13A (Complex Analysis or Complex Methods)

The integrand

F (z) = \frac{e ^{i z^{2} / (4 π)}}{e ^{z /2} - e ^{- z /2}}

has one pole inside the rectangle, at

z = 0

1

. The residue theorem therefore gives

\int_{C} F (z) d z = 2 πi

. The vertical-side integrals tend to zero as

R \to \infty

On the horizontal sides, direct substitution

z = x \pm πi

, with the upper side oppositely oriented, shows that their sum is

i e^{- iπ /4} \int_{- R}^{R} e^{i x^{2} / (4 π)} (\frac{1}{1 + e ^{x}} + \frac{e ^{x}}{1 + e ^{x}}) d x .

Taking the limit and comparing with

2 πi

gives

R \to \infty lim \int_{- R}^{R} e^{i x^{2} / (4 π)} d x = 2 π e^{iπ /4} .

b

Integrate

z e^{iπ z} / (z^{2} + a^{2})

over the upper semicircle. Jordan lemma removes the arc, and the pole at

z = ia

e^{- πa} /2

. Hence

\int_{- \infty}^{\infty} \frac{x e ^{iπ x}}{x ^{2} + a ^{2}} d x = πi e^{- πa} .

The cosine contribution is odd, while the sine contribution is even. Therefore

\int_{0}^{\infty} \frac{x sin ( π x )}{x ^{2} + a ^{2}} d x = \frac{π}{2} e^{- πa} .

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