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Question-202490




Question Number 202490 by Calculusboy last updated on 27/Dec/23
Commented by aleks041103 last updated on 27/Dec/23
is {x} the whole part or the fractional part?
$${is}\:\left\{{x}\right\}\:{the}\:{whole}\:{part}\:{or}\:{the}\:{fractional}\:{part}? \\ $$
Commented by Calculusboy last updated on 28/Dec/23
the fractional part
$$\boldsymbol{{the}}\:\boldsymbol{{fractional}}\:\boldsymbol{{part}} \\ $$
Answered by namphamduc last updated on 28/Dec/23
If you took this from FB, you knew I solved it, the result is γ (Euler Mascheroni constant).
$${If}\:{you}\:{took}\:{this}\:{from}\:{FB},\:{you}\:{knew}\:{I}\:{solved}\:{it},\:{the}\:{result}\:{is}\:\gamma\:\left({Euler}\:{Mascheroni}\:{constant}\right). \\ $$
Answered by witcher3 last updated on 28/Dec/23
let(1/x)=t  =∫_1 ^∞ {t}.((1/(t−1))).(dt/t^2 )=S  =∫_1 ^∞ (({t})/(t^2 (t−1)))dt  =∫_1 ^∞ ((t−[t])/(t^2 (t−1)))dt  =Σ_(k≥1) ∫_k ^(k+1) ((t−k)/(t^2 (t−1)))dt=Σ_(k≥1) ∫_k ^(k+1) ((1−k)/(t−1))+(k/t^2 )+((k−1)/t)dt  =Σ_(k≥1) [(1−k)(ln(k)−ln(k−1)−(k/(k+1))+1)+(k−1)ln(k+1)−(k−1)ln(k)  =Σ_(k≥1) ^N [2(1−k)ln(k)−(1−k)ln(k^2 −1)+(1/(k+1))  =Σ_(k=1) ^N {(1−k)ln(k)+kln(k+1)+(k−1)ln(k−1)−kln((k)  +ln(k)−ln(1+k)+(1/(k+1))}  =Nln(N+1)−ln(2)−Nln(N)+ln(1)−ln(N+1)+ln(2)+H_(N+1) −1  S_N =Nln(1+(1/N))+H_(N+1) −ln(N+1)−1  S=lim_(N→∞) S_N =lim_(N→∞) (Nln(1+(1/N))−1)+lim_(N→∞) (H_(N+1) −ln(N+1)}  =γ
$$\mathrm{let}\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{t} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \left\{\mathrm{t}\right\}.\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{1}}\right).\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }=\mathrm{S} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\left\{\mathrm{t}\right\}}{\mathrm{t}^{\mathrm{2}} \left(\mathrm{t}−\mathrm{1}\right)}\mathrm{dt} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{t}−\left[\mathrm{t}\right]}{\mathrm{t}^{\mathrm{2}} \left(\mathrm{t}−\mathrm{1}\right)}\mathrm{dt} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{t}−\mathrm{k}}{\mathrm{t}^{\mathrm{2}} \left(\mathrm{t}−\mathrm{1}\right)}\mathrm{dt}=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{1}−\mathrm{k}}{\mathrm{t}−\mathrm{1}}+\frac{\mathrm{k}}{\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{k}−\mathrm{1}}{\mathrm{t}}\mathrm{dt} \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{1}} {\sum}\left[\left(\mathrm{1}−\mathrm{k}\right)\left(\mathrm{ln}\left(\mathrm{k}\right)−\mathrm{ln}\left(\mathrm{k}−\mathrm{1}\right)−\frac{\mathrm{k}}{\mathrm{k}+\mathrm{1}}+\mathrm{1}\right)+\left(\mathrm{k}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{k}+\mathrm{1}\right)−\left(\mathrm{k}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{k}\right)\right. \\ $$$$=\underset{\mathrm{k}\geqslant\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\left[\mathrm{2}\left(\mathrm{1}−\mathrm{k}\right)\mathrm{ln}\left(\mathrm{k}\right)−\left(\mathrm{1}−\mathrm{k}\right)\mathrm{ln}\left(\mathrm{k}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right. \\ $$$$=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{N}} {\sum}}\left\{\left(\mathrm{1}−\mathrm{k}\right)\mathrm{ln}\left(\mathrm{k}\right)+\mathrm{kln}\left(\mathrm{k}+\mathrm{1}\right)+\left(\mathrm{k}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{k}−\mathrm{1}\right)−\mathrm{kln}\left(\left(\mathrm{k}\right)\right.\right. \\ $$$$\left.+\mathrm{ln}\left(\mathrm{k}\right)−\mathrm{ln}\left(\mathrm{1}+\mathrm{k}\right)+\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right\} \\ $$$$=\mathrm{Nln}\left(\mathrm{N}+\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{Nln}\left(\mathrm{N}\right)+\mathrm{ln}\left(\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{N}+\mathrm{1}\right)+\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{H}_{\mathrm{N}+\mathrm{1}} −\mathrm{1} \\ $$$$\mathrm{S}_{\mathrm{N}} =\mathrm{Nln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{N}}\right)+\mathrm{H}_{\mathrm{N}+\mathrm{1}} −\mathrm{ln}\left(\mathrm{N}+\mathrm{1}\right)−\mathrm{1} \\ $$$$\mathrm{S}=\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}S}_{\mathrm{N}} =\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{Nln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{N}}\right)−\mathrm{1}\right)+\underset{\mathrm{N}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{H}_{\mathrm{N}+\mathrm{1}} −\mathrm{ln}\left(\mathrm{N}+\mathrm{1}\right)\right\} \\ $$$$=\gamma \\ $$$$ \\ $$$$ \\ $$

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