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Question Number 175244 by infinityaction last updated on 24/Aug/22

p r o v e t h a t

\int_{0}^{a} \int_{0}^{\sqrt{a^{2} - x^{2}}} \frac{d x d y}{(1 + e^{y}) \sqrt{a^{2} - x^{2} - y^{2}}} = \frac{π}{2} \log \frac{2 e^{a}}{1 + e^{a}}

Answered by Ar Brandon last updated on 24/Aug/22

I = \int_{0}^{a} \int_{0}^{\sqrt{a^{2} - y^{2}}} \frac{d x d y}{(1 + e^{y}) \sqrt{a^{2} - x^{2} - y^{2}}}

= \int_{0}^{a} \frac{1}{1 + e^{y}} \int_{0}^{\sqrt{a^{2} - y^{2}}} \frac{d x}{\sqrt{(a^{2} - y^{2}) - x^{2}}} d y

= \int_{0}^{a} \frac{1}{1 + e^{y}} {[\arcsin (\frac{x}{\sqrt{a^{2} - y^{2}}})]}_{0}^{\sqrt{a^{2} - y^{2}}} d y

= \frac{π}{2} \int_{0}^{a} \frac{1}{1 + e^{y}} d y = \frac{π}{2} \int_{0}^{a} \frac{e^{- y}}{e^{- y} + 1} d y

= - \frac{π}{2} {[\ln (e^{- y} + 1)]}_{0}^{a} = \frac{π}{2} (\ln 2 - \ln (e^{- a} + 1))

= \frac{π}{2} \ln (\frac{2}{e^{- a} + 1}) = \frac{π}{2} \ln (\frac{2 e^{a}}{1 + e^{a}})

Commented by infinityaction last updated on 24/Aug/22

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