Question Number 27495 by abdo imad last updated on 07/Jan/18
$${find}\:\alpha\:{and}\:\beta\:{from}\:{R}\:/\int_{\mathrm{0}} ^{\pi} \left(\alpha{t}^{\mathrm{2}} +\beta{t}\right){cos}\left({nt}\right){dt}=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$${for}\:{all}\:{number}\:{n}\:{from}\:{N}^{\ast\:} \:{then}\:{find} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 19/Jan/18
![let put I= ∫_0 ^π (αt^2 +βt)cos(nt)dt I= Re ( ∫_0 ^π (αt^2 +βt)^ e^(int) t) by parts u= αt^2 +βt and v^′ = e^(int) I= [(αt^2 +βt)(e^(int) /(in)) ]_0 ^π − (1/(in))∫_0 ^π (2αt+β)e^(int) dt =(1/(in))((απ^2 +βπ)(−1)^n ) −(1/(in)){[ (1/(in))(2αt+β) e^(int) ]_0 ^π −(1/(in))∫_0 ^π (2α)e^(int) dt} we find I= (1/n^2 ) ((2απ +β)(−1)^n −β) I = (1/n^2 ) ∀n∈N^∗ ⇔ (2απ+β)(−1)^n −β= 1 for alln ⇔ β =−1 and 2απ−1=0 ⇒ α=(1/(2π)) and β=−1 so (1/n^2 )= ∫_0 ^π ((1/(2π)) t^2 −t)cos(nt)dt Σ_(n=1) ^∝ (1/n^2 ) =∫_0 ^π ((1/(2π))t^2 −t)( Σ_(n=1) ^∝ cos(nt))dt but Σ_(n=1) ^∝ cos(nt)=Re( Σ_(n=0) ^∝ (^ e^(it) )^n )−1=Re( (1/(1−e^(it) )))−1 =Re( (1/(2sin^2 ((t/2))−2isin((t/2))cos((t/2))))) =Re ( (1/(−2isin((t/2)) e^(i(t/2)) ))) =Re( (1/2)i((cos((t/2))−i sin((t/2)))/(sin((t/2))))) = (1/2)−1=−(1/2) Σ_(n=1) ^∝ (1/n^2 ) =−(1/2) ∫_0 ^π ((1/(2π))t^2 −t)dt =−(1/(4π)) [ (t^3 /3)]_0 ^π +(1/2) [ (t^2 /2)]_0 ^π = (π^2 /4)−(1/(4π)) (π^3 /3)=(π^2 /4) −(π^2 /(12)) = ((3π^2 −π^2 )/(12)) = (π^2 /6).](https://www.tinkutara.com/question/Q28069.png)
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\pi} \left(\alpha{t}^{\mathrm{2}} \:+\beta{t}\right){cos}\left({nt}\right){dt} \\ $$$${I}=\:{Re}\:\left(\:\int_{\mathrm{0}} ^{\pi} \left(\alpha{t}^{\mathrm{2}} +\beta{t}\right)^{} {e}^{{int}} {t}\right)\:\:{by}\:{parts} \\ $$$${u}=\:\alpha{t}^{\mathrm{2}} +\beta{t}\:\:\:{and}\:\:{v}^{'} =\:{e}^{{int}} \\ $$$${I}=\:\left[\left(\alpha{t}^{\mathrm{2}} +\beta{t}\right)\frac{{e}^{{int}} }{{in}}\:\:\right]_{\mathrm{0}} ^{\pi} \:\:−\:\frac{\mathrm{1}}{{in}}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{2}\alpha{t}+\beta\right){e}^{{int}} {dt} \\ $$$$=\frac{\mathrm{1}}{{in}}\left(\left(\alpha\pi^{\mathrm{2}} +\beta\pi\right)\left(−\mathrm{1}\right)^{{n}} \right)\:−\frac{\mathrm{1}}{{in}}\left\{\left[\:\:\frac{\mathrm{1}}{{in}}\left(\mathrm{2}\alpha{t}+\beta\right)\:{e}^{{int}} \right]_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{{in}}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{2}\alpha\right){e}^{{int}} {dt}\right\} \\ $$$${we}\:{find}\:\:{I}=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\left(\left(\mathrm{2}\alpha\pi\:+\beta\right)\left(−\mathrm{1}\right)^{{n}} −\beta\right) \\ $$$${I}\:=\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\forall{n}\in{N}^{\ast} \:\:\Leftrightarrow\:\left(\mathrm{2}\alpha\pi+\beta\right)\left(−\mathrm{1}\right)^{{n}} \:−\beta=\:\mathrm{1}\:{for}\:{alln} \\ $$$$\Leftrightarrow\:\beta\:=−\mathrm{1}\:{and}\:\mathrm{2}\alpha\pi−\mathrm{1}=\mathrm{0}\:\Rightarrow\:\alpha=\frac{\mathrm{1}}{\mathrm{2}\pi}\:{and}\:\beta=−\mathrm{1}\:{so} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\:\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}}{\mathrm{2}\pi}\:{t}^{\mathrm{2}} −{t}\right){cos}\left({nt}\right){dt} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:=\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}}{\mathrm{2}\pi}{t}^{\mathrm{2}} −{t}\right)\left(\:\sum_{{n}=\mathrm{1}} ^{\propto} \:{cos}\left({nt}\right)\right){dt}\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\propto} \:{cos}\left({nt}\right)={Re}\left(\:\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\left(^{} {e}^{{it}} \right)^{{n}} \right)−\mathrm{1}={Re}\left(\:\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{{it}} }\right)−\mathrm{1} \\ $$$$={Re}\left(\:\:\frac{\mathrm{1}}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)−\mathrm{2}{isin}\left(\frac{{t}}{\mathrm{2}}\right){cos}\left(\frac{{t}}{\mathrm{2}}\right)}\right) \\ $$$$={Re}\:\left(\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{−\mathrm{2}{isin}\left(\frac{{t}}{\mathrm{2}}\right)\:{e}^{{i}\frac{{t}}{\mathrm{2}}} }\right)\:={Re}\left(\:\:\frac{\mathrm{1}}{\mathrm{2}}{i}\frac{{cos}\left(\frac{{t}}{\mathrm{2}}\right)−{i}\:{sin}\left(\frac{{t}}{\mathrm{2}}\right)}{{sin}\left(\frac{{t}}{\mathrm{2}}\right)}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \left(\frac{\mathrm{1}}{\mathrm{2}\pi}{t}^{\mathrm{2}} −{t}\right){dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}\pi}\:\:\left[\:\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\pi} \:+\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\pi} \:\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}\pi}\:\frac{\pi^{\mathrm{3}} }{\mathrm{3}}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$=\:\frac{\mathrm{3}\pi^{\mathrm{2}} \:−\pi^{\mathrm{2}} }{\mathrm{12}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}. \\ $$$$ \\ $$