How should I evaluate $\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$?

Question

$$\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$$

$$\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{1}{\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}x$$

$$\frac1{\pi}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{x^2}{x^4+x^2+1}\ \mathrm{d}x$$

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 $\sum(-1)^kt^k$?

Answer 1

Originally, I recently answered this integral on Quora. I have just copied the same text in this answer.

We want to evaluate the improper integral

$$ I = \int\limits_0^\infty\frac{x^9}{(1+x)(x^{12}+x^6+1)}\,\mathrm dx $$

My approach will be complex analysis based, but before applying that directly, let's simplify the integrand a bit.

Using the substitution $x\mapsto \frac1x$, we get $$ I = \int\limits_0^\infty\frac{x^2}{(1+x)(x^{12}+x^6+1)}\,\mathrm dx $$

Adding the two,

$$\begin{align}I &= \frac12\int\limits_0^\infty \frac{x^2(1+x^7)}{(1+x)(x^{12}+x^6+1)}\,\mathrm dx \\ &=\frac12\int\limits_0^\infty\frac{x^2(x^6-x^5+x^4-x^3+x^2-x+1)} {x^{12}+x^6+1}\,\mathrm dx \\ &= \frac12\underbrace{\int\limits_0^\infty\frac{x^8-x^7+x^6}{x^{12}+x^6+1}\,\mathrm dx }_{x\mapsto \frac1x} + \frac12\int\limits_0^\infty\frac{x^4-x^3+x^2}{x^{12}+x^6+1}\,\mathrm dx-\frac12\underbrace{\int\limits_0^\infty \frac{x^5}{x^{12}+x^6+1}\,\mathrm dx}_{=I_2} \\ &= \int\limits_0^\infty \frac{x^4-x^3+x^2}{x^{12}+x^6+1}\,\mathrm dx -\frac12I_2 \\ &= \sum_{k=2}^4 (-1)^k\underbrace{\int\limits_0^\infty\frac{x^k}{x^{12}+x^6+1}\,\mathrm dx }_{\equiv I_k}-\frac12 I_2 \end{align}$$ $$ I = \sum_{k=2}^4(-1)^kI_k-\frac12I_2 \tag 1$$

As compared to $I_k$, $I_2 $ can be computed with much more ease. Using the substitution $x^6=t$,

$$\begin{align}I_2 &= \frac16\int\limits_0^\infty\frac{\mathrm dt}{t^2+t+1}\\ &= \left. \frac16\cdot\frac2{\sqrt3}\arctan \Big(\frac{2t+1}{\sqrt3}\Big)\right|_0^\infty \\ &= \dfrac\pi{9\sqrt3}\end{align}$$

Now, it's time to evaluate $I_k$. This is going to be quite complicated. Let

$$f(z) = \dfrac{z^k}{z^{12}+z^6+1}$$

Consider the contour $\Gamma = \gamma_1\cup\gamma_2\cup\gamma_3 $ where $\gamma_1 $ runs from $0$ to $R$, $\gamma_2 $ is the circular arc from $R$ to the line $z=Re^{i\pi/3}$ and $\gamma_3 $ is the straight line itself, ending to $0$. For $R=10$, the contour will look like this

Here red, blue and green are $\gamma_1,\gamma_2,\gamma_3 $ respectively. For anyone interested, I used Desmos to get this plot. Now,

$$\begin{align}\oint_\Gamma f(z)\,\mathrm dz &= \int\limits_0^R \frac{x^k}{x^{12}+ x^6+1}\,\mathrm dx + \int\limits _0^{\pi/3}\frac{(R e^{i\theta})^k}{(R e^{i\theta})^{12}+(R e^{i\theta })^ 6+1}\cdot i R e^{i\theta}\, \mathrm d\theta + \int\limits_R^0 \frac{(xe^{i\pi/3})^k}{(xe^{i\pi/3} )^{12}+(xe^{i\pi/3})^6+1} \cdot e^{i\pi/3}\,\mathrm dx \\\oint_\Gamma f(z)\,\mathrm dz &= (1-e^{i\pi(k+1)/3})\int\limits_0^R \frac{x^k}{x^{12}+x^6+1}\,\mathrm dx +i \int\limits_0^{\pi/3}\frac{R^{k+1}e^{i\theta(k+1)}}{R^{12}e^{12i\theta}+R^6e^{6i\theta}+1}\,\mathrm d\theta \end{align}$$

