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-

FacebookTweetPin

Question Number 2675 by prakash jain last updated on 24/Nov/15

$$\mathrm{Bases}\mathrm{on}\mathrm{suggestion}\mathrm{from}\mathrm{Filup}\mathrm{and}\mathrm{some}$$$$\mathrm{discussion}\mathrm{on}\mathrm{that}\mathrm{I}\mathrm{am}\mathrm{suggesting}\mathrm{that}\mathrm{we}$$$$\mathrm{sequence},\mathrm{series}\mathrm{and}\mathrm{related}\mathrm{function}\mathrm{as}\mathrm{a}$$$$\mathrm{topic}\mathrm{for}\mathrm{this}\mathrm{month}.$$$$\zeta \left(x\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}{n}^{-x},x\in \mathbb{R},x>1$$$$\mathrm{Show}\mathrm{that}$$$$\zeta \left(x\right)=\frac{1}{\mathrm{\Gamma }\left(x\right)}{\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{e^t-1}\mathrm{d}t$$

Commented by prakash jain last updated on 24/Nov/15

$$\zeta -\mathrm{Riemann}\mathrm{zeta}\mathrm{function}$$$$\mathrm{\Gamma }-\mathrm{Gamma}\mathrm{function}$$$$\mathrm{These}2\mathrm{functions}\mathrm{were}\mathrm{mentioned}\mathrm{as}\mathrm{part}$$$$\mathrm{of}\mathrm{some}\mathrm{of}\mathrm{the}\mathrm{questions}\mathrm{earlier}.$$$$\mathrm{For}\mathrm{this}\mathrm{question}\mathrm{only}\mathrm{consider}\mathrm{real}\mathrm{numbers}$$$$\mathrm{although}\mathrm{the}\mathrm{functions}\mathrm{are}\mathrm{defined}\mathrm{for}$$$$\mathrm{complex}\mathrm{numbers}\mathrm{as}\mathrm{well}.$$

Commented by Rasheed Soomro last updated on 24/Nov/15

$$\mathcal{I}haveonlyheardthenamesofreimannzetaand$$$$gammafunctions!Anywaynoharmintrying!$$$$Bythewaywhowillbethejudge?$$

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

$$123456\mathrm{is}\mathrm{most}\mathrm{knowledgeable}\mathrm{among}\mathrm{us}.$$$$\left(\mathrm{s}\right)\mathrm{he}\mathrm{will}\mathrm{post}\mathrm{the}\mathrm{challenge}\mathrm{at}\mathrm{some}\mathrm{point}$$$$\mathrm{and}\mathrm{judge}\mathrm{us}.$$$$$$$$\mathrm{In}\mathrm{any}\mathrm{case}\mathrm{regular}\mathrm{Q}\mathrm{and}\mathrm{A}\mathrm{continues}\mathrm{as}$$$$\mathrm{usual}\mathrm{from}\mathrm{everyone}.$$

Commented by Yozzi last updated on 24/Nov/15

$$Cool.$$

Commented by Yozzi last updated on 24/Nov/15

$$Isitdxordtintheexpression{\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{e^t-1}dx?$$

Commented by prakash jain last updated on 24/Nov/15

$$\mathrm{d}t\mathrm{corrected}.$$

Commented by 123456 last updated on 01/Dec/15

$$24/12$$

Answered by Yozzi last updated on 24/Nov/15

$$\frac{t^{x-1}}{e^t-1}=\frac{t^{x-1}{e}^{-t}}{1-e^{-t}}$$$$Fort>0,\mid e^{-t}\mid <1.\frac{1}{1-e^{-t}}isthethe$$$$sumofaninfiniteG.Pwithfirstterm$$$$1andcommonratio{e}^{-t}.$$$$\therefore \frac{1}{1-e^{-t}}=1+e^{-t}+e^{-2t}+e^{-3t}+\dots$$$$\therefore \frac{t^{x-1}{e}^{-t}}{1-e^{-t}}=t^{x-1}{e}^{-t}+t^{x-1}{e}^{-2t}+t^{x-1}{e}^{-3t}+\dots$$$$\frac{t^{x-1}{e}^{-t}}{1-e^{-t}}=\underset{r=1}{\sum ^{\mathrm{\infty }}}{t}^{x-1}{e}^{-rt}$$$$LetI={\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{e^t-1}dt={\int }_0^{\mathrm{\infty }}\left(\underset{r=1}{\sum ^{\mathrm{\infty }}}{t}^{x-1}{e}^{-rt}\right)dt$$$$I=\underset{r=1}{\sum ^{\mathrm{\infty }}}\left({\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-rt}dt\right)$$$$letu=rt\Rightarrow t=\frac{1}{r}u\Rightarrow \frac{1}{r}du=dt$$$$Att=0,u=0.Ast\to \mathrm{\infty },u\to \mathrm{\infty }forr>0.$$$$\therefore t^{x-1}=r^{1-x}{u}^{x-1}$$$$\therefore {\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-rt}dt={\int }_0^{\mathrm{\infty }}{r}^{1-x}{u}^{x-1}{e}^{-u}\times \frac{1}{r}du$$$$=r^{-x}{\int }_0^{\mathrm{\infty }}{u}^{x-1}{e}^{-u}du$$$$Thegammafunction\mathrm{\Gamma }\left(x\right)defined$$$$forx>0isexpressibleas$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt.$$$$\therefore {\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-rt}dt=r^{-x}\mathrm{\Gamma }\left(x\right)$$$$\therefore I=\mathrm{\Gamma }\left(x\right)\times \underset{r=1}{\sum ^{\mathrm{\infty }}}{r}^{-x}$$$$But,\zeta \left(x\right)=\underset{r=1}{\sum ^{\mathrm{\infty }}}{r}^{-x},x\in \mathbb{R},x>1.$$$$\therefore I=\mathrm{\Gamma }\left(x\right)\zeta \left(x\right)\Rightarrow \zeta \left(x\right)=\frac{1}{\mathrm{\Gamma }\left(x\right)}{\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{e^t-1}dt$$$$$$

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

$$2ndline:$$$$Fort>0,\underset{-}{\mid e^{-t}\mid }<1.\frac{1}{1-e^{-t}}isthethe$$$${Couldn}^{\prime }twewrite{e}^{-t}insteadof\mid e^{-t}\mid$$$$Orthereisanyspecificreasonforthis?$$

Commented by Yozzi last updated on 25/Nov/15

$${It}^{\prime }swrittenlikethattospecifically$$$$indicatethattheconvergence$$$$conditionforaninfinitegeometric$$$$seriesissatisfied.Soexpressions$$$$oftheform$$$$\frac{c}{1-dx}$$$$representthesumofaninfinite$$$$geometricseriesunderthecondition$$$$that\mid dx\mid <1\Rightarrow \mid x\mid <1/d(d\ne 0).The$$$$seriesisthendeducedtobeoftheform$$$$c\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(dx\right)}^{n-1}=c\{1+dx+d^2{x}^2+d^3{x}^3+\dots \}$$$$=c+cdx+{cd}^2{x}^2+{cd}^3{x}^3+\dots$$$$Ifdx=0becaused=0orx=0$$$$weget\frac{c}{1-dx}=\frac{c}{1-0}=cwhichisequal$$$$toc\underset{n=1}{\sum ^{\mathrm{\infty }}}{0}^{n-1}=c\{1+0+0+0+\dots \}=c.$$$$$$$$$$$$$$

Commented by Rasheed Soomro last updated on 25/Nov/15

$$\mathcal{T}{h}^{a}nk\mathcal{S}$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin