Question Number 29975 by abdo imad last updated on 14/Feb/18
![let give 0<α<1 1) prove that π coth(πα) −(1/α) = Σ_(n=1) ^∞ ((2α)/(α^2 +n^2 )). 2)by integration on[0,1] find Π_(n=1) ^∞ (1+(1/n^2 )).](Q29975.png)
$$\:{let}\:{give}\:\mathrm{0}<\alpha<\mathrm{1} \\ $$ $$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\pi\:{coth}\left(\pi\alpha\right)\:−\frac{\mathrm{1}}{\alpha}\:=\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{2}\alpha}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }. \\ $$ $$\left.\mathrm{2}\right){by}\:{integration}\:{on}\left[\mathrm{0},\mathrm{1}\right]\:{find}\:\prod_{{n}=\mathrm{1}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right). \\ $$
Commented byabdo imad last updated on 16/Feb/18
![1)let developp the 2π periodic function f(x)=ch(αx) f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) a_n =(2/T) ∫_([T]) f(x)cos(nx)dx= (2/(2π)) ∫_(−π) ^π ch(αx)cos(nx)dx =(2/π) ∫_0 ^π ch(αx)cos(nx)dx⇒(π/2)a_n = ∫_0 ^π ((e^(αx) +e^(−αx) )/2)cos(nx)dx π a_n = ∫_0 ^π e^(αx) cos(nx)dx +∫_0 ^π e^(−αx) cos(nx)dx =I(α) +I(−α) I(α) =Re( ∫_0 ^π e^(αx+inx) dx)=Re( ∫_0 ^π e^((α+in)x) dx)but ∫_0 ^π e^((α+in)x) dx =[(1/(α+in)) e^((α+in)x) ]_0 ^π =(1/(α+in))( e^(απ) (−1)^n −1) =((α−in)/(α^2 +n^2 ))( (−1)^n e^(απ) −1) ⇒ I(α)= ((α( (−1)^n e^(απ) −1))/(α^2 +n^2 )) I(−α)= ((−α((−1)^n e^(−απ) −1))/(α^2 +n^2 )) πa_n =((α(−1)^n (e^(απ) − e^(−απ) ))/(α^2 +n^2 ))=((2α(−1)^n sh(απ))/(α^2 +n^2 ))⇒ a_n =(2/π) α sh(απ) (((−1)^n )/(α^2 +n^2 )) and a_0 =(2/π) ((sh(απ))/α) ⇒ ch(αx)= ((sh(απ))/(πα)) + ((2αsh(απ))/π) Σ_(n=1) ^∞ (((−1)^n )/(α^2 +n^2 ))cos(nx) x=π ⇒ch(πα)=((sh(απ))/(πα)) +((2αsh(απ))/π)Σ_(n=1) ^∞ (1/(α^2 +n^2 )) ⇒ coth(πα)=(1/(πα)) +((2α)/π)Σ_(n=1) ^∞ (1/(α^2 +n^2 )) ⇒ π coth(πα)−(1/α) = Σ_(n=1) ^(∞ ) ((2α)/(α^2 +n^2 )).](Q30100.png)
$$\left.\mathrm{1}\right){let}\:{developp}\:{the}\:\mathrm{2}\pi\:{periodic}\:{function}\:{f}\left({x}\right)={ch}\left(\alpha{x}\right) \\ $$ $${f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \:{cos}\left({nx}\right) \\ $$ $${a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:{f}\left({x}\right){cos}\left({nx}\right){dx}=\:\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{ch}\left(\alpha{x}\right){cos}\left({nx}\right){dx} \\ $$ $$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{ch}\left(\alpha{x}\right){cos}\left({nx}\right){dx}\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:\frac{{e}^{\alpha{x}} \:+{e}^{−\alpha{x}} }{\mathrm{2}}{cos}\left({nx}\right){dx} \\ $$ $$\pi\:{a}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:{e}^{\alpha{x}} \:{cos}\left({nx}\right){dx}\:+\int_{\mathrm{0}} ^{\pi} \:{e}^{−\alpha{x}} {cos}\left({nx}\right){dx}\:={I}\left(\alpha\right)\:+{I}\left(−\alpha\right) \\ $$ $${I}\left(\alpha\right)\:={Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\alpha{x}+{inx}} {dx}\right)={Re}\left(\:\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(\alpha+{in}\right){x}} {dx}\right){but} \\ $$ $$\int_{\mathrm{0}} ^{\pi} \:\:{e}^{\left(\alpha+{in}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{\alpha+{in}}\:{e}^{\left(\alpha+{in}\right){x}} \right]_{\mathrm{0}} ^{\pi} \:\:=\frac{\mathrm{1}}{\alpha+{in}}\left(\:{e}^{\alpha\pi} \left(−\mathrm{1}\right)^{{n}} \:−\mathrm{1}\right) \\ $$ $$=\frac{\alpha−{in}}{\alpha^{\mathrm{2}} +{n}^{\mathrm{2}} }\left(\:\left(−\mathrm{1}\right)^{{n}} \:{e}^{\alpha\pi} −\mathrm{1}\right)\:\Rightarrow\:{I}\left(\alpha\right)=\:\frac{\alpha\left(\:\left(−\mathrm{1}\right)^{{n}} {e}^{\alpha\pi} −\mathrm{1}\right)}{\alpha^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$ $${I}\left(−\alpha\right)=\:\frac{−\alpha\left(\left(−\mathrm{1}\right)^{{n}} {e}^{−\alpha\pi} \:−\mathrm{1}\right)}{\alpha^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$ $$\pi{a}_{{n}} =\frac{\alpha\left(−\mathrm{1}\right)^{{n}} \left({e}^{\alpha\pi} −\:{e}^{−\alpha\pi} \right)}{\alpha^{\mathrm{2}} +{n}^{\mathrm{2}} }=\frac{\mathrm{2}\alpha\left(−\mathrm{1}\right)^{{n}} {sh}\left(\alpha\pi\right)}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\Rightarrow \\ $$ $${a}_{{n}} =\frac{\mathrm{2}}{\pi}\:\alpha\:{sh}\left(\alpha\pi\right)\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:{and}\:\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi}\:\frac{{sh}\left(\alpha\pi\right)}{\alpha}\:\Rightarrow \\ $$ $${ch}\left(\alpha{x}\right)=\:\frac{{sh}\left(\alpha\pi\right)}{\pi\alpha}\:\:+\:\frac{\mathrm{2}\alpha{sh}\left(\alpha\pi\right)}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{cos}\left({nx}\right)\: \\ $$ $${x}=\pi\:\Rightarrow{ch}\left(\pi\alpha\right)=\frac{{sh}\left(\alpha\pi\right)}{\pi\alpha}\:+\frac{\mathrm{2}\alpha{sh}\left(\alpha\pi\right)}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} +{n}^{\mathrm{2}} }\:\Rightarrow \\ $$ $${coth}\left(\pi\alpha\right)=\frac{\mathrm{1}}{\pi\alpha}\:+\frac{\mathrm{2}\alpha}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\Rightarrow \\ $$ $$\pi\:{coth}\left(\pi\alpha\right)−\frac{\mathrm{1}}{\alpha}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\:\frac{\mathrm{2}\alpha}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }. \\ $$
Commented byabdo imad last updated on 16/Feb/18
![we have ∫_0 ^1 (πcoth(πα)−(1/α))dα =Σ_(n=1) ^∞ ∫_0 ^1 ((2α)/(α^2 +n^2 ))dα =Σ_(n=1) ^∞ [ ln(α^2 +n^2 )]_0 ^1 =Σ_(n=1) ^∞ ln(1+n^2 )−ln(n^2 )=Σ_(n=1) ^∞ ln(1+(1/n^2 )) =ln(Π_(n=1) ^∞ (1 +(1/n^2 ))) ⇒ Π_(n=1) ^∞ (1+(1/n^2 ))= e^(∫_0 ^1 (πcoth(πα) −(1/α))dα) but ch.απ=t give ∫_0 ^1 ((απcoth(απ) −1)/α)dα= ∫_0 ^π ((t cotht −1)/(t/π)) (dt/π) = ∫_0 ^π ((tcoth(t)−1)/t)dt= ∫_0 ^π ( ((e^t −e^(−t) )/(e^t +e^(−t) )) −(1/t))dt....be continued... π](Q30103.png)
$${we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\pi{coth}\left(\pi\alpha\right)−\frac{\mathrm{1}}{\alpha}\right){d}\alpha\:=\sum_{{n}=\mathrm{1}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}\alpha}{\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} }{d}\alpha \\ $$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \left[\:{ln}\left(\alpha^{\mathrm{2}} \:+{n}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} {ln}\left(\mathrm{1}+{n}^{\mathrm{2}} \right)−{ln}\left({n}^{\mathrm{2}} \right)=\sum_{{n}=\mathrm{1}} ^{\infty} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$ $$={ln}\left(\prod_{{n}=\mathrm{1}} ^{\infty} \:\left(\mathrm{1}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\right)\:\Rightarrow \\ $$ $$\prod_{{n}=\mathrm{1}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=\:{e}^{\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\pi{coth}\left(\pi\alpha\right)\:−\frac{\mathrm{1}}{\alpha}\right){d}\alpha} \:\:{but}\:\:{ch}.\alpha\pi={t}\:{give} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\alpha\pi{coth}\left(\alpha\pi\right)\:−\mathrm{1}}{\alpha}{d}\alpha=\:\int_{\mathrm{0}} ^{\pi} \:\frac{{t}\:{cotht}\:−\mathrm{1}}{\frac{{t}}{\pi}}\:\frac{{dt}}{\pi} \\ $$ $$=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{tcoth}\left({t}\right)−\mathrm{1}}{{t}}{dt}=\:\int_{\mathrm{0}} ^{\pi} \left(\:\frac{{e}^{{t}} −{e}^{−{t}} }{{e}^{{t}} +{e}^{−{t}} }\:−\frac{\mathrm{1}}{{t}}\right){dt}....{be}\:{continued}... \\ $$ $$\pi \\ $$