# Evaluating $\int_0^{2\pi}\frac{dt}{\sqrt[4]{P(\cos t,\sin t)}}$

$${\LARGE\int}_0^{2\pi}\frac{dt}{\sqrt[{\LARGE 4}]{A\Big(\sin^8t+\cos^8t\Big)+B\Big(\sin^6t\cos^2t+\sin^2t\cos^6t\Big)+C~\sin^4t\cos^4t}}~=~?$$

where $A=0.3$, $B=-3.3$, and $C=10$. Its numerical value is about $12.0165220075768590.$

The Inverse Symbolic Calculator seems baffled. Maple and Mathematica are both unable to

return a closed form. $\bigg($Feel free to choose $A=\dfrac13$ and $B=-\dfrac{10}3$ , if it helps$\bigg)$.

Motivation:

$\qquad\qquad\qquad\quad$ The above shape is given by the implicit polynomial equation $$A\Big(x^8+y^8\Big)+B\Big(x^6y^2+x^2y^6\Big)+C~x^4y^4=R^8,$$ where $A=0.3$, $B=-3.3$, $C=10$, and $R=2$. Letting $x=r\sin t$ and $y=r\cos t$, we are finally able to express it in polar coordinates, since the Cartesian ones seem somewhat inadequate, given its extremely concave shape, which render it quite resistant to being parsed in terms of single-value functions, despite various sectionings and rotations. Thus the afore-mentioned integral is born. But whether it possesses a closed form, even one in terms of special functions, such as elliptic integrals, is beyond me. I don't really see the tangent half-angle substitution going anywhere. Perhaps some complex integration methods are in order ?

• Let's use W|A to do the painful trig simplification. I'll use $A=3/10$, $B=-33/10$, this gives the radicand as $$\frac{1}{320} (45+8 \cos(4 t)+43 \cos(8 t)),$$ which I think you'll agree is a decent improvement. The integral becomes $$\int_0^{2\pi} \frac{2^{3/2} \sqrt[4]{5} \ \mathrm{d}t}{(45+8 \cos(4 t)+43 \cos(8 t))^{1/4}}.$$ – Bennett Gardiner Jun 3 '14 at 11:59
• Another idea is that $$45 + 43 \cos(8 t) = 2 + 43 + 43\cos^2(4t) - 43\sin^2t = 2+86\cos^2t$$ which changes the integral further, into $$\int_0^{2\pi} \frac{2 \sqrt[4]{5} \ \mathrm{d}t}{(1+4 \cos(4 t)+43 \cos^2(4 t))^{1/4}}.$$ – Bennett Gardiner Jun 3 '14 at 12:19
• I didn't use W|A. I managed to reduce it to $$4\cdot (156)^{1/4}\int_0^{\pi/2} \frac{dx}{((13\sin^2x-7)^2+3)^{1/4}}$$ But I have no idea about the next step. – Pranav Arora Jun 3 '14 at 13:04
• @PranavArora: OK, I got it now: Yours fits numerically for the second case, with the simpler values for A and B. – Lucian Jun 3 '14 at 19:57
• @PranavArora: I have no idea. For now, only a couple of ideas come to mind: the Weierstrass tangent half-angle substitution, $t=\sin x$, or $t=\sin^2x$. I'll have to explore all three, and see where they might lead, before forming any opinion on the subject. – Lucian Jun 3 '14 at 20:09

In addition to the several equivalent integrals already published, this one is of interest : $${\LARGE\int}_0^{\infty}\frac{4~dt}{\sqrt[{\LARGE 4}]{4A\cosh^2t+2B\cosh t+C-2A}}$$ Suppose that a closed form exists for this integral. By "closed form" I mean the combination of a limited number of standard functions. This closed form must be valid for particular values of $A,B,C$, for example $A=1/4, B=1/2, C=1/2$, i.e. the integral : $${\LARGE\int}_0^{\infty}\frac{dt}{\sqrt[{\LARGE 4}]{\cosh^2t+\cosh t}}$$ As far as I know, there is no closed form for it, which is in contradiction with the above supposition. That is the reason why I think that it is doubtful that a closed form exists for the first integral. May be, it might exists a closed form involving other special functions not referenced or not today considered as standard function in the publications.

• $(+1)$ for finding a beautiful alternate expression, and for the pertinent insight. Also, see Brad's ongoing work, and feel free to share your thoughts, if you like. – Lucian Jul 7 '14 at 22:50

This post is a work in progress and I still have a lot of work to do in terms of formatting, showing work and the final evaluation of the integral. I have used contour integration to reduce the integral to an integral involving only polynomials. If you have any questions feel free to ask.

