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Question Number 164795 by HongKing last updated on 22/Jan/22

Answered by puissant last updated on 22/Jan/22

Ω=∫_0 ^∞ ((1/(sinhx))−(1/(xcoshx)))dx  =∫_0 ^∞ ((1/((e^x −e^(−x) )/2))−(1/(x((e^x +e^(−x) )/2))))dx  =∫_0 ^∞ ((2/(e^x (1−e^(−2x) )))−(2/(xe^x (1+e^(−2x) ))))dx  = ∫_0 ^∞ (((2e^(−x) )/(1−e^(−2x) ))−((2e^(−x) )/(x(1+e^(−2x) ))))dx..

$$\Omega=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{sinhx}}−\frac{\mathrm{1}}{{xcoshx}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}}−\frac{\mathrm{1}}{{x}\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{2}}{{e}^{{x}} \left(\mathrm{1}−{e}^{−\mathrm{2}{x}} \right)}−\frac{\mathrm{2}}{{xe}^{{x}} \left(\mathrm{1}+{e}^{−\mathrm{2}{x}} \right)}\right){dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{2}{e}^{−{x}} }{\mathrm{1}−{e}^{−\mathrm{2}{x}} }−\frac{\mathrm{2}{e}^{−{x}} }{{x}\left(\mathrm{1}+{e}^{−\mathrm{2}{x}} \right)}\right){dx}.. \\ $$

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