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Prove that

$$\frac{1}{30}<\int_{2}^{\infty}\frac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\,ds<\frac{\sqrt{2}}{20}$$

I think this inequality can $s^5+s^2+1>( ) $

and $\sqrt{s^3-s^2+3}<()$?

StubbornAtom
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math110
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2 Answers2

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For right inequality, write

$$\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\mathrm{d}s < \int_{2}^{\infty}\dfrac{\sqrt{s^3}}{s^5}\mathrm{d}s = \int_{2}^{\infty}s^{-\frac{7}{2}} \mathrm{d}s=\frac{\sqrt{2}}{20}$$

For left one, you may try

$$\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}\mathrm{d}s > \int_{2}^{\infty} \dfrac{\sqrt{7}}{s^5+s^3+s^2+1} \mathrm{d}s$$

Then using the fact that $s^5+s^3+s^2+1 = (s+1)(s^2+1)(s^2-s+1)$, you have

$$\frac{1}{s^5+s^3+s^2+1}=-\frac{2s-1}{3(s^2-s+1)}+\frac{s+1}{2(s^2+1)}+\frac{1}{6(s+1)}$$ Then you can compute the integral,

$$\int_{2}^{\infty} \dfrac{\sqrt{7}}{s^5+s^3+s^2+1} \mathrm{d}s = \sqrt{7} \left(-\frac{\log 5}{4}+\frac{\log 3}{6}+{1\over2}\arctan {1\over2}\right)$$

But it's a bit below $\frac{1}{30}$ ($\approx 0.997 \cdot \frac{1}{30}$) :-(

Jean-Claude Arbaut
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  • oh ,Thank you,but lfet you at last $-0.00008583……$ – math110 Mar 27 '13 at 07:47
  • I mean :$\sqrt{7} \left(-\frac{\log 5}{4}+\frac{\log 3}{6}-\frac{\arctan 2}{2}+\frac{\pi}{4}\right)-\dfrac{1}{30}=-0.00008583<0$ – math110 Mar 27 '13 at 08:01
  • No, left inequality in not necessarily wrong, but my proof does not work. The majoration I choose is too coarse. – Jean-Claude Arbaut Mar 27 '13 at 08:21
  • yes, I mean you left is wrong, so we can use other methods. – math110 Mar 27 '13 at 08:32
  • That's right ;-) – Jean-Claude Arbaut Mar 27 '13 at 08:32
  • i think we might use this and add +2 to 7 after adding x^3 to denominator, the result would still be decreasing. http://math.stackexchange.com/questions/315939/property-of-dfrac-sum-a-i-sum-b-i-when-dfraca-ib-i-is-increasing/316103#316103 – S L Mar 27 '13 at 12:29
  • @experimentX But we must be sure the inequality still holds, and be able to compute the integral. Can you develop on your trick? – Jean-Claude Arbaut Mar 27 '13 at 12:51
  • yes the inequality still holds, we are introducing +2 inside $\sqrt{7 + 2}$ while we are introducing $s^3$ at denominator whose minimum value is $2^3$. consider a decreasing sequence $\frac{a_1}{b_1} > \frac{a_2}{b_2} $, then we can prove $\frac{a_1}{b_1} > \frac{a_1 + a_2}{b_1+b_2} > \frac{\sqrt{a_1 + a_2}}{b_1+b_2}$. which i think is enough – S L Mar 27 '13 at 13:08
  • Your $\sqrt{s^3-s^2+3}<0$ makes no sense, as $\sqrt{x}\geq 0$ for every $x\geq 0$. – Julien Mar 27 '13 at 13:27
  • Not $<0$, but $< ()$ to mean $< (something)$. And it's not so easy to find the good something. – Jean-Claude Arbaut Mar 27 '13 at 15:28
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Denote your integral by $Q$. Then $$Q=\int_2^\infty {1\over s^{7/2}}\ {\sqrt{1-{1\over s}+{3\over s^3}}\over 1+{1\over s^3}+{1\over s^5}}\ ds\ .$$ First off, one has $$\int_2^\infty {1\over s^{7/2}}\ ds={\sqrt{2}\over 20}\ .$$ Furthermore one easily checks that in the interval $2\leq s<\infty$ the following estimates are valid: $$\sqrt{1\over2}\leq\sqrt{1-{1\over s}+{3\over s^3}}=\sqrt{1-{1\over s}\bigl(1-{3\over s^2}\bigr)}\leq 1\ ,$$ $$1\leq 1+{1\over s^3}+{1\over s^5}\leq 1+{1\over 8}+{1\over32}={37\over32}\ .$$ It follows that $${\sqrt{1/ 2}\over 37/32}\cdot {\sqrt{2}\over 20}\leq Q\leq {1\over 1}\cdot {\sqrt{2}\over 20}\ .$$ Since ${32\over 740}>{1\over30}$ the stated inequalities are indeed true.

Christian Blatter
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