A-1-0-e-x-1-x-amp-aA-e-b-0-1-e-x-1-x-than-a-b-a-b-Z-

Question Number 148026 by mnjuly1970 last updated on 25/Jul/21

A := \int_{- 1}^{0} e^{x + \frac{1}{x}} & a A + e^{b} = \int_{0}^{\infty} {(\frac{1}{e})}^{x + \frac{1}{x}}

t h a n : a + b = ? a, b \in Z

Answered by qaz last updated on 25/Jul/21

A = \int_{- 1}^{0} e^{x + \frac{1}{x}} dx = - \int_{1}^{0} e^{- x - \frac{1}{x}} dx = \int_{0}^{1} e^{- x - \frac{1}{x}} dx

aA + e^{b}

= \int_{0}^{\infty} {(\frac{1}{e})}^{x + \frac{1}{x}} dx

= \int_{0}^{1} e^{- x - \frac{1}{x}} dx + \int_{1}^{\infty} e^{- x - \frac{1}{x}} dx

= A + \int_{1}^{0} e^{- x - \frac{1}{x}} d (\frac{1}{x})

= A + \int_{0}^{1} e^{- x} d (e^{- \frac{1}{x}})

= A + e^{- x - \frac{1}{x}} ∣_{0}^{1} + \int_{0}^{1} e^{- \frac{1}{x}} e^{- x} dx

= 2 A + e^{- 2}

\Rightarrow a + b = 0

Commented by mnjuly1970 last updated on 25/Jul/21

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