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Unsolved Challenge A tough Integral

Tony

New member
Apr 9, 2018
17
Given:

A so-called complicate integral has a such a simple closed form, quite amazed me, but how to prove it, is an other story.

$$\int_{0}^{1}\mathrm dt{t-3t^3+t^5\over 1+t^4+t^8}\cdot \ln(-\ln t) dt=\color{red}{{\pi\over 3\sqrt{3}}}\cdot \color{blue}{\ln 2\over 2}$$

Does anyone know to how prove this integral?
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
Given:

A so-called complicate integral has a such a simple closed form, quite amazed me, but how to prove it, is an other story.

$$\int_{0}^{1}\mathrm dt{t-3t^3+t^5\over 1+t^4+t^8}\cdot \ln(-\ln t) dt=\color{red}{{\pi\over 3\sqrt{3}}}\cdot \color{blue}{\ln 2\over 2}$$

Does anyone know to how prove this integral?
Well, I'm not saying it will lead anywhere, but it seems hopeful to factor that denominator and create a partial fraction decomposition of that lovely rational expression.

In case it doesn't strike you: $1 + t^{4} + t^{8} = \left(1 + 2t^{4} + t^{8}\right) - t^{4} = \left(1 + t^{4}\right)^{2} - \left(t^{2}\right)^2$

Can you take it from there? It's possible you can do the same thing again and simplify the mess even more. Of course, you'll end up with four integral pieces, but each one may be more tractable.
 

Greg

Perseverance
Staff member
Feb 5, 2013
1,382
I'm still trying to figure out what the two $dt$'s mean. :confused:
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
I'm still trying to figure out what the two $dt$'s mean. :confused:
haha! I totally missed the second one. In any case, since it doesn't really matter if it's there at all if all we're doing is calculating an integral where the context is clear, I'm not disturbed by it. :)
 

MountEvariste

Well-known member
Jun 29, 2017
79
An observation: it can be written as $\displaystyle \int_0^{\infty}\frac{(e^{-2 x} + e^{2 x}-3) \log{x}}{1 + e^{-4 x} + e^{4 x}}\,dx$

It may be worth trying to write it as a series -- perhaps even twice.

Where's this integral from? It's one of the trickiest integrals I've seen.
 

Theia

Well-known member
Mar 30, 2016
92
Another observation: The integral can be written as

\(\displaystyle I = \int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1} \mathrm{d}x - \frac{1}{2} \int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1} \mathrm{d}x\),

but I'm not sure how to continue. It looks like these integrals can't be evaluated in terms of elementary functions and series (or other tricks) should be used, as June29 wrote.
 

MountEvariste

Well-known member
Jun 29, 2017
79
Taking the equation given by Theia , define the functions $\mathcal{I}(\lambda)$ and $\mathcal{J}(\lambda)$ as follows:

$\displaystyle \mathcal{I}(\lambda)= \int_0^1 \frac{\log(-\lambda \log(x))}{1+x+x^2}\mathrm{d}x, ~~ \mathcal{J}(\lambda)= -\frac{1}{2}\int_0^1 \frac{\log(-\lambda \log(x))}{1-x+x^2}\mathrm{d}x$,

We have $\displaystyle \mathcal{I}(\lambda) = \frac{\pi \log(\lambda)}{3\sqrt{3}}+\mathcal{I}(1)$ and $\displaystyle \mathcal{J}(\lambda) = -\frac{\pi \log(\lambda)}{3\sqrt{3}}+\mathcal{J}(1)$

I feel like I'm missing something. The constants are so close to the value of the integral (for $\lambda = 2$ ).

But it seems all it's saying is that $\mathcal{I}(\lambda)+\mathcal{J}(\lambda) =\mathcal{I}(1)+\mathcal{J}(1)$ (i.e. the sum $\mathcal{I}+\mathcal{J}$ is independent of $\lambda$).
 
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MountEvariste

Well-known member
Jun 29, 2017
79
Another observation: The integral can be written as

\(\displaystyle I = \int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1} \mathrm{d}x - \frac{1}{2} \int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1} \mathrm{d}x\),

but I'm not sure how to continue. It looks like these integrals can't be evaluated in terms of elementary functions and series (or other tricks) should be used, as June29 wrote.
Forgot to post this, but someone pointed out to me letting $x \mapsto x^2$ in either integral gives us the answer!
 

zgsqcy

New member
Dec 16, 2018
1
Another observation: The integral can be written as

\(\displaystyle I = \int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1} \mathrm{d}x - \frac{1}{2} \int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1} \mathrm{d}x\),

but I'm not sure how to continue. It looks like these integrals can't be evaluated in terms of elementary functions and series (or other tricks) should be used, as June29 wrote.
\[\int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1}\,\mathrm{d}x=\frac{\pi}{\sqrt{3}}\ln\left(\frac{\Gamma(2/3)}{\Gamma(1/3)}\sqrt[3]{2\pi}\right),\]
\[\int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1}\,\mathrm{d}x=\frac{2\pi}{\sqrt{3}}\ln\left(\frac{\sqrt[6]{32\pi^5}}{\Gamma(1/6)}\right).\]
 

MountEvariste

Well-known member
Jun 29, 2017
79
\[\int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1}\,\mathrm{d}x=\frac{\pi}{\sqrt{3}}\ln\left(\frac{\Gamma(2/3)}{\Gamma(1/3)}\sqrt[3]{2\pi}\right),\]
\[\int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1}\,\mathrm{d}x=\frac{2\pi}{\sqrt{3}}\ln\left(\frac{\sqrt[6]{32\pi^5}}{\Gamma(1/6)}\right).\]
It's easier than that. Call the integrals $I$ and $J$ and by letting $x \mapsto x^2$ we have:

$\begin{aligned} I & = \int_{0}^{1}\frac{\log(-\log x)}{x^2+x+1}\,dx \\& = \int_{0}^{1}\frac{2x\left[\log 2+\log(-\log x)\right]}{x^4+x^2+1}\,dx \\& =\frac{\pi\log 2}{3\sqrt{3}}+\int_{0}^{1}\log(-\log x)\left[\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}\right]\,dx \\& = \frac{\pi\log 2}{3\sqrt{3}}+\int_{0}^{1}\frac{\log(-\log x)}{x^2-x+1}\,dx-\int_{0}^{1}\frac{\log(-\log x)}{x^2+x+1}\,dx\end{aligned}$

Thus $\displaystyle 2I = \frac{\pi\log 2}{3\sqrt{3}}+J.$ Hence $\displaystyle I-\frac{1}{2}J = \frac{\pi\log 2}{6\sqrt{3}}.$