How would you prove $int^8_0frac1{sqrt{x+frac1{sqrt{x}}}}dx<4-frac1{2019}$?












12












$begingroup$


We would like to prove the following inequality.




$$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dx<4-frac{1}{2019}tag{1}$$




What I've tried is using the AM-GM inequality, $$x+frac{1}{sqrt{x}}ge 2,sqrt{x timesfrac{1}{sqrt{x}}}=2x^{frac14} $$ then $$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dxlefrac{1}{sqrt{2}}int^{8}_{0}frac{1}{sqrt[8]{x}}, dx=4.98cdots<color{red}{5}-frac{1}{2019}.tag{2}$$How would you prove $(1)$?



It is tempting to upper-bound the integrand by an easier function.



$color{white}{11110811197115115116117100101110116}$






calculus integration inequality definite-integrals nested-radicals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This looks like a contest task; if this is actually the case I would recommend to delete the question hence posting of active contest questions is not allowed here on MSE.
    $endgroup$
    – mrtaurho
    Dec 14 '18 at 11:23






  • 5




    $begingroup$
    @mrtaurho No contest task here, just our passionate teacher motivating his students!
    $endgroup$
    – Urbanmaths
    Dec 14 '18 at 13:34






  • 1




    $begingroup$
    If you apply AM-GM just on the denominator $sqrt{x cdot 1}$ of the fraction $frac{1}{sqrt{x}}$, you will have that $$I leq int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}} approx 4.07, $$ which is a better result than yours. I think $int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}}$ has a closed form. Not sure!
    $endgroup$
    – Alex Silva
    Dec 14 '18 at 13:56








  • 2




    $begingroup$
    @Urbanmaths What exactly do you mean?
    $endgroup$
    – mrtaurho
    Dec 18 '18 at 10:12






  • 2




    $begingroup$
    I mean I'm just hoping some comments don't prevent people from answering the above question. Thanks.
    $endgroup$
    – Urbanmaths
    Dec 18 '18 at 18:04


















12












$begingroup$


We would like to prove the following inequality.




$$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dx<4-frac{1}{2019}tag{1}$$




What I've tried is using the AM-GM inequality, $$x+frac{1}{sqrt{x}}ge 2,sqrt{x timesfrac{1}{sqrt{x}}}=2x^{frac14} $$ then $$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dxlefrac{1}{sqrt{2}}int^{8}_{0}frac{1}{sqrt[8]{x}}, dx=4.98cdots<color{red}{5}-frac{1}{2019}.tag{2}$$How would you prove $(1)$?



It is tempting to upper-bound the integrand by an easier function.



$color{white}{11110811197115115116117100101110116}$






calculus integration inequality definite-integrals nested-radicals







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This looks like a contest task; if this is actually the case I would recommend to delete the question hence posting of active contest questions is not allowed here on MSE.
    $endgroup$
    – mrtaurho
    Dec 14 '18 at 11:23






  • 5




    $begingroup$
    @mrtaurho No contest task here, just our passionate teacher motivating his students!
    $endgroup$
    – Urbanmaths
    Dec 14 '18 at 13:34






  • 1




    $begingroup$
    If you apply AM-GM just on the denominator $sqrt{x cdot 1}$ of the fraction $frac{1}{sqrt{x}}$, you will have that $$I leq int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}} approx 4.07, $$ which is a better result than yours. I think $int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}}$ has a closed form. Not sure!
    $endgroup$
    – Alex Silva
    Dec 14 '18 at 13:56








  • 2




    $begingroup$
    @Urbanmaths What exactly do you mean?
    $endgroup$
    – mrtaurho
    Dec 18 '18 at 10:12






  • 2




    $begingroup$
    I mean I'm just hoping some comments don't prevent people from answering the above question. Thanks.
    $endgroup$
    – Urbanmaths
    Dec 18 '18 at 18:04
















12












12








12


8



$begingroup$


We would like to prove the following inequality.




$$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dx<4-frac{1}{2019}tag{1}$$




What I've tried is using the AM-GM inequality, $$x+frac{1}{sqrt{x}}ge 2,sqrt{x timesfrac{1}{sqrt{x}}}=2x^{frac14} $$ then $$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dxlefrac{1}{sqrt{2}}int^{8}_{0}frac{1}{sqrt[8]{x}}, dx=4.98cdots<color{red}{5}-frac{1}{2019}.tag{2}$$How would you prove $(1)$?



