Find-0-x-x-x-1-dx-x-fractional-part-x-full-part-

Question Number 209918 by hardmath last updated on 25/Jul/24

Find : \int_{0}^{\infty} \frac{{x}^{[x]}}{[x] + 1} dx = ?

{x} \to fractional part

[x] \to full part

Answered by MM42 last updated on 25/Jul/24

{x} = x - [x]

I_{n} = \int_{0}^{1} d x + \int_{1}^{2} \frac{(x - 1)}{2} d x + \int_{2}^{3} \frac{{(x - 2)}^{2}}{3} d x + \dots + \int_{n - 1}^{n} \frac{{(x - n + 1)}^{n - 1}}{n} d x

{= {x]_{0}^{1} {+ \frac{{(x - 1)}^{2}}{4}]}_{1}^{2} + \frac{{(x - 2)}^{3}}{9}]}_{2}^{3} + \dots + \frac{{(x - n + 1)}^{n}}{n^{2}}]}_{n - 1}^{n}

= 1 + \frac{1}{4} + \frac{1}{9} + \dots + \frac{1}{n^{2}}

\Rightarrow I = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} ✓

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