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Question Number 2675 by prakash jain last updated on 24/Nov/15
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. ζ(x)=Σ_(n=1) ^∞ n^(−x) , x∈R, x>1 Show that ζ(x)=(1/(Γ(x)))∫_0 ^∞ (t^(x−1) /(e^t −1))dt
$$\mathrm{Bases}\:\mathrm{on}\:\mathrm{suggestion}\:\mathrm{from}\:\mathrm{Filup}\:\mathrm{and}\:\mathrm{some} \\ $$$$\mathrm{discussion}\:\mathrm{on}\:\mathrm{that}\:\mathrm{I}\:\mathrm{am}\:\mathrm{suggesting}\:\mathrm{that}\:\mathrm{we} \\ $$$$\mathrm{sequence},\:\mathrm{series}\:\mathrm{and}\:\mathrm{related}\:\mathrm{function}\:\mathrm{as}\:\mathrm{a} \\ $$$$\mathrm{topic}\:\mathrm{for}\:\mathrm{this}\:\mathrm{month}. \\ $$$$\zeta\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{−{x}} ,\:{x}\in\mathbb{R},\:{x}>\mathrm{1} \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\zeta\left({x}\right)=\frac{\mathrm{1}}{\Gamma\left({x}\right)}\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}\mathrm{d}{t} \\ $$
Commented by prakash jain last updated on 24/Nov/15
ζ−Riemann zeta function Γ−Gamma function These 2 functions were mentioned as part of some of the questions earlier. For this question only consider real numbers although the functions are defined for complex numbers as well.
$$\zeta−\mathrm{Riemann}\:\mathrm{zeta}\:\mathrm{function} \\ $$$$\Gamma−\mathrm{Gamma}\:\mathrm{function} \\ $$$$\mathrm{These}\:\mathrm{2}\:\mathrm{functions}\:\mathrm{were}\:\mathrm{mentioned}\:\mathrm{as}\:\mathrm{part} \\ $$$$\mathrm{of}\:\mathrm{some}\:\mathrm{of}\:\mathrm{the}\:\mathrm{questions}\:\mathrm{earlier}. \\ $$$$\mathrm{For}\:\mathrm{this}\:\mathrm{question}\:\mathrm{only}\:\mathrm{consider}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{although}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{are}\:\mathrm{defined}\:\mathrm{for} \\ $$$$\mathrm{complex}\:\mathrm{numbers}\:\mathrm{as}\:\mathrm{well}. \\ $$
Commented by Rasheed Soomro last updated on 24/Nov/15
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?
$$\mathcal{I}\:{have}\:{only}\:{heard}\:{the}\:{names}\:{of}\:{reimann}\:{zeta}\:{and} \\ $$$${gamma}\:{functions}!\:{Anyway}\:{no}\:{harm}\:{in}\:{trying}! \\ $$$${By}\:{the}\:{way}\:{who}\:{will}\:{be}\:{the}\:{judge}? \\ $$
Commented by prakash jain last updated on 24/Nov/15
This question was not meant to be a challenge just learning details of a few things that got mentioned earlier.
$$\mathrm{This}\:\mathrm{question}\:\mathrm{was}\:\mathrm{not}\:\mathrm{meant}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a} \\ $$$$\mathrm{challenge}\:\mathrm{just}\:\mathrm{learning}\:\mathrm{details}\:\mathrm{of}\:\mathrm{a}\:\mathrm{few} \\ $$$$\mathrm{things}\:\mathrm{that}\:\mathrm{got}\:\mathrm{mentioned}\:\mathrm{earlier}. \\ $$
Commented by prakash jain last updated on 24/Nov/15
123456 is most knowledgeable among us. (s)he will post the challenge at some point and judge us. In any case regular Q and A continues as usual from everyone.
$$\mathrm{123456}\:\mathrm{is}\:\mathrm{most}\:\mathrm{knowledgeable}\:\mathrm{among}\:\mathrm{us}. \\ $$$$\left(\mathrm{s}\right)\mathrm{he}\:\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{challenge}\:\mathrm{at}\:\mathrm{some}\:\mathrm{point} \\ $$$$\mathrm{and}\:\mathrm{judge}\:\mathrm{us}. \\ $$$$ \\ $$$$\mathrm{In}\:\mathrm{any}\:\mathrm{case}\:\mathrm{regular}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{A}\:\mathrm{continues}\:\mathrm{as} \\ $$$$\mathrm{usual}\:\mathrm{from}\:\mathrm{everyone}. \\ $$
Commented by Yozzi last updated on 24/Nov/15
Cool.
$${Cool}. \\ $$
Commented by Yozzi last updated on 24/Nov/15
Is it dx or dt in the expression ∫_0 ^∞ (t^(x−1) /(e^t −1))dx?
$${Is}\:{it}\:{dx}\:{or}\:{dt}\:{in}\:{the}\:{expression}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dx}? \\ $$
Commented by prakash jain last updated on 24/Nov/15
dt corrected.
