Question Number 80312 by TawaTawa last updated on 02/Feb/20
Commented by mr W last updated on 02/Feb/20
$${what}\:{is}\:\left\{…\right\}\:{here}\:? \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\left(\frac{\mathrm{x}\:−\:\mathrm{1}}{\mathrm{x}\:+\:\mathrm{1}}\right)\:\mathrm{sir} \\ $$
Commented by mr W last updated on 02/Feb/20
$${whether}\:\left\{{x}\right\}=\left({x}\right)\:{or}\:\left\{{x}\right\}={x}−\left[{x}\right],\:{have} \\ $$$${you}\:{ever}\:{tried}\:{to}\:{solve}\:{it}\:{by}\:{yourself}? \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{I}\:\mathrm{did}\:\mathrm{it}\:\mathrm{like}\:\mathrm{normal}\:\mathrm{bracket}.\:\mathrm{So}\:\mathrm{i}\:\mathrm{thought}\:\mathrm{i}\:\mathrm{could}\:\mathrm{confirm}\:\mathrm{the} \\ $$$$\mathrm{answers}.\:\mathrm{But}\:\mathrm{now}\:\mathrm{it}\:\mathrm{is}\:\mathrm{different}\:\mathrm{definition}. \\ $$$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{that}\:\mathrm{one}. \\ $$
Commented by mr W last updated on 02/Feb/20
$${with}\:\left\{\frac{\mathrm{x}\:−\:\mathrm{1}}{\mathrm{x}\:+\:\mathrm{1}}\right\}=\left(\frac{\mathrm{x}\:−\:\mathrm{1}}{\mathrm{x}\:+\:\mathrm{1}}\right)\:{the}\:{question} \\ $$$${makes}\:{not}\:{much}\:{sense}.\:{i}\:{think}\:{it} \\ $$$${means}\:{fraction}\:{part},\:{i}.{e}. \\ $$$$\left\{{x}\right\}={x}−\left[{x}\right] \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{Maybe}\:\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{just}\:\mathrm{guess}\:\mathrm{sir}. \\ $$$$\mathrm{Since}\:\:\left\{\:\:\right\}\:\:\mathrm{has}\:\mathrm{meaning}.\:\:\mathrm{Help}\:\mathrm{me}\:\mathrm{sir}. \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{Sir},\:\mathrm{it}\:\mathrm{means}\:\mathrm{the}\:\mathrm{fractional}\:\mathrm{part}. \\ $$
Commented by john santu last updated on 02/Feb/20
$$\left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}=\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\left[\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right]=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\left[\mathrm{1}−\frac{\mathrm{2}}{{x}+\mathrm{1}}\right] \\ $$$$=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{1}=\:\frac{{x}−\mathrm{1}−{x}−\mathrm{1}}{{x}+\mathrm{1}}=\frac{−\mathrm{2}}{{x}+\mathrm{1}} \\ $$$$\ast\:\Omega=\:\int\:\left(\frac{−\mathrm{2}}{{x}+\mathrm{1}}\right).\frac{{xdx}}{\mathrm{1}−{x}}\:=\: \\ $$$$=\:\int\:\frac{\:\:\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:=\:\int\:\frac{{d}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Omega=\:{ln}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{c} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by john santu last updated on 02/Feb/20
$${my}\:{ans}\:{is}\:{right}? \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{sir}.\:\:\mathrm{Sir}\:\:\mathrm{mrW}\:\:\mathrm{can}\:\mathrm{confirm}. \\ $$
Commented by john santu last updated on 02/Feb/20
$${ok}\:{miss}.\:{thank}\:{you}.\:{god}\:{bless}\: \\ $$$${you} \\ $$
Commented by behi83417@gmail.com last updated on 02/Feb/20
$$\Omega=\int\:\:\frac{−\mathrm{2}\boldsymbol{\mathrm{x}}\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}=−\mathrm{ln}\mid\mathrm{x}^{\mathrm{2}} −\mathrm{1}\mid+\mathrm{const}. \\ $$
Commented by mathmax by abdo last updated on 02/Feb/20
