0-e-x-2-e-x-x-dx-k-find-k-Euler-constant-

Question Number 140399 by mnjuly1970 last updated on 07/May/21

ξ := \int_{0}^{\infty} \frac{e^{- x^{2}} - e^{- x}}{x} d x = k . γ

f i n d^{″} k^{″} \dots

γ := E u l e r c o n s t a n t \dots .

Answered by qaz last updated on 07/May/21

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

- γ = \int_{0}^{\infty} (e^{- x^{2}} - \frac{1}{1 + x^{2}}) \frac{d (x^{2})}{x^{2}} = 2 \int_{0}^{\infty} (e^{- x^{2}} - \frac{1}{1 + x^{2}}) \frac{d x}{x}

\Rightarrow - \frac{γ}{2} = \int_{0}^{\infty} (e^{- x^{2}} - \frac{1}{1 + x^{2}}) \frac{d x}{x}

ξ = \int_{0}^{\infty} (e^{- x^{2}} - e^{- x}) \frac{d x}{x}

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

= - \frac{γ}{2} + γ + \int_{0}^{\infty} (\frac{1}{1 + x^{2}} - \frac{1}{1 + x}) \frac{d x}{x}

= \frac{γ}{2} + \int_{0}^{\infty} (\frac{1}{1 + x} - \frac{x}{1 + x^{2}}) d x

= \frac{γ}{2} + l n \frac{1 + x}{\sqrt{1 + x^{2}}} ∣_{0}^{\infty}

= \frac{γ}{2}

\Rightarrow k = \frac{1}{2}

Commented by mnjuly1970 last updated on 07/May/21

b r a v o \dots m r p a y a n \dots