Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that $$\dfrac{1}{30}<\displaystyle\int_{2}^{\infty}\dfrac{\sqrt{s^3-s^2+3}}{s^5+s^2+1}ds<\dfrac{\sqrt{2}}{20}$$

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

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

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

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.

share|improve this answer
very nice,Thank you, –  math110 Mar 27 '13 at 13:22
+1, nice proof ;-) –  Jean-Claude Arbaut Mar 27 '13 at 15:27
add comment

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 $\frac{1}{30}$ ($\approx 0.997 \cdot \frac{1}{30}$) :-(

share|improve this answer
oh ,Thank you,but lfet you at last $-0.00008583……$ –  math110 Mar 27 '13 at 7:47
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 8:01
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 8:21
yes, I mean you left is wrong, so we can use other methods. –  math110 Mar 27 '13 at 8:32
That's right ;-) –  Jean-Claude Arbaut Mar 27 '13 at 8:32
show 5 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.