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# Bases-on-suggestion-from-Filup-and-some-discussion-on-that-I-am-suggesting-that-we-sequence-series-and-related-function-as-a-topic-for-this-month-x-n-1-n-x-x-R-x-gt-1-Show-that-x-

Question Number 2675 by prakash jain last updated on 24/Nov/15
$$\mathrm{Bases}\:\mathrm{on}\:\mathrm{suggestion}\:\mathrm{from}\:\mathrm{Filup}\:\mathrm{and}\:\mathrm{some} \\$$$$\mathrm{discussion}\:\mathrm{on}\:\mathrm{that}\:\mathrm{I}\:\mathrm{am}\:\mathrm{suggesting}\:\mathrm{that}\:\mathrm{we} \\$$$$\mathrm{sequence},\:\mathrm{series}\:\mathrm{and}\:\mathrm{related}\:\mathrm{function}\:\mathrm{as}\:\mathrm{a} \\$$$$\mathrm{topic}\:\mathrm{for}\:\mathrm{this}\:\mathrm{month}. \\$$$$\zeta\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{−{x}} ,\:{x}\in\mathbb{R},\:{x}>\mathrm{1} \\$$$$\mathrm{Show}\:\mathrm{that} \\$$$$\zeta\left({x}\right)=\frac{\mathrm{1}}{\Gamma\left({x}\right)}\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}\mathrm{d}{t} \\$$
Commented by prakash jain last updated on 24/Nov/15
$$\zeta−\mathrm{Riemann}\:\mathrm{zeta}\:\mathrm{function} \\$$$$\Gamma−\mathrm{Gamma}\:\mathrm{function} \\$$$$\mathrm{These}\:\mathrm{2}\:\mathrm{functions}\:\mathrm{were}\:\mathrm{mentioned}\:\mathrm{as}\:\mathrm{part} \\$$$$\mathrm{of}\:\mathrm{some}\:\mathrm{of}\:\mathrm{the}\:\mathrm{questions}\:\mathrm{earlier}. \\$$$$\mathrm{For}\:\mathrm{this}\:\mathrm{question}\:\mathrm{only}\:\mathrm{consider}\:\mathrm{real}\:\mathrm{numbers} \\$$$$\mathrm{although}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{are}\:\mathrm{defined}\:\mathrm{for} \\$$$$\mathrm{complex}\:\mathrm{numbers}\:\mathrm{as}\:\mathrm{well}. \\$$
Commented by Rasheed Soomro last updated on 24/Nov/15
$$\mathcal{I}\:{have}\:{only}\:{heard}\:{the}\:{names}\:{of}\:{reimann}\:{zeta}\:{and} \\$$$${gamma}\:{functions}!\:{Anyway}\:{no}\:{harm}\:{in}\:{trying}! \\$$$${By}\:{the}\:{way}\:{who}\:{will}\:{be}\:{the}\:{judge}? \\$$
Commented by prakash jain last updated on 24/Nov/15
$$\mathrm{This}\:\mathrm{question}\:\mathrm{was}\:\mathrm{not}\:\mathrm{meant}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a} \\$$$$\mathrm{challenge}\:\mathrm{just}\:\mathrm{learning}\:\mathrm{details}\:\mathrm{of}\:\mathrm{a}\:\mathrm{few} \\$$$$\mathrm{things}\:\mathrm{that}\:\mathrm{got}\:\mathrm{mentioned}\:\mathrm{earlier}. \\$$
Commented by prakash jain last updated on 24/Nov/15
$$\mathrm{123456}\:\mathrm{is}\:\mathrm{most}\:\mathrm{knowledgeable}\:\mathrm{among}\:\mathrm{us}. \\$$$$\left(\mathrm{s}\right)\mathrm{he}\:\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{challenge}\:\mathrm{at}\:\mathrm{some}\:\mathrm{point} \\$$$$\mathrm{and}\:\mathrm{judge}\:\mathrm{us}. \\$$$$\\$$$$\mathrm{In}\:\mathrm{any}\:\mathrm{case}\:\mathrm{regular}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{A}\:\mathrm{continues}\:\mathrm{as} \\$$$$\mathrm{usual}\:\mathrm{from}\:\mathrm{everyone}. \\$$
Commented by Yozzi last updated on 24/Nov/15
$${Cool}. \\$$
Commented by Yozzi last updated on 24/Nov/15
$${Is}\:{it}\:{dx}\:{or}\:{dt}\:{in}\:{the}\:{expression}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dx}? \\$$
Commented by prakash jain last updated on 24/Nov/15
$$\mathrm{d}{t}\:\mathrm{corrected}. \\$$
Commented by 123456 last updated on 01/Dec/15
$$\mathrm{24}/\mathrm{12} \\$$
Answered by Yozzi last updated on 24/Nov/15
$$\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}=\frac{{t}^{{x}−\mathrm{1}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} } \\$$$${For}\:{t}>\mathrm{0},\:\:\mid{e}^{−{t}} \mid<\mathrm{1}.\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} }\:{is}\:{the}\:{the} \\$$$${sum}\:{of}\:{an}\:{infinite}\:{G}.