Past exam of the integration course of the University of Oxford / 2021 / 2 / b / i

Solution

e^{- a x}

, whose integral is

1/ a

\int_{0}^{\infty} e^{(- a + ib) x} d x = [\frac{e ^{(- a + ib) x}}{- a + ib}]_{0}^{\infty} = \frac{1}{a - ib} = \frac{a + ib}{a ^{2} + b ^{2}} .

Taking the real part, justified by absolute integrability, gives

\int_{0}^{\infty} e^{- a x} cos (b x) d x = \frac{a}{a ^{2} + b ^{2}} .

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