I came across while solving some problem that I need the exact/ approximate value of following integral. I need an upper bound of $-0.4$ or $-0.5$. Is there some way or programming tools.

$$\int_0^{\sqrt{2}} \ln\left(\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}\right)\,\mathrm dx$$

This is not homework problem. Just while solving some calculus problem, it came out

  • $\begingroup$ It does have an elementary antiderivative and closed form solution provided by Wolfram for example (although it seems horrible). $\endgroup$ Nov 29 '19 at 11:20
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    $\begingroup$ A very simple bound is found by seeing that its monotonely increasing and is $0$ at $x=\sqrt{2}$ so its bounded below by $\log(4/6)\cdot\sqrt{2}\approx -0.57$ and bounded above by $0\cdot \sqrt{2} = 0$. $\endgroup$
    – Winther
    Nov 29 '19 at 11:25
  • $\begingroup$ $$4 \text{RootSum}\left[\text{$\#$1}^6+8 \text{$\#$1}^4+16 \text{$\#$1}^2+4\&,\frac{2 \text{$\#$1}^4 \log (x-\text{$\#$1})+8 \text{$\#$1}^2 \log (x-\text{$\#$1})+3 \log (x-\text{$\#$1})}{3 \text{$\#$1}^5+16 \text{$\#$1}^3+16 \text{$\#$1}}\&\right]+x \log \left(\frac{x^6+8 x^4+16 x^2+4}{x^6+8 x^4+15 x^2+6}\right)-2 \sqrt{2} \tan ^{-1}\left(\frac{x}{\sqrt{2}}\right)-\left(\sqrt{6}-2\right) \sqrt{2 \left(3+\sqrt{6}\right)} \tan ^{-1}\left(\frac{x}{\sqrt{3-\sqrt{6}}}\right)-\sqrt{6-2 \sqrt{6}} \left(2+\sqrt{6}\right) \tan ^{-1}\left(\frac{x}{\sqrt{3+\sqrt{6}}}\right)$$ $\endgroup$ Nov 29 '19 at 11:34
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    $\begingroup$ @MaximilianJanisch Finally it comes down to the fact that closed form solution is an ambiguous definition... It requires a list of acceptable functions, and you are going one step further by putting conditions on how constants are actually computed. $\endgroup$ Nov 29 '19 at 12:14
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    $\begingroup$ @PierreCarre This is a very nice comment $\endgroup$ Nov 29 '19 at 12:23

Write numerator and denominator as $$\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}=\frac{(x^2+a)(x^2+b)(x^2+c)}{(x^2+d)(x^2+e)(x^2+f)}$$ where $a,b,c,d,e,f$ are all positive numbers.

Expand the logarithm to face $$I(p)=\int_0^{\sqrt 2} \log(x^2+p)=\sqrt{2} (\log (p+2)-2)+2 \sqrt{p} \tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{p}}\right)$$

If I am not mistaken, $a,b,c$ are given by $$\frac{8}{3}-\frac{8}{3} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(\frac{5}{32}\right)\right)\qquad \text{with} \qquad k=0,1,2$$ and $c,d,e$ by $$\frac{8}{3}-\frac{2}{3} \sqrt{19} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(-\frac{53}{19 \sqrt{19}}\right)\right)\qquad \text{with} \qquad k=0,1,2$$


If we just want an approximate value of the integral, using the simplest Padé approximant built around $x=1$ (more or less the midpoint of the integration range), we have $$\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}\sim \frac{\frac{3389 (x-1)^2}{4520}+\frac{4691 (x-1)}{3390}+\frac{29}{30}}{\frac{1799 (x-1)^2}{2260}+\frac{2177 (x-1)}{1695}+1}$$

Thus, completing the squares, we have $$\log\left(\dfrac{x^6+8x^4+16x^2+4}{x^6+8x^4+15x^2+6}\right)\,dx \sim\log\left(\dfrac{a (x-b)^2+c}{d (x-e)^2+f}\right)\,dx$$ and $$J=\int \log \left(a (x-b)^2+c\right)\,dx$$ is given by $$J=(x-b) \log \left(a (x-b)^2+c\right)+\frac{2 \sqrt{c} \tan ^{-1}\left(\frac{\sqrt{a} }{\sqrt{c}}(x-b)\right)}{\sqrt{a}}-2 (x-b)$$ This should lead to a value equal to $-0.212276$ to be compared to the exact $−0.211212$ given by Wolfram Alpha (this makes a relative error of $0.5$%.

  • $\begingroup$ $d, e, f$ should be $$\frac 8 3 + \frac {2 \sqrt {19}} 3 \cos \left( \frac {2 \pi k} 3 - \frac 1 3 \arccos \frac {26} {19 \sqrt {19}} \right).$$ $\endgroup$
    – Maxim
    Nov 29 '19 at 15:08

Ask Wolfram alpha or a computer algebra package at the university...

It gives $-0.211212$ as an approximation.


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