Proving Integral: $$t-3t^3+t^5\over 1+t^4+t^8$$ $\&$ $$\ln(-\ln t)$$

[Tony1]

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]

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]

I'm still trying to figure out what the two $dt$'s mean.

 

[tkhunny]

I'm still trying to figure out what the two $dt$'s mean.

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]

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]

Another observation: The integral can be written as

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

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]

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 $\lambda$).

 

[MountEvariste]

Another observation: The integral can be written as

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

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]

Another observation: The integral can be written as

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

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]

\[\int _0 ^1 \frac{\ln (- \ln x)}{x^2 + x + 1}\,\mathrm{d}x=\frac{\pi}{\sqrt{3}}\ln\left(\frac{\Gamma(2/3)}{\Gamma(1/3)}\sqrt[3]{2\pi}\right),\]
\[\int _0 ^1 \frac{\ln (- \ln x)}{x^2 - x + 1}\,\mathrm{d}x=\frac{2\pi}{\sqrt{3}}\ln\left(\frac{\sqrt[6]{32\pi^5}}{\Gamma(1/6)}\right).\]

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