$\int_{0}^{\infty}{\left\lbrack \frac{\lbrack x + 1\rbrack}{e^{x}} \right\rbrack dx}$ (where $\lbrack \bullet \rbrack$ denotes greatest integer function) is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
$\frac{x + 1}{e^{x}} < 1$ …(i)
$\Rightarrow \ \frac{\lbrack x + 1brack}{e^{x}} < 1$ …(ii)
(as $\lbrack x + 1brack < x + 1$)
$	herefore\ \left\lbrack \frac{\lbrack x + 1brack}{e^{x}} ightbrack = 0$ [from Eqs.(i) and (ii)]

Q. $\int_{0}^{\infty}{\left\lbrack \frac{\lbrack x + 1\rbrack}{e^{x}} \right\rbrack dx}$ (where $\lbrack \bullet \rbrack$ denotes greatest integer function) is equal to

Explanation

Reviewed by Dr. Abubacker

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