Question-151145

Question Number 151145 by mnjuly1970 last updated on 18/Aug/21

Answered by qaz last updated on 18/Aug/21

∫_0 ^∞ (x^2 /(cosh^2 (x^2 )))dx =2∫_0 ^∞ (x^2 /(1+cosh (2x^2 )))dx =∫_0 ^∞ ((√x)/(1+cosh (2x)))dx =∫_0 ^∞ ((2(√x)e^(2x) )/((1+e^(2x) )^2 ))dx =−((√x)/(1+e^(2x) ))∣_0 ^∞ +(1/2)∫_0 ^∞ (x^(−1/2) /(1+e^(2x) ))dx =(1/2)∫_0 ^∞ ((x^(−1/2) e^(−2x) )/(1+e^(−2x) ))dx =(1/2)Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ x^(−1/2) e^(−(2n+2)x) dx =((√π)/2)Σ_(n=0) ^∞ (((−1)^n )/((2n+2)^(1/2) )) =((√π)/(2(√2)))Σ_(n=1) ^∞ (((−1)^(n−1) )/n^(1/2) ) =((√π)/(2(√2)))η((1/2)) =((√π)/(2(√2)))(1−2^(1−(1/2)) )ζ((1/2)) ⇒k=((√π)/(2(√2)))(1−(√2))

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{cosh}\:^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{cosh}\:\left(\mathrm{2x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{\mathrm{x}}}{\mathrm{1}+\mathrm{cosh}\:\left(\mathrm{2x}\right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}\sqrt{\mathrm{x}}\mathrm{e}^{\mathrm{2x}} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{2x}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\sqrt{\mathrm{x}}}{\mathrm{1}+\mathrm{e}^{\mathrm{2x}} }\mid_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{−\mathrm{1}/\mathrm{2}} }{\mathrm{1}+\mathrm{e}^{\mathrm{2x}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{−\mathrm{1}/\mathrm{2}} \mathrm{e}^{−\mathrm{2x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\infty} \mathrm{x}^{−\mathrm{1}/\mathrm{2}} \mathrm{e}^{−\left(\mathrm{2n}+\mathrm{2}\right)\mathrm{x}} \mathrm{dx} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\eta\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} \right)\zeta\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{k}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right) \\ $$

Commented by mnjuly1970 last updated on 18/Aug/21

$$\:\:{very}\:{nice}\:{mr}\:{qaz}… \\ $$

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