$${if}\:\left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}\:{means}\:\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)\:\Rightarrow\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}×\frac{{xdx}}{\mathrm{1}−{x}} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{xdx}}{{x}+\mathrm{1}}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}+\mathrm{1}−\mathrm{1}}{{x}+\mathrm{1}}{dx}\:=−\mathrm{1}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}+\mathrm{1}}\:=−\mathrm{1}+{ln}\left(\mathrm{2}\right) \\ $$$${if}\:\left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}\:{means}\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\left[\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right]\:{changement}\:{x}=\frac{\mathrm{1}}{{t}}\:{give} \\ $$$$\Omega=−\int_{\mathrm{1}} ^{+\infty} \left\{\frac{\frac{\mathrm{1}}{{t}}−\mathrm{1}}{\frac{\mathrm{1}}{{t}}+\mathrm{1}}\right\}×\frac{\mathrm{1}}{{t}\left(\mathrm{1}−\frac{\mathrm{1}}{{t}}\right)}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \left\{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right\}\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:=\int_{\mathrm{1}} ^{+\infty} \left\{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}−\left[\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right]\right)\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)} \\ $$$$=−\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)}\:−\int_{\mathrm{1}} ^{+\infty} \left[\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:{let}\:{decompose}\: \\ $$$${f}\left({t}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} \left({t}+\mathrm{1}\right)}\:\Rightarrow{f}\left({t}\right)=\frac{{a}}{{t}}\:+\frac{{b}}{{t}^{\mathrm{2}} }\:+\frac{{c}}{{t}+\mathrm{1}} \\ $$$${b}={t}^{\mathrm{2}} {f}\left({t}\right)\mid_{{t}=\mathrm{0}} =\mathrm{1}\:\:{and}\:{c}=\left({t}+\mathrm{1}\right){f}\left({t}\right)\mid_{{t}=−\mathrm{1}} =\mathrm{1}\:\Rightarrow \\ $$$${f}\left({t}\right)=\frac{{a}}{{t}}+\frac{\mathrm{1}}{{t}+\mathrm{1}}\:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:={a}+\frac{\mathrm{1}}{\mathrm{2}}\:+\mathrm{1}\:\:\Rightarrow{a}=−\mathrm{1}\:\Rightarrow{f}\left({t}\right)=−\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:{f}\left({t}\right){dt}\:=\left[−{ln}\mid{t}\mid+{ln}\mid{t}+\mathrm{1}\mid−\frac{\mathrm{1}}{{t}}\right]_{\mathrm{1}} ^{+\infty} =\left[{ln}\mid\frac{{t}+\mathrm{1}}{{t}}\mid−\frac{\mathrm{1}}{{t}}\right]_{\mathrm{1}} ^{+\infty} \\ $$$$=−{ln}\left(\mathrm{2}\right)+\mathrm{1}\:{also} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\left[\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:=\int_{\mathrm{1}} ^{+\infty} \left[−\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \left[−\frac{{t}+\mathrm{1}−\mathrm{2}}{{t}+\mathrm{1}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:=\int_{\mathrm{1}} ^{+\infty} \left[−\mathrm{1}+\frac{\mathrm{2}}{{t}+\mathrm{1}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)} \\ $$$$=−\int_{\mathrm{1}} ^{+\infty} \frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:+\int_{\mathrm{1}} ^{+\infty} \:\left[\frac{\mathrm{2}}{{t}+\mathrm{1}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:=_{{t}−\mathrm{1}={u}} \:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{du}}{{u}\left({u}+\mathrm{1}\right)^{\mathrm{2}\:} }\:\:{this}\:{integral}\:{is}\:{divergent}..! \\ $$$${also}\:\int_{\mathrm{1}} ^{+\infty} \left[\frac{\mathrm{2}}{{t}+\mathrm{1}}\right]\frac{{dt}}{{t}^{\mathrm{2}} \left({t}−\mathrm{1}\right)}\:=_{{t}+\mathrm{1}={u}} \:\:\int_{\mathrm{2}} ^{+\infty} \left[\frac{\mathrm{2}}{{u}}\right]\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} \left({u}−\mathrm{2}\right)} \\ $$$${for}\:{u}>\mathrm{2}\:\Rightarrow\mathrm{0}<\frac{\mathrm{1}}{{u}}<\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{0}<\frac{\mathrm{2}}{{u}}<\mathrm{1}\:\Rightarrow\left[\frac{\mathrm{2}}{{u}}\right]=\mathrm{0}\:\:{any}\:{way}\:\Omega\:{diverges}… \\ $$$$ \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{I}\:\mathrm{appreciate} \\ $$
Commented by mathmax by abdo last updated on 04/Feb/20
$${you}\:{are}\:{welcome}\:. \\ $$
Answered by mr W last updated on 02/Feb/20
$$\left\{{x}\right\}={x}−\left[{x}\right] \\ $$$$\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}=\mathrm{1}−\frac{\mathrm{2}}{{x}+\mathrm{1}} \\ $$$${for}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}: \\ $$$$−\mathrm{1}\leqslant\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\leqslant\mathrm{0} \\ $$$${concerning}\:\left[{x}\right]\:{for}\:{x}<\mathrm{0}\:{there}\:{are}\:{two} \\ $$$${different}\:{ways}\:{for}\:{the}\:{definition}: \\ $$$${way}\:\mathrm{1}:\:\left[{x}\right]=\lfloor{x}\rfloor,\:{i}.{e}.\:\left[−\mathrm{0}.\mathrm{5}\right]=−\mathrm{1} \\ $$$${way}\:\mathrm{2}:\:\left[{x}\right]=\lceil{x}\rceil,\:{i}.{e}.\:\left[−\mathrm{0}.\mathrm{5}\right]=\mathrm{0} \\ $$$$ \\ $$$${if}\:{we}\:{take}\:{definition}\:{way}\:\mathrm{1}: \\ $$$${since}\:−\mathrm{1}\leqslant\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\leqslant\mathrm{0},\: \\ $$$$\:\left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\left(−\mathrm{1}\right)=\frac{\mathrm{2}{x}}{{x}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}\frac{{xdx}}{\mathrm{1}−{x}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2}{x}}{{x}−\mathrm{1}}×\frac{{xdx}}{\mathrm{1}−{x}} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left({x}−\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{d}\left({x}−\mathrm{1}\right) \\ $$$$=−\mathrm{2}\int_{−\mathrm{1}} ^{\:\mathrm{0}} \frac{\left({u}+\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=−\mathrm{2}\int_{−\mathrm{1}} ^{\:\mathrm{0}} \left(\mathrm{1}+\frac{\mathrm{2}}{{u}}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right){du} \\ $$$$=−\mathrm{2}\left[{u}+\mathrm{2ln}\:{u}−\frac{\mathrm{1}}{{u}}\right]_{−\mathrm{1}} ^{\mathrm{0}} \\ $$$$=\infty \\ $$$$ \\ $$$${if}\:{we}\:{take}\:{definition}\:{way}\:\mathrm{2}: \\ $$$${since}\:−\mathrm{1}\leqslant\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\leqslant\mathrm{0},\: \\ $$$$\:\left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{0}=\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right\}\frac{{xdx}}{\mathrm{1}−{x}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}×\frac{{xdx}}{\mathrm{1}−{x}} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}}{{x}+\mathrm{1}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){dx} \\ $$$$=−\left[{x}−\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{ln}\:\mathrm{2}−\mathrm{1} \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by TawaTawa last updated on 02/Feb/20
$$\mathrm{I}\:\mathrm{appreciate} \\ $$