advanced-calculus-prove-that-0-x-2-cosh-2-x-2-dx-2-2-4-pi-1-2-

Question Number 140715 by mnjuly1970 last updated on 11/May/21

......advanced calculus...... prove that: 𝛗:=∫_0 ^( ∞) (x^2 /(cosh^2 (x^2 )))dx=^? (((√2) −2)/4) (√π) ζ ( (1/2) ) ..............

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:……{advanced}\:\:{calculus}…… \\ $$$$\:\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}{dx}\overset{?} {=}\frac{\sqrt{\mathrm{2}}\:−\mathrm{2}}{\mathrm{4}}\:\sqrt{\pi}\:\zeta\:\left(\:\frac{\mathrm{1}}{\mathrm{2}}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:………….. \\ $$

Answered by Dwaipayan Shikari last updated on 11/May/21

∫_0 ^∞ (x^2 /(cosh^2 (x^2 )))dx=(1/2)∫_0 ^∞ ((√u)/(cosh^2 (u)))du =∫_0 ^∞ ((√u)/(1+cosh(2u)))=2∫_0 ^∞ ((e^(−2u) (√u))/((1+e^(−2u) )^2 ))du Σ_(n=1) ^∞ (−1)^(n+1) e^(−2nu) =(1/(1+e^(−2u) ))⇒Σ_(n=1) ^∞ (−1)^(n+1) ne^(−2nu) =((−e^(2u) )/((1+e^(−2u) )^2 )) =−2∫_0 ^∞ Σ_(n=1) ^∞ (−1)^(n+1) (√u) ne^(−2nu) du=2Σ_(n=1) ^∞ (−1)^n n∫_0 ^∞ (√u) e^(−2nu) du 2nu=t ⇒Σ_(n=1) ^∞ (−1)^n ∫_0 ^∞ ((√t)/( (√(2n))))e^(−t) dt=−Σ_(n=1) ^∞ ((√π)/( (√2)(√n)))(−1)^(n+1) =−(√(π/2)) ζ((1/2))(1−(1/2^((1/2)−1) ))=(√(π/2)) ((√2)−1)ζ((1/2))

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{{cosh}^{\mathrm{2}} \left({x}^{\mathrm{2}} \right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{u}}}{{cosh}^{\mathrm{2}} \left({u}\right)}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{u}}}{\mathrm{1}+{cosh}\left(\mathrm{2}{u}\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{2}{u}} \sqrt{{u}}}{\left(\mathrm{1}+{e}^{−\mathrm{2}{u}} \right)^{\mathrm{2}} }{du} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{−\mathrm{2}{nu}} =\frac{\mathrm{1}}{\mathrm{1}+{e}^{−\mathrm{2}{u}} }\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {ne}^{−\mathrm{2}{nu}} =\frac{−{e}^{\mathrm{2}{u}} }{\left(\mathrm{1}+{e}^{−\mathrm{2}{u}} \right)^{\mathrm{2}} }\:\: \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \sqrt{{u}}\:{ne}^{−\mathrm{2}{nu}} {du}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {n}\int_{\mathrm{0}} ^{\infty} \sqrt{{u}}\:{e}^{−\mathrm{2}{nu}} {du} \\ $$$$\mathrm{2}{nu}={t}\:\:\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{t}}}{\:\sqrt{\mathrm{2}{n}}}{e}^{−{t}} {dt}=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{2}}\sqrt{{n}}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$$=−\sqrt{\frac{\pi}{\mathrm{2}}}\:\zeta\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }\right)=\sqrt{\frac{\pi}{\mathrm{2}}}\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\zeta\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 12/May/21

$${grateful}\:…. \\ $$

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