advanced-calculus-prove-that-F-1-0-e-x-e-1-x-1-x-dx-

Question Number 141417 by mnjuly1970 last updated on 18/May/21

\dots \dots . . a d v a n c e d \dots \dots \dots c a l c u l u s \dots \dots .

p r o v e t h a t ::

F := \int_{- 1}^{0} \frac{e^{x} + e^{\frac{1}{x}} - 1}{x} d x \overset{? ?}{=} γ

Answered by mnjuly1970 last updated on 27/May/21

:= \int_{- 1}^{0} \frac{e^{\frac{1}{x}}}{x} d x \overset{\frac{1}{x} = - y}{=} \int_{1}^{\infty} - {y e}^{- y} \frac{d y}{y^{2}}

:= - \int_{1}^{\infty} \frac{e^{- y}}{y} d y = - \int_{1}^{\infty} e^{- y} l n (y) - {[l n (y) e^{- y}]}_{1}^{\infty}

:= - \int_{1}^{\infty} e^{- y} l n (y) d y \Rightarrow - \int_{0}^{\infty} e^{- y} l n (y) d y + \int_{0}^{1} e^{- y} l n (y) d y

\int_{- 1}^{0} \frac{e^{\frac{1}{x}}}{x} d x := γ + \int_{0}^{1} e^{- x} l n (x) d x (★)

:= γ + \int_{0}^{1} l n (x) d (1 - e^{- x})

:= γ + {[l n (x) (1 - e^{- x})]}_{0}^{1} - \int_{0}^{1} (\frac{1 - e^{- x}}{x}) d x

:= γ - \int_{0}^{1} \frac{1 - e^{- x}}{x} d x = γ + \int_{0}^{- 1} \frac{1 - e^{x}}{- x} d x

:= γ + \int_{- 1}^{0} \frac{1 - e^{x}}{x} d x

(★) :: \int_{- 1}^{0} \frac{e^{\frac{1}{x}} + e^{x} - 1}{x} = γ \dots . . ✓

\dots \dots . . F := \int_{- 1}^{0} \frac{e^{x} + e^{\frac{1}{x}} - 1}{x} d x = γ \dots \dots

Answered by mindispower last updated on 19/May/21

x = \frac{1}{t} \Rightarrow F = \int_{- 1}^{0} \frac{e^{x} - 1}{x} d x + \int_{- 1}^{0} \frac{e^{\frac{1}{x}}}{x} d x = A + B

{A = [e^{x} - 1] l n (- x)]}_{- 1}^{0} - \int_{- 1}^{0} e^{x} l n (- x)

= - \int_{- 1}^{0} e^{x} l n (- x) d x = \int_{1}^{0} e^{- t} l n (t) d t

B = {[l n (- x) e^{\frac{1}{x}}]}_{- 1}^{0} - \int_{- 1}^{0} \frac{e^{\frac{1}{x}}}{x^{2}} l n (- x) d x

= \int_{- 1}^{0} - \frac{e^{\frac{1}{x}}}{x^{2}} l n (- x), t = - \frac{1}{x} \Rightarrow

B = - \int_{1}^{\infty} e^{- t} l n (t)

F = A + B = - \int_{0}^{\infty} l n (t) e^{- t} d t = \partial_{x} (- \int_{0}^{\infty} e^{- t} t^{x - 1} d t) ∣_{x = 1}

= \partial_{x} . - Γ (x + 1) ∣_{x = 0} = - Γ^{'} (1) = - Γ (1) Ψ (1) = - 1 . - γ = γ

\Rightarrow \int_{- 1}^{0} \frac{e^{x} + e^{\frac{1}{x}} - 1}{x} d x = γ