0-dx-1-e-x-2-by-two-method-

Question Number 134546 by mohammad17 last updated on 05/Mar/21

\int_{0}^{\infty} \frac{d x}{1 + e^{x^{2}}} b y t w o m e t h o d

Answered by Dwaipayan Shikari last updated on 05/Mar/21

\int_{0}^{\infty} \frac{d x}{1 + e^{x^{2}}} x^{2} = u

= \frac{1}{2} \int_{0}^{\infty} \frac{u^{- \frac{1}{2}}}{1 + e^{u}} d u = \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{\infty} e^{- n u} u^{- \frac{1}{2}} d u

= \frac{1}{2} \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{Γ (\frac{1}{2})}{n^{\frac{1}{2}}} = \frac{1}{2} ζ (\frac{1}{2}) (1 - \frac{1}{2^{\frac{1}{2} - 1}}) \sqrt{π} = \frac{\sqrt{π}}{2} ζ (\frac{1}{2}) (1 - \sqrt{2})

Commented by mohammad17 last updated on 05/Mar/21

s i r w h a t s t h e v a l u e o r f o r m u l l a o f ζ

Commented by Dwaipayan Shikari last updated on 05/Mar/21

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n^{s}} = ζ (s) (1 - \frac{1}{2^{s - 1}})

Answered by mathmax by abdo last updated on 05/Mar/21

Φ = \int_{0}^{\infty} \frac{dx}{1 + e^{x^{2}}} we do the changement x^{2} = t \Rightarrow

Φ = \int_{0}^{\infty} \frac{dt}{2 \sqrt{t} (1 + e^{t})} = \frac{1}{2} \int_{0}^{\infty} \frac{t^{- \frac{1}{2}}}{1 + e^{t}} dt = \frac{1}{2} \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} e^{- t}}{1 + e^{- t}} dt

= \frac{1}{2} \int_{0}^{\infty} t^{- \frac{1}{2}} e^{- t} \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- nt} dt

= \frac{1}{2} \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} t^{- \frac{1}{2}} e^{- (n + 1) t} dt

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

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

= \frac{Γ (\frac{1}{2})}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{\frac{1}{2}}} = - \frac{1}{2} \sqrt{π} δ (\frac{1}{2})

δ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{x}} = (2^{1 - x} - 1) ξ (x) \Rightarrow δ (\frac{1}{2}) = (2^{\frac{1}{2}} - 1) ξ (\frac{1}{2})

= (\sqrt{2} - 1) ξ (\frac{1}{2}) \dots .