It is tempting to upper-bound the integrand by an easier function.



$color{white}{11110811197115115116117100101110116}$






calculus integration inequality definite-integrals nested-radicals







share|cite|improve this question











$endgroup$




We would like to prove the following inequality.




$$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dx<4-frac{1}{2019}tag{1}$$




What I've tried is using the AM-GM inequality, $$x+frac{1}{sqrt{x}}ge 2,sqrt{x timesfrac{1}{sqrt{x}}}=2x^{frac14} $$ then $$int^{8}_{0}frac{1}{sqrt{x+frac{1}{sqrt{x}}}},dxlefrac{1}{sqrt{2}}int^{8}_{0}frac{1}{sqrt[8]{x}}, dx=4.98cdots<color{red}{5}-frac{1}{2019}.tag{2}$$How would you prove $(1)$?



It is tempting to upper-bound the integrand by an easier function.



$color{white}{11110811197115115116117100101110116}$






calculus integration inequality definite-integrals nested-radicals




calculus integration inequality definite-integrals nested-radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 16:00







Urbanmaths

















asked Dec 14 '18 at 11:11









UrbanmathsUrbanmaths

714




714








  • 2




    $begingroup$
    This looks like a contest task; if this is actually the case I would recommend to delete the question hence posting of active contest questions is not allowed here on MSE.
    $endgroup$
    – mrtaurho
    Dec 14 '18 at 11:23






  • 5




    $begingroup$
    @mrtaurho No contest task here, just our passionate teacher motivating his students!
    $endgroup$
    – Urbanmaths
    Dec 14 '18 at 13:34






  • 1




    $begingroup$
    If you apply AM-GM just on the denominator $sqrt{x cdot 1}$ of the fraction $frac{1}{sqrt{x}}$, you will have that $$I leq int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}} approx 4.07, $$ which is a better result than yours. I think $int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}}$ has a closed form. Not sure!
    $endgroup$
    – Alex Silva
    Dec 14 '18 at 13:56








  • 2




    $begingroup$
    @Urbanmaths What exactly do you mean?
    $endgroup$
    – mrtaurho
    Dec 18 '18 at 10:12






  • 2




    $begingroup$
    I mean I'm just hoping some comments don't prevent people from answering the above question. Thanks.
    $endgroup$
    – Urbanmaths
    Dec 18 '18 at 18:04
















  • 2




    $begingroup$
    This looks like a contest task; if this is actually the case I would recommend to delete the question hence posting of active contest questions is not allowed here on MSE.
    $endgroup$
    – mrtaurho
    Dec 14 '18 at 11:23






  • 5




    $begingroup$
    @mrtaurho No contest task here, just our passionate teacher motivating his students!
    $endgroup$
    – Urbanmaths
    Dec 14 '18 at 13:34






  • 1




    $begingroup$
    If you apply AM-GM just on the denominator $sqrt{x cdot 1}$ of the fraction $frac{1}{sqrt{x}}$, you will have that $$I leq int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}} approx 4.07, $$ which is a better result than yours. I think $int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}}$ has a closed form. Not sure!
    $endgroup$
    – Alex Silva
    Dec 14 '18 at 13:56








  • 2




    $begingroup$
    @Urbanmaths What exactly do you mean?
    $endgroup$
    – mrtaurho
    Dec 18 '18 at 10:12






  • 2




    $begingroup$
    I mean I'm just hoping some comments don't prevent people from answering the above question. Thanks.
    $endgroup$
    – Urbanmaths
    Dec 18 '18 at 18:04










2




2




$begingroup$
This looks like a contest task; if this is actually the case I would recommend to delete the question hence posting of active contest questions is not allowed here on MSE.
$endgroup$
– mrtaurho
Dec 14 '18 at 11:23




$begingroup$
This looks like a contest task; if this is actually the case I would recommend to delete the question hence posting of active contest questions is not allowed here on MSE.
$endgroup$
– mrtaurho
Dec 14 '18 at 11:23




5




5




$begingroup$
@mrtaurho No contest task here, just our passionate teacher motivating his students!
$endgroup$
– Urbanmaths
Dec 14 '18 at 13:34




$begingroup$
@mrtaurho No contest task here, just our passionate teacher motivating his students!
$endgroup$
– Urbanmaths
Dec 14 '18 at 13:34




1




1




$begingroup$
If you apply AM-GM just on the denominator $sqrt{x cdot 1}$ of the fraction $frac{1}{sqrt{x}}$, you will have that $$I leq int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}} approx 4.07, $$ which is a better result than yours. I think $int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}}$ has a closed form. Not sure!
$endgroup$
– Alex Silva
Dec 14 '18 at 13:56






$begingroup$
If you apply AM-GM just on the denominator $sqrt{x cdot 1}$ of the fraction $frac{1}{sqrt{x}}$, you will have that $$I leq int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}} approx 4.07, $$ which is a better result than yours. I think $int_{0}^{8}sqrt{frac{x+1}{x^2+x+2}}$ has a closed form. Not sure!
$endgroup$
– Alex Silva
Dec 14 '18 at 13:56






2




2




$begingroup$
@Urbanmaths What exactly do you mean?
$endgroup$
– mrtaurho
Dec 18 '18 at 10:12




$begingroup$
@Urbanmaths What exactly do you mean?
$endgroup$
– mrtaurho
Dec 18 '18 at 10:12




2




2




$begingroup$
I mean I'm just hoping some comments don't prevent people from answering the above question. Thanks.
$endgroup$
– Urbanmaths
Dec 18 '18 at 18:04






$begingroup$
I mean I'm just hoping some comments don't prevent people from answering the above question. Thanks.
$endgroup$
– Urbanmaths
Dec 18 '18 at 18:04












1 Answer
1






active

oldest

votes


















6












$begingroup$

For $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag1
end{align}
then, using
begin{align}
small{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}>2sqrt{x}}
end{align}
one gets




$$int_1^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}. tag2
$$




By the change of variable $u=2sqrt{x}$, $du=frac{dx}{sqrt{x}}$, setting $varphi=frac{1+sqrt{5}}2$, one has
begin{align}
&int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}
\\&=int_1^{8}frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}:frac{dx}{sqrt{x}}
\\&=int_2^{4sqrt{2}}frac{u^6+6u^3+4}{u^6+10u^3+20}:du, qquad u=2^{1/3}5^{1/6}t,
\\&=4sqrt{2}-2-2^{4/3}5^{-1/3}varphi int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+varphi}-frac{2^{4/3}5^{2/3}}{varphi}int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+frac1{varphi}}.
end{align}
Then, using the evaluation
$$
int_0^bfrac{dt}{t^3+c^3}=frac{sqrt{3}pi}{18c^2}+frac{sqrt{3}}{3c^2}arctan left(frac{2b-c}{sqrt{3}c}right)-frac{1}{6c^2}ln left(frac{(2b-c)^2+3c^2}{4(b+c)^2}right),,c>0,,b>0,
$$ one gets




$$
int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}=3.3199color{red}{58928}cdots. tag3
$$




In a like manner, for $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{1}{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag4
end{align}
then, using
begin{align}
small{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}>frac{2}{x^{1/4}}}
end{align}
one gets




$$int_0^1frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}. tag5
$$




By the change of variable $u=sqrt[4]{x}$, $dx=4u^3du$, setting $varphi=frac{1+sqrt{5}}2$, one has



begin{align}
&int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}
\\&=int_0^{1}frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}:sqrt[4]{x}:dx
\\&=4int_0^1frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}:u^4du, qquad u=2^{1/3}5^{-1/12}t,
\\&=frac4{25}+frac{6}{sqrt{5}}+2^{-2/3}5^{-1/3}varphi int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+varphi}-frac{2^{4/3}5^{-1/3}}{varphi}int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+frac1{varphi}}.
end{align}



