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Question Number 190622 by mnjuly1970 last updated on 07/Apr/23
prove : ∫_(−∞) ^( ∞) ((( x)/( ⋮ )^2 dx= Σ_(k=1) ^∞ (1/( k^2 )) ⋖))
$$ \\ $$$$\:\:\:{prove}\:: \\ $$$$\:\:\int_{−\infty} ^{\:\infty} \:\:\:\left(\frac{\:{x}}{\left.\:\underline{\vdots} \right)^{\mathrm{2}} \mathrm{d}{x}=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\:{k}\:^{\mathrm{2}} }\:\:\:\lessdot}\right. \\ $$
Answered by leodera last updated on 17/May/23
cosh^2 (x) = (((e^x +e^(−x) )/2))^2 = (e^(2x) /4)(1+e^(−2x) )^2 ∫_(−∞) ^∞ (x^2 /(cosh^2 (x)))dx = 8∫_0 ^∞ ((x^2 e^(−2x) )/((1+e^(−2x) )^2 ))dx = 8∫_0 ^∞ x^2 e^(−2x) Σ_(k=0) ^∞ e^(−2kx) (−1)^k (1+k)dx = 8Σ_(k=0) ^∞ (−1)^k (1+k)∫_0 ^∞ x^2 e^(−(2+2k)x) dx = 16Σ_(k=0) ^∞ (((−1)^k (1+k))/(2^3 (1+k)^3 )) = 2Σ_(k=0) ^∞ (−1)^k (1/((1+k)^2 )) 2Σ_(k=1) ^∞ (−1)^(k−1) (1/k^2 ) = 2η(2) = 2(1−2^(1−2) )ζ(2) = ζ(2) = Σ_(k=1) ^∞ (1/k^2 ) = (π^2 /6)
$$ \\ $$$$\mathrm{cosh}\:^{\mathrm{2}} \left({x}\right)\:=\:\left(\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}\right)^{\mathrm{2}} =\:\frac{{e}^{\mathrm{2}{x}} }{\mathrm{4}}\left(\mathrm{1}+{e}^{−\mathrm{2}{x}} \right)^{\mathrm{2}} \\ $$$$\int_{−\infty} ^{\infty} \frac{{x}^{\mathrm{2}} }{\mathrm{cosh}\:^{\mathrm{2}} \left({x}\right)}{dx}\:=\:\mathrm{8}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} {e}^{−\mathrm{2}{x}} }{\left(\mathrm{1}+{e}^{−\mathrm{2}{x}} \right)^{\mathrm{2}} }{dx} \\ $$$$=\:\mathrm{8}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−\mathrm{2}{x}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}{kx}} \left(−\mathrm{1}\right)^{{k}} \left(\mathrm{1}+{k}\right){dx} \\ $$$$=\:\mathrm{8}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{1}+{k}\right)\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−\left(\mathrm{2}+\mathrm{2}{k}\right){x}} {dx} \\ $$$$=\:\mathrm{16}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{1}+{k}\right)}{\mathrm{2}^{\mathrm{3}} \left(\mathrm{1}+{k}\right)^{\mathrm{3}} }\:=\:\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\left(\mathrm{1}+{k}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\:\mathrm{2}\eta\left(\mathrm{2}\right)\:=\:\mathrm{2}\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\mathrm{2}} \right)\zeta\left(\mathrm{2}\right) \\ $$$$=\:\zeta\left(\mathrm{2}\right)\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$

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