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it's problem found with the help of WA :

$$\int_{0}^{e}\operatorname{W^{2^{-n}}(x)}\leq \int_{0}^{e}\operatorname{sin^{2^{-n}}(x)}$$

And $n\geq 2$ a natural number.

$\operatorname{W(x)}$ is the Lambert's function.

In fact it would be very easy if we have (on $[0,e]$): $$\operatorname{W(x)}\leq \sin(x)$$

But it's really not the case .Furthermore I don't think that calculate the different integral are a good idea maybe .I think there is a big trick behind all of this but I don't know wich one it is .

My last idea was to use power series and compare each coefficient but it seems it's not a good idea again.

My question:

How to solve the integral inequality ?

Many thanks .

Ps:Cut the integral ?

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Let me try with $$I=\int_0^e W^a(x)\,dx \qquad \text{and} \qquad J=\int_0^e \sin^a(x)\,dx$$

The first thing I did was to expand as series the logarithms of the functions $$\log (W(x))=\log (x)-x+x^2-\frac{3 x^3}{2}+\frac{8 x^4}{3}-\frac{125 x^5}{24}+O\left(x^6\right)$$ $$\log (\sin(x))=\log (x)-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$

Multiply each expansion by $a$ and continue with Taylor series using $y=e^{\log(y)}$.

This gives $$W^a(x)=x^a \left(1-a x+\frac{a (a+2)}{2} x^2-\frac{a (a+3)^2}{6} x^3+\frac{a (a+4)^3}{24} x^4+O\left(x^5\right)\right)$$ $$\sin^a(x)=x^a \left(1-\frac{a }{6}x^2+\frac{a (5 a-2)}{360} x^4+O\left(x^6\right)\right)$$ $$\sin^a(x)-W^a(x)={ a x^{a+1}}\left(1-\frac{3 a+7}{6} x+\frac{a^2+6 a+9}{6} x^2+O\left(x^{3}\right)\right)$$ Now, compute $$J-I=\int_0^e\left(\sin^a(x)-W^a(x) \right)\,dx$$ to get $$J-I=\frac{a e^{a+2} \left(e^2 (a+2) (a+3)^3+6 (a+4) (a+3)-e (a+2) (a+4) (3 a+7)\right)}{6 (a+2) (a+3) (a+4)}$$ which is positive.

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  • $\begingroup$ Thanks for the answer but we need $a\geq 0$ no ? Or we have implicitly $a=2^{-n}$?And the last result does not invoke special function which is surprisingly elegant or it is an inequality ? $\endgroup$
    – Erik Satie
    Mar 28 '20 at 9:57

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