Contour-integral derivation of theta-function and Gauss-sum identities

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Problem 5485 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.3 "Residues and Their Applications"

Setting, by definition,

f(u)=∫γe−πiz2+2πiuzeπiz−e−πiz dz, f(u) = \int _{\gamma } \frac{e^{-\pi i z^{2}+2\pi i uz}}{e^{\pi i z}-e^{-\pi i z}}\, dz,

where the integration proceeds from bottom to top along the entire line x+y=12x+y = \frac{1}{2}, verify the identity:

f(u+1)−f(u)=eπi(u+12)2∫γe−πiz2 dz. f\left(u+1\right)-f(u) = e^{\pi i \left(u+\frac{1}{2}\right)^{2}}\int _{\gamma } e^{-\pi i z^{2}}\, dz.
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Problem 5486 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.3 "Residues and Their Applications"

Shifting, in the previous problem, the contour of integration to the line x+y=−12x+y = -\frac{1}{2}, where f(u)=∫γe−πiz2+2πiuzeπiz−e−πiz dzf(u) = \int_{\gamma } \dfrac {e^{-\pi i z^{2}+2\pi i uz}}{e^{\pi i z}-e^{-\pi i z}}\, dz (problem 5485), prove the identity:

f(u)=e−2πiuf(u+1)+1. f(u) = e^{-2\pi i u}f(u+1)+1.
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Problem 5487 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.3 "Residues and Their Applications"

Prove that ∫γe−πiz2 dz=e3πi4\displaystyle \int_{\gamma } e^{-\pi i z^{2}}\, dz = e^{\frac{3\pi i}{4}} (where γ\gamma is the line x+y=12x+y = \frac{1}{2}, integration from bottom to top, problem 5485), and, using the preceding results, derive Riemann's formula:

∫γe−πiz2+2πiuzeπiz−e−πiz dz=eπiu−eπiu2eπiu−e−πiu. \int _{\gamma } \frac{e^{-\pi i z^{2}+2\pi i uz}}{e^{\pi i z}-e^{-\pi i z}}\, dz = \frac{e^{\pi i u}-e^{\pi i u^{2}}}{e^{\pi i u}-e^{-\pi i u}}.
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Problem 5488 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.3 "Residues and Their Applications"

Consider the integral of the function e2πiz2n1−e2πiz\dfrac {e^{\frac{2\pi i z^{2}}{n}}}{1-e^{2\pi i z}}, where nn is a positive integer, along the boundary of the domain

0<x<n2,−ω<y<ω,∣z∣>ρ,∣z−n2∣>π. 0 < x < \frac{n}{2}, \quad -\omega < y < \omega , \quad \left|z\right| > \rho , \quad \left|z-\frac{n}{2}\right| > \pi .

By passing to the limit as ρ→0\rho \to 0, ω→∞\omega \to \infty, obtain the formula for the Gauss sum:

∑ν=0n−1e2πiν2n=1+(−i)n1−i⋅n. \sum _{\nu =0}^{n-1} e^{\frac{2\pi i \nu ^{2}}{n}} = \frac{1+(-i)^{n}}{1-i}\cdot \sqrt{n}.
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References

[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.