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Question Number 60534 by aliesam last updated on 21/May/19

Commented by maxmathsup by imad last updated on 22/May/19

we have ∫_(−∞) ^(+∞) (dx/(1+e^x^2 )) =2 ∫_0 ^(+∞) (e^(−x^2 ) /(1+e^(−x^2 ) ))dx =2 ∫_0 ^∞ e^(−x^2 ) (Σ_(n=0) ^∞ (−1)^n e^(−nx^2 ) )dx=2 Σ_(=0) ^(∞ ) (−1)^n ∫_0 ^∞ e^(−(n+1)x^2 ) dx =_((√(n+1))x =u) 2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ e^(−u^2 ) (du/(√(n+1))) =2 ((√π)/2) Σ_(n=0) ^∞ (((−1)^n )/(√(n+1))) =(√π)Σ_(n=0) ^∞ (((−1)^n )/(√(n+1))) we have Σ_(n=0) ^∞ (((−1)^n )/(√(n+1))) =Σ_(n=1) ^∞ (((−1)^(n−1) )/(√n)) =−Σ_(n=1) ^∞ (1/(√(2n))) +Σ_(n=0) ^∞ (1/(√(2n+1))) =−(1/(√2))ξ((1/2))+Σ_(n=0) ^∞ (1/(√(2n+1))) Σ_(n=1) ^∞ (1/(√n)) =Σ_(n=1) ^∞ (1/(√(2n))) +Σ_(n=0) ^∞ (1/(√(2n+1))) ⇒Σ_(n=0) ^∞ (1/(√(2n+1))) =ξ((1/2))−(1/(√2))ξ((1/2)) =(1−(1/(√2)))ξ((1/2)) ⇒ Σ_(n=0) ^∞ (((−1)^n )/(√(n+1))) =−(1/(√2))ξ((1/2)) +(1−(1/(√2)))ξ((1/2)) =(1−(√2))ξ((1/2)) ⇒ ∫_(−∞) ^(+∞) (dx/(1+e^x^2 )) =(√π)(1−(√2))ξ((1/2)) the equality is proved .

$${we}\:{have}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{e}^{{x}^{\mathrm{2}} } }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } }{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}^{\mathrm{2}} } \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−{nx}^{\mathrm{2}} } \right){dx}=\mathrm{2}\:\sum_{=\mathrm{0}} ^{\infty\:} \left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−\left({n}+\mathrm{1}\right){x}^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{{n}+\mathrm{1}}{x}\:={u}} \:\:\:\:\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:\:\:\frac{\boldsymbol{{du}}}{\sqrt{\boldsymbol{{n}}+\mathrm{1}}}\:=\mathrm{2}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\sqrt{{n}+\mathrm{1}}} \\ $$$$=\sqrt{\pi}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\sqrt{{n}+\mathrm{1}}}\:\:\:\:\:{we}\:{have}\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\sqrt{{n}+\mathrm{1}}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} }{\sqrt{\boldsymbol{{n}}}} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{n}}}\:\:\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{n}+\mathrm{1}}}\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{n}+\mathrm{1}}} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\sqrt{{n}}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{n}}}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{n}+\mathrm{1}}}\:\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\sqrt{\mathrm{2}{n}+\mathrm{1}}}\:=\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\sqrt{{n}+\mathrm{1}}}\:=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{e}^{{x}^{\mathrm{2}} } }\:\:=\sqrt{\pi}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\xi\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${the}\:{equality}\:{is}\:{proved}\:. \\ $$

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