How should I evaluate $\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$? $$\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$$
$$\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{1}{\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}x$$
$$\frac1{\pi}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{x^2}{x^4+x^2+1}\ \mathrm{d}x$$
then substituting $t^3=x$
$$\frac3{\pi}\int_0^{\infty}\frac{t^9}{(1+t)(t^{12}+t^6+1)}\ \mathrm{d}t$$
what should i do after this should i write $\frac{1}{1+t}$ as $\sum(-1)^kt^k$?
 A: If, as @RAHUL commented, you use partial fraction decomposition, you should be able to prove that the result is the smallest positive root of
$$19683 y^6-94041 y^4+105786 y^2-2809=0$$
Solving the cubic equation in $y^2$ and taking the square root of it, the result is
$$\sqrt{\frac{1}{27} \left(43-2 \sqrt{543} \cos \left(\frac{1}{3} \cos
   ^{-1}\left(\frac{739}{362}\sqrt{\frac{3}{181}}\right)\right)\right)}=0.164948\cdots$$ which is confirmed by numerical integration.
Now, just prove it.
The partial fraction decomposition
$$\frac{t^9}{(t+1)(t^{12}+t^6+1)}=\frac{t^3}{1+t}\times\frac{t^6}{(t^{12}+t^6+1)}$$
$$\frac{t^6}{(t^{12}+t^6+1)}=\frac 1{a-b}\Big[\frac{a}{t^6-a}-\frac{b}{t^6-b} \Big]$$ where
$$a=-\frac{1}{2}-i\frac{ \sqrt{3}}{2}\qquad \text{and}\qquad b=-\frac{1}{2}+i\frac{ \sqrt{3}}{2}$$
So now, we face two terms looking like
$$\frac{t^3}{(t+1)(t^6-c)}=\frac 1{c-1}\Bigg[\frac 1{t+1}+\frac{c t^2-c t+c-t^5+t^4-t^3}{t^6-c} \Bigg]$$
$$I_k=\int \frac {t^k}{t^6-c}\,dt=-\frac{t^{k+1} }{c (k+1)}\,
   _2F_1\left(1,\frac{k+1}{6};\frac{k+7}{6};\frac{t^6}{c}\right)$$ All these integrals are easily computed. The simplest are
$$I_2=-\frac{1}{3 \sqrt{c}}\tanh ^{-1}\left(\frac{t^3}{\sqrt{c}}\right)\qquad \qquad I_5=\frac{1}{6} \log \left(1-\frac{t^6}{c}\right)$$ The other ones lead to sums of logarithms.
On the other side
$$J_k=\int_0^\infty \frac {t^k}{t^6-c}\,dt=\frac{\pi}{6}   \left(-\frac{1}{c}\right)^{\frac{5-k}{6}} \sec
   \left(\frac{(k-2)\pi}{6} \right)$$
To make the link with @Laxmi Narayan Bhandari's answer, notice that
$$a^{1/6}=\cos \left(\frac{\pi }{9}\right)-i \sin \left(\frac{\pi }{9}\right)\qquad \qquad b^{1/6}=\cos \left(\frac{\pi }{9}\right)+i \sin \left(\frac{\pi }{9}\right)$$
A: After some truly horrific partial fraction decomposition (I'm trusting WolframAlpha (check near the bottom of the page) for now, until I can verify it), we get
\begin{align}
\frac{ t^3 }{3 (t^6 - t^3 + 1)} + \frac{t}{6 (t^6 - t^3 + 1)} - \frac{t}{2 (t^6 + t^3 + 1)} - \frac{1}{6 (t^6 - t^3 + 1)} + \frac{1}{2 (t^6 + t^3 + 1)} + \frac{t^5}{3 (t^6 - t^3 + 1)} - \frac{t^4}{3 (t^6 - t^3 + 1)} - \frac{t^2}{6 (t^6 - t^3 + 1)} + \frac{t^2}{2 (t^6 + t^3 + 1)} - \frac{1}{3 (t + 1)}.
\end{align}
If I plug this guy into FriCAS, it simplifies to
\begin{align}
\frac{t^5 - t^4 + t^3 + t^2 - t}{3}.
\end{align}
This still looks like it'll diverge, so something else funky is going on here.