Let $\gamma$ be the unit$\require{autoload-all}$ circle.

$$I = \int_0^{2\pi}\!\!\frac{\mathrm{d}t}{\sqrt[4]{A(\sin^8(t)+\cos^8(t))+B(\sin^6(t)\cos^2(t)+\sin^2(t)\cos^6(t))+C\sin^4(t)\cos^4(t)}}$$

\toggle{ \text{Set} \; x = e^{it}\quad\enclose{roundedbox}{\text{ Click for Information }} }{ \begin{align} x &= e^{it}\\ \sin(t) &= \frac{1}{2i}\left(x-\frac{1}{x}\right)\\ \cos(t) &=\frac{1}{2}\left(x+\frac{1}{x}\right)\\ \mathrm{d}t &= \frac{-i \, \mathrm{d}x}{x} \end{align} }\endtoggle

$$I = \int_\gamma\!\frac{-i\,\mathrm{d}x}{x\sqrt[4]{A((x-\frac{1}{x})^8+(x+\frac{1}{x})^8)-B((x-\frac{1}{x})^6(x+\frac{1}{x})^2+(x-\frac{1}{x})^2(x+\frac{1}{x})^6)+C(x^2-\frac{1}{x^2})^4}}$$

We will let $$P'(x) = A\left(\left(\!x-\frac{1}{x}\!\right)^8\!\!+\left(\!x+\frac{1}{x}\!\right)^8\right) \!-B\left(\!x^2\!-\!\frac{1}{x^2}\!\right)^2\!\left(\left(\!x-\frac{1}{x}\!\right)^4\!\!+\left(\!x+\frac{1}{x}\!\right)^4\right)\! + C\left(\!x^2\!-\!\frac{1}{x^2}\!\right)^4$$

and $$P(x) = 256\cdot P'(x)$$

$$I = \int_\gamma \! \frac{-4i}{x\sqrt[4]{P(x)}} \mathrm{d}x$$

We will now proceed to evaluate the integral using contour methods. A careful note suggests that $P(x)$ can be solved like a quartic in $x^4$. Let the roots of this polynomial be $\xi_j$ for $1\leq j \leq 16$. Note that the function we are integrating tends to $0$ as $x\to 0$.

For any given sets of parameters, find the roots and use the theory of contour integration to evaluate the integral. This could be possible in the general case but most likely not worth pursuing.

With this we should be able to find a somewhat closed form for your specific case.

For your parameters, \begin{align}P(x) =\frac{86 x^8}{5}+\frac{86}{5 x^8}+\frac{16 x^4}{5}+\frac{16}{5 x^4}+36 \end{align}

Because it shares the same roots as a quartic in $x^4$ it admits roots expressible in terms of radicals. \toggle{\enclose{roundedbox}{\text{ Click for Analysis of Roots }} }{ \text{For simplicity, define}\\ \Omega_1 = \Re\left(-\frac{2}{43}-\frac{i \sqrt{39}}{43}+\frac{1}{2} \sqrt{-\frac{7536}{1849}+\frac{16 i \sqrt{39}}{1849}}\right) \approx -0.0398198141\\ \Omega_2 =\Im\left(-\frac{2}{43}-\frac{i \sqrt{39}}{43}+\frac{1}{2} \sqrt{-\frac{7536}{1849}+\frac{16 i \sqrt{39}}{1849}}\right) \approx 0.8642098747 \\ \Omega_3 = \Re\left(-\frac{2}{43}+\frac{i \sqrt{39}}{43}+\frac{1}{2} \sqrt{-\frac{7536}{1849}-\frac{16 i \sqrt{39}}{1849}}\right) \approx -0.0532034417\\ \Omega_4 = \Im\left(-\frac{2}{43}+\frac{i \sqrt{39}}{43}+\frac{1}{2} \sqrt{-\frac{7536}{1849}-\frac{16 i \sqrt{39}}{1849}}\right) \approx 1.1546748979 \\ \text{The roots are given by}\\ \begin{align} \zeta_{1-4} &= (\Omega_1+i\Omega_2 )^{1/4}e^{k\pi i/2}\\ \zeta_{5-8} &= (\Omega_1-i\Omega_2 )^{1/4}e^{k\pi i/2}\\ \zeta_{9-12} &= (\Omega_3+i\Omega_4 )^{1/4}e^{k\pi i/2}\\ \zeta_{13-16} &= (\Omega_3-i\Omega_4 )^{1/4}e^{k\pi i/2}\\ \end{align}\\ \text{for}\; k = 0,1,2,3. }\endtoggle

It is clear that $8$ of these roots ($\zeta_{1-8}\!$) are inside the contour.

Now consider $$I = \int_\gamma \! \frac{-4i}{x\sqrt[4]{P(x)}} \mathrm{d}x$$

We are now looking for a different way of evaluating this integral.

This integral will be split into many contours that include, the unit circle, small semi-circles around each root of $P(x)$ inside the unit circle, lines connecting the unit circle to the roots and lines connecting the roots to the origin. For a better idea of what contour I am describing, see the image at the bottom of the post.