{P}\:{with}\:{first}\:{term} \\$$$$\mathrm{1}\:{and}\:{common}\:{ratio}\:{e}^{−{t}} . \\$$$$\therefore\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} }=\mathrm{1}+{e}^{−{t}} +{e}^{−\mathrm{2}{t}} +{e}^{−\mathrm{3}{t}} +… \\$$$$\therefore\frac{{t}^{{x}−\mathrm{1}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }={t}^{{x}−\mathrm{1}} {e}^{−{t}} +{t}^{{x}−\mathrm{1}} {e}^{−\mathrm{2}{t}} +{t}^{{x}−\mathrm{1}} {e}^{−\mathrm{3}{t}} +… \\$$$$\frac{{t}^{{x}−\mathrm{1}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{t}^{{x}−\mathrm{1}} {e}^{−{rt}} \\$$$${Let}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dt}=\int_{\mathrm{0}} ^{\infty} \left(\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{t}^{{x}−\mathrm{1}} {e}^{−{rt}} \right){dt} \\$$$${I}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{rt}} {dt}\right) \\$$$${let}\:{u}={rt}\Rightarrow{t}=\frac{\mathrm{1}}{{r}}{u}\Rightarrow\frac{\mathrm{1}}{{r}}{du}={dt} \\$$$${At}\:{t}=\mathrm{0},{u}=\mathrm{0}.\:{As}\:{t}\rightarrow\infty,\:{u}\rightarrow\infty\:{for}\:{r}>\mathrm{0}. \\$$$$\therefore{t}^{{x}−\mathrm{1}} ={r}^{\mathrm{1}−{x}} {u}^{{x}−\mathrm{1}} \\$$$$\therefore\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{rt}} {dt}=\int_{\mathrm{0}} ^{\infty} {r}^{\mathrm{1}−{x}} {u}^{{x}−\mathrm{1}} {e}^{−{u}} ×\frac{\mathrm{1}}{{r}}{du} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={r}^{−{x}} \int_{\mathrm{0}} ^{\infty} {u}^{{x}−\mathrm{1}} {e}^{−{u}} {du} \\$$$${The}\:{gamma}\:{function}\:\Gamma\left({x}\right)\:{defined} \\$$$${for}\:{x}>\mathrm{0}\:{is}\:{expressible}\:{as} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}. \\$$$$\therefore\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{rt}} {dt}={r}^{−{x}} \Gamma\left({x}\right) \\$$$$\therefore{I}=\Gamma\left({x}\right)×\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{r}^{−{x}} \\$$$${But},\:\zeta\left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{r}^{−{x}} \:\:,{x}\in\mathbb{R},{x}>\mathrm{1}. \\$$$$\therefore{I}=\Gamma\left({x}\right)\zeta\left({x}\right)\Rightarrow\zeta\left({x}\right)=\frac{\mathrm{1}}{\Gamma\left({x}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dt}\: \\$$$$\\$$
Commented by 123456 last updated on 24/Nov/15
$$\mathrm{nice} \\$$
Commented by Filup last updated on 25/Nov/15
$${Awesome}!\:\mathrm{I}\:\mathrm{had}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{how}\:\mathrm{to}\:\mathrm{do}\:\mathrm{this} \\$$$$\mathrm{one}!\:{S}\mathrm{o}\:\mathrm{simple}! \\$$
Commented by Rasheed Soomro last updated on 25/Nov/15
$$\mathrm{2}{nd}\:{line}: \\$$$${For}\:{t}>\mathrm{0},\:\:\underset{−} {\mid{e}^{−{t}} \mid}<\mathrm{1}.\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} }\:{is}\:{the}\:{the} \\$$$${Couldn}'{t}\:{we}\:{write}\:{e}^{−{t}} \:{instead}\:{of}\:\mid{e}^{−{t}} \mid \\$$$${Or}\:{there}\:{is}\:{any}\:{specific}\:{reason}\:{for}\:{this}? \\$$
Commented by Yozzi last updated on 25/Nov/15
$${It}'{s}\:{written}\:{like}\:{that}\:{to}\:{specifically} \\$$$${indicate}\:{that}\:{the}\:{convergence} \\$$$${condition}\:{for}\:{an}\:{infinite}\:{geometric} \\$$$${series}\:{is}\:{satisfied}.\:{So}\:{expressions} \\$$$${of}\:{the}\:{form}\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{c}}{\mathrm{1}−{dx}}\:\: \\$$$${represent}\:{the}\:{sum}\:{of}\:{an}\:{infinite}\: \\$$$${geometric}\:{series}\:{under}\:{the}\:{condition} \\$$$${that}\:\mid{dx}\mid<\mathrm{1}\Rightarrow\mid{x}\mid<\mathrm{1}/{d}\:\:\left({d}\neq\mathrm{0}\right).\:{The} \\$$$${series}\:{is}\:{then}\:{deduced}\:{to}\:{be}\:{of}\:{the}\:{form} \\$$$${c}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({dx}\right)^{{n}−\mathrm{1}} ={c}\left\{\mathrm{1}+{dx}+{d}^{\mathrm{2}} {x}^{\mathrm{2}} +{d}^{\mathrm{3}} {x}^{\mathrm{3}} +…\right\} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={c}+{cdx}+{cd}^{\mathrm{2}} {x}^{\mathrm{2}} +{cd}^{\mathrm{3}} {x}^{\mathrm{3}} +… \\$$$${If}\:{dx}=\mathrm{0}\:{because}\:{d}=\mathrm{0}\:{or}\:{x}=\mathrm{0} \\$$$${we}\:{get}\:\frac{{c}}{\mathrm{1}−{dx}}=\frac{{c}}{\mathrm{1}−\mathrm{0}}={c}\:{which}\:{is}\:{equal} \\$$$${to}\:{c}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{0}^{{n}−\mathrm{1}} ={c}\left\{\mathrm{1}+\mathrm{0}+\mathrm{0}+\mathrm{0}+…\right\}={c}. \\$$$$\:\:\:\:\:\:\:\:\:\: \\$$$$\\$$$$\\$$
Commented by Rasheed Soomro last updated on 25/Nov/15
$$\mathcal{T}{h}^{{a}} {nk}\mathcal{S} \\$$