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Question Number 89867 by M±th+et£s last updated on 19/Apr/20

find the integration  ∫ln⌊x⌋ dx      ; x>2

$${find}\:{the}\:{integration} \\ $$ $$\int{ln}\lfloor{x}\rfloor\:{dx}\:\:\:\:\:\:;\:{x}>\mathrm{2} \\ $$

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  ∫ln ⌊x⌋dx=ln n ∫df=(ln n)f+C  =(ln ⌊x⌋)(x−⌊x⌋)+C

$${let}\:{x}={n}+{f} \\ $$ $${dx}={df} \\ $$ $$\int\mathrm{ln}\:\lfloor{x}\rfloor{dx}=\mathrm{ln}\:{n}\:\int{df}=\left(\mathrm{ln}\:{n}\right){f}+{C} \\ $$ $$=\left(\mathrm{ln}\:\lfloor{x}\rfloor\right)\left({x}−\lfloor{x}\rfloor\right)+{C} \\ $$

Commented byM±th+et£s last updated on 20/Apr/20

sir why ∫ln⌊x⌋dx=ln(n)∫df  and how  did you get ln(⌊x⌋)(x−⌊x⌋)

$${sir}\:{why}\:\int{ln}\lfloor{x}\rfloor{dx}={ln}\left({n}\right)\int{df}\:\:{and}\:{how} \\ $$ $${did}\:{you}\:{get}\:{ln}\left(\lfloor{x}\rfloor\right)\left({x}−\lfloor{x}\rfloor\right) \\ $$

Commented bymr W last updated on 21/Apr/20

x=n+f with n=⌊x⌋, f=x−⌊x⌋  in x=n+f, n is constant, therefore  dx=df  ∫ln ⌊x⌋dx=∫(ln n)df=ln (n)∫df  =ln (n)f+C  =(ln ⌊x⌋)(x−⌊x⌋)+C

$${x}={n}+{f}\:{with}\:{n}=\lfloor{x}\rfloor,\:{f}={x}−\lfloor{x}\rfloor \\ $$ $${in}\:{x}={n}+{f},\:{n}\:{is}\:{constant},\:{therefore} \\ $$ $${dx}={df} \\ $$ $$\int\mathrm{ln}\:\lfloor{x}\rfloor{dx}=\int\left(\mathrm{ln}\:{n}\right){df}=\mathrm{ln}\:\left({n}\right)\int{df} \\ $$ $$=\mathrm{ln}\:\left({n}\right){f}+{C} \\ $$ $$=\left(\mathrm{ln}\:\lfloor{x}\rfloor\right)\left({x}−\lfloor{x}\rfloor\right)+{C} \\ $$

Commented byM±th+et£s last updated on 21/Apr/20

thank you for explaing that for me

$${thank}\:{you}\:{for}\:{explaing}\:{that}\:{for}\:{me} \\ $$

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

$${i}\:{try}\:{a}\:{lot}\:{and}\:{i}\:{get}\:{that} \\ $$ $${ln}\left[{x}\right]{dx}={ln}\left(\mathrm{2}\right)\:\:\:\:{x}\in\left(\mathrm{2},\mathrm{3}\right) \\ $$ $${f}\left({x}\right)={ln}\left[{x}\right]\:{x}\in\left({n},{n}+\mathrm{1}\right) \\ $$ $$\int{ln}\left[{x}\right]{dx}=\underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({x}\right)\:+\:\int_{{n}=\left[{x}\right]} ^{{x}} {ln}\left[{x}\right]{dx} \\ $$ $$\underset{{k}=\mathrm{2}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left({k}\right)+{ln}\left({n}\right)\left({x}−\left[{x}\right]\right)+{c} \\ $$

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