calculate-0-xe-x-2-x-dx-

Question Number 91603 by mathmax by abdo last updated on 01/May/20

c a l c u l a t e \int_{0}^{\infty} {x e}^{- x^{2} - [x]} d x

Commented by mathmax by abdo last updated on 02/May/20

I = \int_{0}^{\infty} x e^{- x^{2} - [x]} d x \Rightarrow I = \sum_{n = 0}^{\infty} \int_{n}^{n + 1} {x e}^{- x^{2} - n} d x

= \sum_{n = 0}^{\infty} e^{- n} \int_{n}^{n + 1} {x e}^{- x^{2}} d x = \sum_{n = 0}^{\infty} e^{- n} {[- \frac{1}{2} e^{- x^{2}}]}_{n}^{n + 1}

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

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