Then, using the evaluation
begin{align}
&6cint_0^bfrac{t^4dt}{t^6+c^6}qquad (c>0,,b>0)
\\&=arctan left(sqrt{3}+frac{2b}{c}right)-arctan left(sqrt{3}-frac{2b}{c}right)+2arctan left(frac{b}{c}right)-frac{sqrt{3}}{2}ln left(frac{b^2+sqrt{3}bc+c^2}{b^2+sqrt{3}bc+c^2}right)
end{align} one obtains




$$
int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}=0.6739color{red}{21415}cdots. tag6
$$




Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields




$$int_0^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <color{red}{3.99}38803cdots<color{red}{4}-frac1{2019}=color{red}{3.99}950cdots tag7
$$




as desired.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1. Of course it's the man himself.
    $endgroup$
    – dezdichado
    Jan 6 at 2:37






  • 1




    $begingroup$
    That is ingenious.
    $endgroup$
    – TheSimpliFire
    Jan 6 at 11:55










  • $begingroup$
    This is too hard for us...
    $endgroup$
    – Urbanmaths
    Jan 10 at 15:47











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

For $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag1
end{align}
then, using
begin{align}
small{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}>2sqrt{x}}
end{align}
one gets




$$int_1^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}. tag2
$$




By the change of variable $u=2sqrt{x}$, $du=frac{dx}{sqrt{x}}$, setting $varphi=frac{1+sqrt{5}}2$, one has
begin{align}
&int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}
\\&=int_1^{8}frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}:frac{dx}{sqrt{x}}
\\&=int_2^{4sqrt{2}}frac{u^6+6u^3+4}{u^6+10u^3+20}:du, qquad u=2^{1/3}5^{1/6}t,
\\&=4sqrt{2}-2-2^{4/3}5^{-1/3}varphi int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+varphi}-frac{2^{4/3}5^{2/3}}{varphi}int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+frac1{varphi}}.
end{align}
Then, using the evaluation
$$
int_0^bfrac{dt}{t^3+c^3}=frac{sqrt{3}pi}{18c^2}+frac{sqrt{3}}{3c^2}arctan left(frac{2b-c}{sqrt{3}c}right)-frac{1}{6c^2}ln left(frac{(2b-c)^2+3c^2}{4(b+c)^2}right),,c>0,,b>0,
$$ one gets




$$
int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}=3.3199color{red}{58928}cdots. tag3
$$




In a like manner, for $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{1}{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag4
end{align}
then, using
begin{align}
small{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}>frac{2}{x^{1/4}}}
end{align}
one gets




$$int_0^1frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}. tag5
$$




By the change of variable $u=sqrt[4]{x}$, $dx=4u^3du$, setting $varphi=frac{1+sqrt{5}}2$, one has



begin{align}
&int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}
\\&=int_0^{1}frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}:sqrt[4]{x}:dx
\\&=4int_0^1frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}:u^4du, qquad u=2^{1/3}5^{-1/12}t,
\\&=frac4{25}+frac{6}{sqrt{5}}+2^{-2/3}5^{-1/3}varphi int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+varphi}-frac{2^{4/3}5^{-1/3}}{varphi}int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+frac1{varphi}}.
end{align}



Then, using the evaluation
begin{align}
&6cint_0^bfrac{t^4dt}{t^6+c^6}qquad (c>0,,b>0)
\\&=arctan left(sqrt{3}+frac{2b}{c}right)-arctan left(sqrt{3}-frac{2b}{c}right)+2arctan left(frac{b}{c}right)-frac{sqrt{3}}{2}ln left(frac{b^2+sqrt{3}bc+c^2}{b^2+sqrt{3}bc+c^2}right)
end{align} one obtains




$$
int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}=0.6739color{red}{21415}cdots. tag6
$$




Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields




$$int_0^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <color{red}{3.99}38803cdots<color{red}{4}-frac1{2019}=color{red}{3.99}950cdots tag7
$$




as desired.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1. Of course it's the man himself.
    $endgroup$
    – dezdichado
    Jan 6 at 2:37






  • 1




    $begingroup$
    That is ingenious.
    $endgroup$
    – TheSimpliFire
    Jan 6 at 11:55










  • $begingroup$
    This is too hard for us...
    $endgroup$
    – Urbanmaths
    Jan 10 at 15:47
















6












$begingroup$

For $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag1
end{align}
then, using
begin{align}
small{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}>2sqrt{x}}
end{align}
one gets




$$int_1^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}. tag2
$$




By the change of variable $u=2sqrt{x}$, $du=frac{dx}{sqrt{x}}$, setting $varphi=frac{1+sqrt{5}}2$, one has
begin{align}
&int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}
\\&=int_1^{8}frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}:frac{dx}{sqrt{x}}
\\&=int_2^{4sqrt{2}}frac{u^6+6u^3+4}{u^6+10u^3+20}:du, qquad u=2^{1/3}5^{1/6}t,
\\&=4sqrt{2}-2-2^{4/3}5^{-1/3}varphi int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+varphi}-frac{2^{4/3}5^{2/3}}{varphi}int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+frac1{varphi}}.
end{align}
Then, using the evaluation
$$
int_0^bfrac{dt}{t^3+c^3}=frac{sqrt{3}pi}{18c^2}+frac{sqrt{3}}{3c^2}arctan left(frac{2b-c}{sqrt{3}c}right)-frac{1}{6c^2}ln left(frac{(2b-c)^2+3c^2}{4(b+c)^2}right),,c>0,,b>0,
$$ one gets




$$
int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}=3.3199color{red}{58928}cdots. tag3
$$




In a like manner, for $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{1}{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag4
end{align}
then, using
begin{align}
small{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}>frac{2}{x^{1/4}}}
end{align}
one gets




$$int_0^1frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}. tag5
$$




By the change of variable $u=sqrt[4]{x}$, $dx=4u^3du$, setting $varphi=frac{1+sqrt{5}}2$, one has



begin{align}
&int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}
\\&=int_0^{1}frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}:sqrt[4]{x}:dx
\\&=4int_0^1frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}:u^4du, qquad u=2^{1/3}5^{-1/12}t,
\\&=frac4{25}+frac{6}{sqrt{5}}+2^{-2/3}5^{-1/3}varphi int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+varphi}-frac{2^{4/3}5^{-1/3}}{varphi}int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+frac1{varphi}}.
end{align}



Then, using the evaluation
begin{align}
&6cint_0^bfrac{t^4dt}{t^6+c^6}qquad (c>0,,b>0)
\\&=arctan left(sqrt{3}+frac{2b}{c}right)-arctan left(sqrt{3}-frac{2b}{c}right)+2arctan left(frac{b}{c}right)-frac{sqrt{3}}{2}ln left(frac{b^2+sqrt{3}bc+c^2}{b^2+sqrt{3}bc+c^2}right)
end{align} one obtains




$$
int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}=0.6739color{red}{21415}cdots. tag6
$$




Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields




$$int_0^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <color{red}{3.99}38803cdots<color{red}{4}-frac1{2019}=color{red}{3.99}950cdots tag7
$$




as desired.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1. Of course it's the man himself.
    $endgroup$
    – dezdichado
    Jan 6 at 2:37






  • 1




    $begingroup$
    That is ingenious.
    $endgroup$
    – TheSimpliFire
    Jan 6 at 11:55










  • $begingroup$
    This is too hard for us...
    $endgroup$
    – Urbanmaths
    Jan 10 at 15:47














