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Prove-0-1-2-2-x-x-dx-ln2-

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Question Number 219316 by Nicholas666 last updated on 23/Apr/25
Prove; ∫^( ∞) _( 0) (1/(2^2^(⌊x⌋) + {x})) dx = ln2
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:\:\:\:\:\underset{\:\mathrm{0}} {\int}^{\:\infty} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{\lfloor\boldsymbol{{x}}\rfloor} } \:+\:\left\{{x}\right\}}\:{dx}\:=\:{ln}\mathrm{2}\:\:\: \\ $$
Answered by vnm last updated on 23/Apr/25
(1+(1/2^1 ))(1+(1/2^2 ))(1+(1/2^4 ))...(1+(1/2^2^n ))= 1+(1/2^1 )+(1/2^2 )+(1/2^3 )+...+(1/2^(1+2+4+...+2^n ) ) →_(n→∞) 2 ∫_0 ^∞ (1/(2^2^(⌊x⌋) +{x}))dx=Σ_(n=0) ^∞ ∫_0 ^1 (dt/(2^2^n +t))= Σ_(n=0) ^∞ ln(2^2^n +t)∣_0 ^1 =Σ_(n=0) ^∞ ln((2^2^n +1)/2^2^n )= lim_(n→∞) ln(((2^1 +1)/2^1 )∙((2^2 +1)/2^2 )∙((2^4 +1)/2^4 )∙...∙((2^2^n +1)/2^2^n ))=ln 2.
$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\right)…\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{{n}} } }\right)= \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+…+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}+\mathrm{2}+\mathrm{4}+…+\mathrm{2}^{{n}} } }\:\:\underset{{n}\rightarrow\infty} {\rightarrow}\mathrm{2} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{\lfloor{x}\rfloor} } +\left\{{x}\right\}}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dt}}{\mathrm{2}^{\mathrm{2}^{{n}} } +{t}}= \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{ln}\left(\mathrm{2}^{\mathrm{2}^{{n}} } +{t}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{ln}\frac{\mathrm{2}^{\mathrm{2}^{{n}} } +\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{{n}} } }= \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}ln}\left(\frac{\mathrm{2}^{\mathrm{1}} +\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{2}^{\mathrm{4}} +\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\centerdot…\centerdot\frac{\mathrm{2}^{\mathrm{2}^{{n}} } +\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{{n}} } }\right)=\mathrm{ln}\:\mathrm{2}. \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
tank you sir nice step by step solution
$${tank}\:{you}\:{sir} \\ $$$${nice}\:{step}\:{by}\:{step}\:{solution} \\ $$
Commented by Nicholas666 last updated on 23/Apr/25
I forgot to tell you that you made a mistake in some steps
$${I}\:{forgot}\:{to}\:{tell}\:{you}\:{that}\:{you}\:{made}\:{a}\:{mistake} \\ $$$${in}\:{some}\:{steps} \\ $$$$ \\ $$

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