It is good to note that the contribution from the small circles around each root is zero because the function tends to $\frac{1}{\sqrt[4]{z}}$ around the roots. Additionally, the total contribution of the lines connecting the unit circle to the root will be zero because it is the same path traversed in the opposite direction. This is not the case for the other lines because of the branch cut.

We can now write that the final result, $I$, is equal to the sum of the integrals that connect the roots to the origin. For a simplification,

\begin{align} \Omega_1 &= -\frac{2}{43}-\frac{i \sqrt{39}}{43}+\frac{1}{2} \sqrt{-\frac{7536}{1849}+\frac{16 i \sqrt{39}}{1849}}\\[.2cm] \xi &= \Omega_1^{1/4} \approx 0.8867082411 + 0.3793025249 i\end{align}

$\xi$ is one of the roots of $P(x)$. Using symmetry, which will be justified later, we find that

$$\large J = \int_0^1\!\!\! \frac{32 i}{t \sqrt[4]{\frac{86}{5}t^8\xi^8+\frac{86}{5 t^8 \xi^8}+\frac{16}{5} t^4\xi^4+\frac{16}{5 t^4 \xi^4}+36}} \mathrm{d}t$$

$$\large I = 12.01652200\ldots = \boxed{\Re({J}) + \Im({J})}$$

For numerical evidence

In :=

w= N[(-(2/43)-(I Sqrt[39])/43+1/2 Sqrt[-(7536/1849)+(16 I Sqrt[39])/1849])^(1/4),100]

J= NIntegrate[(32 I)/(t (36+86/(5 t^8 w^8)+16/(5 t^4 w^4)+16/5 t^4 w^4+86/5 t^8 w^8)^(1/4)),{t,0,1},WorkingPrecision->50];

Re[J]+Im[J]

Out := 12.016522007576859017247019246788042118792983861776

The next step is to find a simpler form for $J$. The most natural substitution is that of $x = t\xi$. With this, write

\begin{align} J &= 32 i\int_0^\xi\!\!\! \frac{ \mathrm{d}x}{x\sqrt[4]{\frac{86}{5}x^8+\frac{86}{5x^8}+\frac{16}{5} x^4+\frac{16}{5x^4}+36}} \\[.2cm] &= 32i \cdot 5^{1/4} \int_0^\xi \!\!\! \frac{x}{\sqrt[4]{86x^{16}+16x^{12}+180x^8+16x^4+86}}\mathrm{d}x\end{align}

This result may not seem satisfactory but I have turned the trigonometric integral into one involving only polynomials. This post is a work in progress so I may be able to find a nicer form for the integral in the coming time.

For the reasoning behind my choice of contour, see this picture. The $16$ roots are located at the tips of the white lines with the white lines being the branch cuts. Due to aforementioned simplification, the final integral is equal to the sum of the integrals along the white lines in the unit circle. Symmetry can be used to simplify the final evaluation.

Here is the code used to generate this image (stolen from this post).

GraphicsRow[Table[ContourPlot[g[f],{x,-1.5,1.5},{y,-1.5,1.5},ColorFunction->"GrayYellowTones",Epilog->{Cyan,Thick,Circle[{0,0}],Red,PointSize[0.02],Point[{Re@#,Im@#}&/@{}]},Contours->100],{g,{Re,Im}}]]

• If the mathematica code gives any trouble try replacing the arrow with ->. – Brad Jul 7 '14 at 21:55
• $(+1)$ for the great work! The grace period, starting in $3$ hours, lasts for an entire day, so there's still plenty of time. – Lucian Jul 7 '14 at 23:00
• $y=x^2$ seems like an obvious substitution in the last integral. – Lucian Jul 8 '14 at 2:37
• This seems way too complicated. Notice $$\sqrt[4]{P(t)} = |\sin t\cos t| \sqrt[4]{A \left( \tan^4 t + \frac{1}{\tan^4 t} \right) + B \left( \tan^2 t + \frac{1}{\tan^2 t}\right) + C}$$ Since $\displaystyle\;\frac{d\tan t}{\tan t} = \frac{dt}{\sin t\cos t}\;$, if one define $\;z = \tan t\;$ for $t \in [0,\frac{\pi}{2})$, we have $$\int_0^{2\pi} \frac{dt}{\sqrt[4]{P(t)}} = 4 \int_0^{\pi/2} \frac{dt}{\sqrt[4]{P(t)}} = 4 \int_0^\infty \frac{dz}{z\sqrt[4]{A (z^4 + z^{-4} ) + B(z^2 + z^{-2}) + C}}\\ = 4 \int_0^\infty \frac{dz}{\sqrt[4]{A (z^8 + 1 ) + B(z^6 + z^2) + Cz^4}}$$ – achille hui Jul 8 '14 at 2:51
• @achillehui That is a very slick way of finding the integral. I regret that the thought of contour integration came to me before a tangent substitution. Do you think that either my or your integral is likely to have a closed form? – Brad Jul 8 '14 at 3:13