6












6








6





$begingroup$

For $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag1
end{align}
then, using
begin{align}
small{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}>2sqrt{x}}
end{align}
one gets




$$int_1^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}. tag2
$$




By the change of variable $u=2sqrt{x}$, $du=frac{dx}{sqrt{x}}$, setting $varphi=frac{1+sqrt{5}}2$, one has
begin{align}
&int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}
\\&=int_1^{8}frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}:frac{dx}{sqrt{x}}
\\&=int_2^{4sqrt{2}}frac{u^6+6u^3+4}{u^6+10u^3+20}:du, qquad u=2^{1/3}5^{1/6}t,
\\&=4sqrt{2}-2-2^{4/3}5^{-1/3}varphi int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+varphi}-frac{2^{4/3}5^{2/3}}{varphi}int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+frac1{varphi}}.
end{align}
Then, using the evaluation
$$
int_0^bfrac{dt}{t^3+c^3}=frac{sqrt{3}pi}{18c^2}+frac{sqrt{3}}{3c^2}arctan left(frac{2b-c}{sqrt{3}c}right)-frac{1}{6c^2}ln left(frac{(2b-c)^2+3c^2}{4(b+c)^2}right),,c>0,,b>0,
$$ one gets




$$
int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}=3.3199color{red}{58928}cdots. tag3
$$




In a like manner, for $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{1}{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag4
end{align}
then, using
begin{align}
small{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}>frac{2}{x^{1/4}}}
end{align}
one gets




$$int_0^1frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}. tag5
$$




By the change of variable $u=sqrt[4]{x}$, $dx=4u^3du$, setting $varphi=frac{1+sqrt{5}}2$, one has



begin{align}
&int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}
\\&=int_0^{1}frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}:sqrt[4]{x}:dx
\\&=4int_0^1frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}:u^4du, qquad u=2^{1/3}5^{-1/12}t,
\\&=frac4{25}+frac{6}{sqrt{5}}+2^{-2/3}5^{-1/3}varphi int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+varphi}-frac{2^{4/3}5^{-1/3}}{varphi}int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+frac1{varphi}}.
end{align}



Then, using the evaluation
begin{align}
&6cint_0^bfrac{t^4dt}{t^6+c^6}qquad (c>0,,b>0)
\\&=arctan left(sqrt{3}+frac{2b}{c}right)-arctan left(sqrt{3}-frac{2b}{c}right)+2arctan left(frac{b}{c}right)-frac{sqrt{3}}{2}ln left(frac{b^2+sqrt{3}bc+c^2}{b^2+sqrt{3}bc+c^2}right)
end{align} one obtains




$$
int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}=0.6739color{red}{21415}cdots. tag6
$$




Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields




$$int_0^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <color{red}{3.99}38803cdots<color{red}{4}-frac1{2019}=color{red}{3.99}950cdots tag7
$$




as desired.






share|cite|improve this answer









$endgroup$



For $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag1
end{align}
then, using
begin{align}
small{sqrt{x}+sqrt {x+frac{1}{sqrt{x}}}>2sqrt{x}}
end{align}
one gets




$$int_1^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}. tag2
$$




By the change of variable $u=2sqrt{x}$, $du=frac{dx}{sqrt{x}}$, setting $varphi=frac{1+sqrt{5}}2$, one has
begin{align}
&int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}
\\&=int_1^{8}frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}:frac{dx}{sqrt{x}}
\\&=int_2^{4sqrt{2}}frac{u^6+6u^3+4}{u^6+10u^3+20}:du, qquad u=2^{1/3}5^{1/6}t,
\\&=4sqrt{2}-2-2^{4/3}5^{-1/3}varphi int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+varphi}-frac{2^{4/3}5^{2/3}}{varphi}int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}frac{dt}{t^3+frac1{varphi}}.
end{align}
Then, using the evaluation
$$
int_0^bfrac{dt}{t^3+c^3}=frac{sqrt{3}pi}{18c^2}+frac{sqrt{3}}{3c^2}arctan left(frac{2b-c}{sqrt{3}c}right)-frac{1}{6c^2}ln left(frac{(2b-c)^2+3c^2}{4(b+c)^2}right),,c>0,,b>0,
$$ one gets




$$
int_1^8frac{dx}{sqrt{x}+frac {1}{2x + frac 1{2sqrt{x} +frac{1}{2x+frac{1}{2sqrt{x}}}}}}=3.3199color{red}{58928}cdots. tag3
$$