$$\mathrm{d}{t}\:\mathrm{corrected}. \\ $$
Commented by 123456 last updated on 01/Dec/15
24/12
$$\mathrm{24}/\mathrm{12} \\ $$
Answered by Yozzi last updated on 24/Nov/15
(t^(x−1) /(e^t −1))=((t^(x−1) e^(−t) )/(1−e^(−t) )) For t>0, ∣e^(−t) ∣<1. (1/(1−e^(−t) )) is the the sum of an infinite G.P with first term 1 and common ratio e^(−t) . ∴ (1/(1−e^(−t) ))=1+e^(−t) +e^(−2t) +e^(−3t) +... ∴((t^(x−1) e^(−t) )/(1−e^(−t) ))=t^(x−1) e^(−t) +t^(x−1) e^(−2t) +t^(x−1) e^(−3t) +... ((t^(x−1) e^(−t) )/(1−e^(−t) ))=Σ_(r=1) ^∞ t^(x−1) e^(−rt) Let I=∫_0 ^∞ (t^(x−1) /(e^t −1))dt=∫_0 ^∞ (Σ_(r=1) ^∞ t^(x−1) e^(−rt) )dt I=Σ_(r=1) ^∞ (∫_0 ^∞ t^(x−1) e^(−rt) dt) let u=rt⇒t=(1/r)u⇒(1/r)du=dt At t=0,u=0. As t→∞, u→∞ for r>0. ∴t^(x−1) =r^(1−x) u^(x−1) ∴∫_0 ^∞ t^(x−1) e^(−rt) dt=∫_0 ^∞ r^(1−x) u^(x−1) e^(−u) ×(1/r)du =r^(−x) ∫_0 ^∞ u^(x−1) e^(−u) du The gamma function Γ(x) defined for x>0 is expressible as Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt. ∴∫_0 ^∞ t^(x−1) e^(−rt) dt=r^(−x) Γ(x) ∴I=Γ(x)×Σ_(r=1) ^∞ r^(−x) But, ζ(x)=Σ_(r=1) ^∞ r^(−x) ,x∈R,x>1. ∴I=Γ(x)ζ(x)⇒ζ(x)=(1/(Γ(x)))∫_0 ^∞ (t^(x−1) /(e^t −1))dt
$$\frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}=\frac{{t}^{{x}−\mathrm{1}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} } \\ $$$${For}\:{t}>\mathrm{0},\:\:\mid{e}^{−{t}} \mid<\mathrm{1}.\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} }\:{is}\:{the}\:{the} \\ $$$${sum}\:{of}\:{an}\:{infinite}\:{G}.{P}\:{with}\:{first}\:{term} \\ $$$$\mathrm{1}\:{and}\:{common}\:{ratio}\:{e}^{−{t}} . \\ $$$$\therefore\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} }=\mathrm{1}+{e}^{−{t}} +{e}^{−\mathrm{2}{t}} +{e}^{−\mathrm{3}{t}} +… \\ $$$$\therefore\frac{{t}^{{x}−\mathrm{1}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }={t}^{{x}−\mathrm{1}} {e}^{−{t}} +{t}^{{x}−\mathrm{1}} {e}^{−\mathrm{2}{t}} +{t}^{{x}−\mathrm{1}} {e}^{−\mathrm{3}{t}} +… \\ $$$$\frac{{t}^{{x}−\mathrm{1}} {e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{t}^{{x}−\mathrm{1}} {e}^{−{rt}} \\ $$$${Let}\:{I}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dt}=\int_{\mathrm{0}} ^{\infty} \left(\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{t}^{{x}−\mathrm{1}} {e}^{−{rt}} \right){dt} \\ $$$${I}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{rt}} {dt}\right) \\ $$$${let}\:{u}={rt}\Rightarrow{t}=\frac{\mathrm{1}}{{r}}{u}\Rightarrow\frac{\mathrm{1}}{{r}}{du}={dt} \\ $$$${At}\:{t}=\mathrm{0},{u}=\mathrm{0}.\:{As}\:{t}\rightarrow\infty,\:{u}\rightarrow\infty\:{for}\:{r}>\mathrm{0}. \\ $$$$\therefore{t}^{{x}−\mathrm{1}} ={r}^{\mathrm{1}−{x}} {u}^{{x}−\mathrm{1}} \\ $$$$\therefore\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{rt}} {dt}=\int_{\mathrm{0}} ^{\infty} {r}^{\mathrm{1}−{x}} {u}^{{x}−\mathrm{1}} {e}^{−{u}} ×\frac{\mathrm{1}}{{r}}{du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={r}^{−{x}} \int_{\mathrm{0}} ^{\infty} {u}^{{x}−\mathrm{1}} {e}^{−{u}} {du} \\ $$$${The}\:{gamma}\:{function}\:\Gamma\left({x}\right)\:{defined} \\ $$$${for}\:{x}>\mathrm{0}\:{is}\:{expressible}\:{as} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{t}} {dt}. \\ $$$$\therefore\int_{\mathrm{0}} ^{\infty} {t}^{{x}−\mathrm{1}} {e}^{−{rt}} {dt}={r}^{−{x}} \Gamma\left({x}\right) \\ $$$$\therefore{I}=\Gamma\left({x}\right)×\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{r}^{−{x}} \\ $$$${But},\:\zeta\left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{r}^{−{x}} \:\:,{x}\in\mathbb{R},{x}>\mathrm{1}. \\ $$$$\therefore{I}=\Gamma\left({x}\right)\zeta\left({x}\right)\Rightarrow\zeta\left({x}\right)=\frac{\mathrm{1}}{\Gamma\left({x}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{x}−\mathrm{1}} }{{e}^{{t}} −\mathrm{1}}{dt}\: \\ $$$$ \\ $$
Commented by 123456 last updated on 24/Nov/15
nice
$$\mathrm{nice} \\ $$
Commented by Filup last updated on 25/Nov/15
Awesome! I had no idea how to do this one! So simple!
