Question Number 89867 by M±th+et£s last updated on 19/Apr/20
$${find}\:{the}\:{integration} \\ $$$$\int{ln}\lfloor{x}\rfloor\:{dx}\:\:\:\:\:\:;\:{x}>\mathrm{2} \\ $$
Commented by ~blr237~ last updated on 20/Apr/20
Answered by mr W last updated on 20/Apr/20
$${let}\:{x}={n}+{f} \\ $$$${dx}={df} \\ $$$$\int\mathrm{ln}\:\lfloor{x}\rfloor{dx}=\mathrm{ln}\:{n}\:\int{df}=\left(\mathrm{ln}\:{n}\right){f}+{C} \\ $$$$=\left(\mathrm{ln}\:\lfloor{x}\rfloor\right)\left({x}−\lfloor{x}\rfloor\right)+{C} \\ $$
Commented by M±th+et£s last updated on 20/Apr/20
$${sir}\:{why}\:\int{ln}\lfloor{x}\rfloor{dx}={ln}\left({n}\right)\int{df}\:\:{and}\:{how} \\ $$$${did}\:{you}\:{get}\:{ln}\left(\lfloor{x}\rfloor\right)\left({x}−\lfloor{x}\rfloor\right) \\ $$
Commented by mr W last updated on 21/Apr/20
$${x}={n}+{f}\:{with}\:{n}=\lfloor{x}\rfloor,\:{f}={x}−\lfloor{x}\rfloor \\ $$$${in}\:{x}={n}+{f},\:{n}\:{is}\:{constant},\:{therefore} \\ $$$${dx}={df} \\ $$$$\int\mathrm{ln}\:\lfloor{x}\rfloor{dx}=\int\left(\mathrm{ln}\:{n}\right){df}=\mathrm{ln}\:\left({n}\right)\int{df} \\ $$$$=\mathrm{ln}\:\left({n}\right){f}+{C} \\ $$$$=\left(\mathrm{ln}\:\lfloor{x}\rfloor\right)\left({x}−\lfloor{x}\rfloor\right)+{C} \\ $$
Commented by M±th+et£s last updated on 21/Apr/20
$${thank}\:{you}\:{for}\:{explaing}\:{that}\:{for}\:{me} \\ $$
Answered by M±th+et£s last updated on 22/Apr/20
![i try a lot and i get that ln[x]dx=ln(2) x∈(2,3) f(x)=ln[x] x∈(n,n+1) ∫ln[x]dx=Σ_(k=2) ^(n−1) ln(x) + ∫_(n=[x]) ^x ln[x]dx Σ_(k=2) ^(n−1) ln(k)+ln(n)(x−[x])+c](https://www.tinkutara.com/question/Q90207.png)
$${i}\:{try}\:{a}\:{lot}\:{and}\:{i}\:{get}\:{that} \\ $$$${ln}\left[{x}\right]{dx}={ln}\left(\mathrm{2}\right)\:\:\:\:{x}\in\left(\mathrm{2},\mathrm{3}\right) \\ $$$${f}\left({x}\right)={ln}\left[{x}\right]\:{x}\in\left({n},{n}+\mathrm{1}\right) \\ $$$$\int{ln}\left[{x}\right]{dx}=\underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({x}\right)\:+\:\int_{{n}=\left[{x}\right]} ^{{x}} {ln}\left[{x}\right]{dx} \\ $$$$\underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({k}\right)+{ln}\left({n}\right)\left({x}−\left[{x}\right]\right)+{c} \\ $$