In a like manner, for $x>0$, one may check that
begin{align}
small{sqrt {x+frac{1}{sqrt{x}}}}&=frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{1}{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}}}}} tag4
end{align}
then, using
begin{align}
small{frac{1}{x^{1/4}}+sqrt {x+frac{1}{sqrt{x}}}>frac{2}{x^{1/4}}}
end{align}
one gets




$$int_0^1frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}. tag5
$$




By the change of variable $u=sqrt[4]{x}$, $dx=4u^3du$, setting $varphi=frac{1+sqrt{5}}2$, one has



begin{align}
&int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}
\\&=int_0^{1}frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}:sqrt[4]{x}:dx
\\&=4int_0^1frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}:u^4du, qquad u=2^{1/3}5^{-1/12}t,
\\&=frac4{25}+frac{6}{sqrt{5}}+2^{-2/3}5^{-1/3}varphi int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+varphi}-frac{2^{4/3}5^{-1/3}}{varphi}int_0^{2^{-1/3}5^{1/12}}!!frac{t^4dt}{t^6+frac1{varphi}}.
end{align}



Then, using the evaluation
begin{align}
&6cint_0^bfrac{t^4dt}{t^6+c^6}qquad (c>0,,b>0)
\\&=arctan left(sqrt{3}+frac{2b}{c}right)-arctan left(sqrt{3}-frac{2b}{c}right)+2arctan left(frac{b}{c}right)-frac{sqrt{3}}{2}ln left(frac{b^2+sqrt{3}bc+c^2}{b^2+sqrt{3}bc+c^2}right)
end{align} one obtains




$$
int_0^1frac{dx}{frac{1}{x^{1/4}}+frac {1}{frac{2}{x^{5/4}} + frac 1{frac{2}{x^{1/4}} +frac{1}{frac{2}{x^{5/4}}+frac{x^{1/4}}2}}}}=0.6739color{red}{21415}cdots. tag6
$$




Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields




$$int_0^8frac{dx}{sqrt {x+frac{1}{sqrt{x}}}} <color{red}{3.99}38803cdots<color{red}{4}-frac1{2019}=color{red}{3.99}950cdots tag7
$$




as desired.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 9:52









Olivier OloaOlivier Oloa

108k17177294




108k17177294












  • $begingroup$
    +1. Of course it's the man himself.
    $endgroup$
    – dezdichado
    Jan 6 at 2:37






  • 1




    $begingroup$
    That is ingenious.
    $endgroup$
    – TheSimpliFire
    Jan 6 at 11:55










  • $begingroup$
    This is too hard for us...
    $endgroup$
    – Urbanmaths
    Jan 10 at 15:47


















  • $begingroup$
    +1. Of course it's the man himself.
    $endgroup$
    – dezdichado
    Jan 6 at 2:37






  • 1




    $begingroup$
    That is ingenious.
    $endgroup$
    – TheSimpliFire
    Jan 6 at 11:55










  • $begingroup$
    This is too hard for us...
    $endgroup$
    – Urbanmaths
    Jan 10 at 15:47
















$begingroup$
+1. Of course it's the man himself.
$endgroup$
– dezdichado
Jan 6 at 2:37




$begingroup$
+1. Of course it's the man himself.
$endgroup$
– dezdichado
Jan 6 at 2:37




1




1




$begingroup$
That is ingenious.
$endgroup$
– TheSimpliFire
Jan 6 at 11:55




$begingroup$
That is ingenious.
$endgroup$
– TheSimpliFire
Jan 6 at 11:55












$begingroup$
This is too hard for us...
$endgroup$
– Urbanmaths
Jan 10 at 15:47




$begingroup$
This is too hard for us...
$endgroup$
– Urbanmaths
Jan 10 at 15:47


















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