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Question Number 127420 by liberty last updated on 29/Dec/20

∫_0 ^( ∞) ((x−1)/( (√(2^x −1)) ln (2^x −1))) dx ?

$$\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}\:\mathrm{ln}\:\left(\mathrm{2}^{{x}} −\mathrm{1}\right)}\:{dx}\:? \\ $$

Answered by mnjuly1970 last updated on 29/Dec/20

solution: t=(√(2^x −1)) ⇒t^2 +1=2^x Ω=(1/(ln^2 (2)))∫_0 ^( ∞) (((ln(t^2 +1)−ln(2))/(2tln(t)))) ((2tdt)/((t^2 +1))) =(1/(ln^2 (2)))∫_0 ^( ∞) ((ln(t^2 +1)−ln(2))/(ln(t)(t^2 +1)))dt Ω=^(t=(1/y)) (1/(ln^2 (2)))∫_0 ^( ∞) ((ln(y^2 +1)−2ln(y)−ln(2))/(−ln(y)(1+y^2 )))dy =−Ω+(2/(ln^2 (2)))∫_0 ^( ∞) (dy/(1+y^2 )) 2Ω=(2/(ln^2 (2)))((π/2))⇒Ω=(π/(2ln^2 (2))) ✓

$${solution}: \\ $$ $$\:\:{t}=\sqrt{\mathrm{2}^{{x}} −\mathrm{1}}\:\Rightarrow{t}^{\mathrm{2}} +\mathrm{1}=\mathrm{2}^{{x}} \\ $$ $$\:\:\:\Omega=\frac{\mathrm{1}}{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{ln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left(\mathrm{2}\right)}{\mathrm{2}{tln}\left({t}\right)}\right)\:\frac{\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$=\frac{\mathrm{1}}{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)−{ln}\left(\mathrm{2}\right)}{{ln}\left({t}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$ $$\Omega\overset{{t}=\frac{\mathrm{1}}{{y}}} {=}\frac{\mathrm{1}}{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\:\infty} \:\frac{{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}{ln}\left({y}\right)−{ln}\left(\mathrm{2}\right)}{−{ln}\left({y}\right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}{dy} \\ $$ $$=−\Omega+\frac{\mathrm{2}}{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\int_{\mathrm{0}} ^{\:\infty} \frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$ $$\mathrm{2}\Omega=\frac{\mathrm{2}}{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\left(\frac{\pi}{\mathrm{2}}\right)\Rightarrow\Omega=\frac{\pi}{\mathrm{2}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}\:\checkmark \\ $$

Commented byliberty last updated on 30/Dec/20

sabiħ ... grazzi għat-tlestija tiegħek

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