈Rt \in \R, then Im⁡f(t)=sin⁡t2\Im{f\p{t}} = \sin{t^2}. For R>0R > 0, let γR\gamma_R be the contour

[0,R]∪{Reiθ∣0≤θ≤π4}∪[0,Reiπ/4].\br{0, R} \cup \set{Re^{i\theta} \mid 0 \leq \theta \leq \frac{\pi}{4}} \cup \br{0, Re^{i\pi/4}}.

On the arc CRC_R, we have

∣∫CRf(z) dz∣≤∫0π/4∣iReiθeiR2ei2θ∣ dθ=∫0π/4Re−R2sin⁡2θ dθ≤∫0π/4Re−4R2θ/π dθ(concavity of sin⁡θ)=−πR4R2(e−R2−1)→R→∞0.\begin{aligned} \abs{\int_{C_R} f\p{z} \,\diff{z}} &\leq \int_0^{\pi/4} \abs{iRe^{i\theta} e^{iR^2e^{i2\theta}}} \,\diff\theta \\ &= \int_0^{\pi/4} Re^{-R^2\sin{2\theta}} \,\diff\theta \\ &\leq \int_0^{\pi/4} Re^{-4R^2\theta/\pi} \,\diff\theta && \p{\text{concavity of }\sin{\theta}} \\ &= -\frac{\pi R}{4R^2} \p{e^{-R^2} - 1} \xrightarrow{R\to\infty} 0. \end{aligned}

On the segment LR=[0,Reiπ/4]L_R = \br{0, Re^{i\pi/4}},

∫LRf(z) dz=−eiπ/4∫0Reit2eiπ/2 dt=−eiπ/4∫0Re−t2 dt→R→∞−eiπ/4π2.\begin{aligned} \int_{L_R} f\p{z} \,\diff{z} &= -e^{i\pi/4} \int_0^R e^{it^2e^{i\pi/2}} \,\diff{t} \\ &= -e^{i\pi/4} \int_0^R e^{-t^2} \,\diff{t} \xrightarrow{R\to\infty} -\frac{e^{i\pi/4}\sqrt{\pi}}{2}. \end{aligned}

Thus, by Cauchy's theorem, sending R→∞R \to \infty, and taking imaginary parts,

∫0∞f(x) dx=eiπ/4π2  ⟹  ∫0∞sin⁡t2 dt=2π4.\int_0^\infty f\p{x} \,\diff{x} = \frac{e^{i\pi/4}\sqrt{\pi}}{2} \implies \int_0^\infty \sin{t^2} \,\diff{t} = \frac{\sqrt{2\pi}}{4}.