integrate-0-e-x-2-1-e-x-dx-

Question Number 204707 by Mummyjay last updated on 25/Feb/24

i n t e g r a t e \int_{0}^{\infty} \frac{e^{- x^{2}}}{1 + e^{x}} d x

Answered by witcher3 last updated on 26/Feb/24

Ω = \int_{0}^{\infty} \frac{e^{- x^{2} - x}}{1 + e^{- x}} dx = \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{\infty} e^{- (x^{2} + (1 + n) x)} dx

x = (1 + n) y

= \sum_{n ⩾ 0} {(- 1)}^{n} (n + 1) \int_{0}^{\infty} e^{- {(1 + n)}^{2} ({(y + \frac{1}{2})}^{2} - \frac{1}{4})} dy; (1 + n) (y + \frac{1}{2}) = z

= \sum_{n ⩾ 0} {(- 1)}^{n} e^{{(\frac{1 + n}{2})}^{2}} \int_{\frac{n + 1}{2}}^{\infty} e^{- z^{2}} dz

\frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} dt . erfc (x)

= \sum_{n ⩾ 0} {(- 1)}^{n} e^{{(\frac{1 + n}{2})}^{2}} . (\frac{\sqrt{π}}{2} - \frac{\sqrt{π}}{2} erfc (\frac{n + 1}{2}))

= \frac{\sqrt{π}}{2} \sum_{n ⩾ 0} {(- 1)}^{n} e^{{(\frac{1 + n}{2})}^{2}} (1 - erfc (\frac{n + 1}{2}))

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