Solve-1-pi-1-2-ln-1-x-dx-

Question Number 115026 by jm2bok last updated on 23/Sep/20

Solve : \int_{1 / π}^{1 / 2} \ln ⌊ \frac{1}{x} ⌋ d x

Answered by PRITHWISH SEN 2 last updated on 23/Sep/20

when

\frac{1}{π} ⩽ x ⩽ \frac{1}{3} \Rightarrow 3 ⩽ \frac{1}{x} ⩽ π \Rightarrow \ln ⌊ \frac{1}{x} ⌋ = \ln 3

\frac{1}{3} < x ⩽ \frac{1}{2} \Rightarrow 2 ⩽ x < 3 \Rightarrow \ln ⌊ \frac{1}{x} ⌋ = \ln 2

∴ the integration

= \int_{\frac{1}{π}}^{\frac{1}{3}} \ln (3) dx + \int_{\frac{1}{3}}^{\frac{1}{2}} \ln (2) dx

= (\frac{1}{3} - \frac{1}{π}) \ln 3 + (\frac{1}{2} - \frac{1}{3}) \ln 2 ∽ 0.13202947375

Answered by mathmax by abdo last updated on 23/Sep/20

I = \int_{\frac{1}{π}}^{\frac{1}{2}} \ln {[\frac{1}{x}]} dx we do the changement \frac{1}{x} = t \Rightarrow

I = \int_{π}^{2} \ln {[t]} (- \frac{dt}{t^{2}}) = \int_{2}^{π} \frac{\ln [t]}{t^{2}} dt (π \sim 3, 14) \Rightarrow

I = \int_{2}^{3} \frac{\ln [t]}{t^{2}} dt + \int_{3}^{π} \frac{\ln [t]}{t^{2}} dt = 2 \int_{2}^{3} \frac{dt}{t^{2}} + 3 \int_{3}^{π} \frac{dt}{t^{2}}

= 2 {[- \frac{1}{t}]}_{2}^{3} + 3 {[- \frac{1}{t}]}_{3}^{π} = 2 (\frac{1}{2} - \frac{1}{3}) + 3 (\frac{1}{3} - \frac{1}{π})

= 1 - \frac{2}{3} + 1 - \frac{3}{π} = 2 - \frac{2}{3} - \frac{3}{π} = \frac{4}{3} - \frac{3}{π}