Clearly, as $R\to \infty$, the second integral tends to $0$. So,

$$\lim_{R\to \infty} \oint_\Gamma f(z)\,\mathrm dz = (1-e^{i\pi(k+1)/3})I_k$$ $$ I_k = \mathfrak R \Big[\frac1{1-e^{i\pi(k+1)/3}}\lim_{R\to \infty}\oint_\Gamma f(z)\,\mathrm dz \Big] \tag 2$$

The result follows due to the fact that $I_k$ is real. Now, the denominator has $12$ zeroes out of which $2$ are inside our contour, namely $z=e^{i\pi/9},e^{2i\pi/9}$, which will be the poles of our integrand. To obtain this result, I used Wolfram|Alpha.

I used Wolfram|Alpha to calculate the residues, as simplifying the $11$ factors of the denominator which will be left while taking the limit is a tedious job.

$$\begin{align} \text{Res}_{z=e^{i\pi/9}} f(z) &= \frac1{6\sqrt3}\exp\Big(\frac{i\pi}{18}(2k+17)\Big) \\ \text{Res}_{z=e^{2i\pi/9}} f(z) &= \frac1{6\sqrt3}\exp\Big(\frac{i\pi}{18}(4k+25)\Big) \end{align}$$

So,

$$\begin{align}\oint_\Gamma f(z)\,\mathrm dz &= 2\pi i \Big[ \frac1{6\sqrt3 } \exp\Big(\frac{i\pi}{18}(4k+25)\Big) +\frac1 {6\sqrt3}\exp \Big(\frac{i\pi}{18}(2k+17)\Big) \Big] \\ &= \frac{\pi i}{3\sqrt 3} \Big[ \exp\Big(\frac{i\pi}{18}(4k+25)\Big) +\exp\Big(\frac{i\pi}{18}(2k+17)\Big) \Big] \end{align}$$

Finally using $(2)$,

$$I_k = -\frac\pi{6\sqrt3}\bigg\{ \sin\Big(\frac\pi{18}(2k+17)\Big)+\sin\Big(\frac\pi{18}(4k+25)\Big)+ \frac{\sin(\frac\pi3(k+1))}{1-\cos(\frac\pi3(k+1))}\Big[ \cos \Big(\frac\pi{18}(2k+17)\Big)+\cos \Big(\frac\pi{18}(4k+25)\Big)\Big]\bigg\} $$

Now, using $(1)$, we conclude that

$$\boxed{\boxed{\int\limits_0^\infty\frac{x^9}{(x+1)(x^{12}+x^6+1)}\,\mathrm dx = \frac\pi{9\sqrt3}+\frac\pi{6\sqrt3}\sin\frac\pi{18}-\frac{4\pi}9\sin\frac\pi9-\frac\pi{18}\cos\frac\pi{18}+\frac\pi{3\sqrt3}\cos\frac\pi9\approx 0.172733}}$$

Using Wolfram|Alpha, the difference between this answer and the numerical value of the integral is $-8.83515\times 10^{-13}$.

Answer 2

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 $$a^{1/6}=\cos \left(\frac{\pi }{9}\right)-i \sin \left(\frac{\pi }{9}\right)\qquad \qquad b^{1/6}=\cos \left(\frac{\pi }{9}\right)+i \sin \left(\frac{\pi }{9}\right)$$

Answer 3

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.