$${Awesome}!\:\mathrm{I}\:\mathrm{had}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{how}\:\mathrm{to}\:\mathrm{do}\:\mathrm{this} \\ $$$$\mathrm{one}!\:{S}\mathrm{o}\:\mathrm{simple}! \\ $$
Commented by Rasheed Soomro last updated on 25/Nov/15
2nd line: For t>0, ∣e^(−t) ∣_(−) <1. (1/(1−e^(−t) )) is the the Couldn′t we write e^(−t) instead of ∣e^(−t) ∣ Or there is any specific reason for this?
$$\mathrm{2}{nd}\:{line}: \\ $$$${For}\:{t}>\mathrm{0},\:\:\underset{−} {\mid{e}^{−{t}} \mid}<\mathrm{1}.\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} }\:{is}\:{the}\:{the} \\ $$$${Couldn}'{t}\:{we}\:{write}\:{e}^{−{t}} \:{instead}\:{of}\:\mid{e}^{−{t}} \mid \\ $$$${Or}\:{there}\:{is}\:{any}\:{specific}\:{reason}\:{for}\:{this}? \\ $$
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 (c/(1−dx)) represent the sum of an infinite geometric series under the condition that ∣dx∣<1⇒∣x∣<1/d (d≠0). The series is then deduced to be of the form cΣ_(n=1) ^∞ (dx)^(n−1) =c{1+dx+d^2 x^2 +d^3 x^3 +...} =c+cdx+cd^2 x^2 +cd^3 x^3 +... If dx=0 because d=0 or x=0 we get (c/(1−dx))=(c/(1−0))=c which is equal to cΣ_(n=1) ^∞ 0^(n−1) =c{1+0+0+0+...}=c.
$${It}'{s}\:{written}\:{like}\:{that}\:{to}\:{specifically} \\ $$$${indicate}\:{that}\:{the}\:{convergence} \\ $$$${condition}\:{for}\:{an}\:{infinite}\:{geometric} \\ $$$${series}\:{is}\:{satisfied}.\:{So}\:{expressions} \\ $$$${of}\:{the}\:{form}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{c}}{\mathrm{1}−{dx}}\:\: \\ $$$${represent}\:{the}\:{sum}\:{of}\:{an}\:{infinite}\: \\ $$$${geometric}\:{series}\:{under}\:{the}\:{condition} \\ $$$${that}\:\mid{dx}\mid<\mathrm{1}\Rightarrow\mid{x}\mid<\mathrm{1}/{d}\:\:\left({d}\neq\mathrm{0}\right).\:{The} \\ $$$${series}\:{is}\:{then}\:{deduced}\:{to}\:{be}\:{of}\:{the}\:{form} \\ $$$${c}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({dx}\right)^{{n}−\mathrm{1}} ={c}\left\{\mathrm{1}+{dx}+{d}^{\mathrm{2}} {x}^{\mathrm{2}} +{d}^{\mathrm{3}} {x}^{\mathrm{3}} +…\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={c}+{cdx}+{cd}^{\mathrm{2}} {x}^{\mathrm{2}} +{cd}^{\mathrm{3}} {x}^{\mathrm{3}} +… \\ $$$${If}\:{dx}=\mathrm{0}\:{because}\:{d}=\mathrm{0}\:{or}\:{x}=\mathrm{0} \\ $$$${we}\:{get}\:\frac{{c}}{\mathrm{1}−{dx}}=\frac{{c}}{\mathrm{1}−\mathrm{0}}={c}\:{which}\:{is}\:{equal} \\ $$$${to}\:{c}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{0}^{{n}−\mathrm{1}} ={c}\left\{\mathrm{1}+\mathrm{0}+\mathrm{0}+\mathrm{0}+…\right\}={c}. \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 25/Nov/15
Th^a nkS
$$\mathcal{T}{h}^{{a}} {nk}\mathcal{S} \\ $$

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