0-2-1-e-x-2-1-dx-x-is-fractional-part-of-x-

Question Number 144849 by Dwaipayan Shikari last updated on 29/Jun/21

\int_{0}^{2} \frac{1}{e^{{x}^{2}} + 1} d x {x} i s f r a c t i o n a l p a r t o f x

Answered by mathmax by abdo last updated on 29/Jun/21

Φ = \int_{0}^{2} \frac{dx}{e^{{x}^{2}} + 1} we have x = [x] + {x} \Rightarrow {x} = x - [x] \Rightarrow

Φ = \int_{0}^{1} \frac{dx}{e^{{(x - [x])}^{2}} + 1} + \int_{1}^{2} \frac{dx}{e^{{(x - [x])}^{2}} + 1}

= \int_{0}^{1} \frac{dx}{1 + e^{x^{2}}} + \int_{1}^{2} \frac{dx}{1 + e^{{(x - 1)}^{2}}} (\to x - 1 = t)

= \int_{0}^{1} \frac{dx}{1 + e^{x^{2}}} + \int_{0}^{1} \frac{dt}{1 + e^{t^{2}}} = 2 \int_{0}^{1} \frac{dx}{1 + e^{x^{2}}}

\int_{0}^{1} \frac{dx}{1 + e^{x^{2}}} = \int_{0}^{1} \frac{e^{- x^{2}}}{1 + e^{- x^{2}}} dx = \int_{0}^{1} e^{- x^{2}} \sum_{n = 0}^{\infty} e^{- {nx}^{2}} dx

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

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

Commented by mathmax by abdo last updated on 29/Jun/21

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

Commented by Dwaipayan Shikari last updated on 29/Jun/21

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