# How to solve this integral/better way to approach?

$$\int_{0}^{c} dy \sqrt{\frac{c-1/2y^2+1/3y^3}{1+2y}}$$ where c is a constant. This is coming from trying to find the area $$\int_{U \le c} dq_1dq_2$$ where $$U=\frac{1}{2}(q_1^2+q_2^2)-\frac{1}{3}q_2^3+q_1^2q_2$$ bounded by energy $$c=U(q_1,q_2)$$.

• What have you done to try and solve it? Commented Nov 23, 2018 at 23:43
• Tried to solve for the cubic term and then realized I didn’t want to deal with all the solutions, so solve for the quadratic $q_1$ And that leads me directly to this integral that I don’t know how to solve Commented Nov 23, 2018 at 23:58
• @Некто That helps a lot. Also, to reiterate Yuri's question, are you using $e$ to denote the base of the natural logarithm, or are you using to denote an arbitrary constant? If it's the latter, WHYYYYY would you ever do such a thing? Commented Nov 25, 2018 at 6:56
• The domain of integration in the original problem is not convex and is not compact. So there are doubts that the proposed integral solves the problem of finding the area. Therefore, it makes sense to analyse cubic equation in the polar coordinates. This approach is detalized in my answer. Commented Dec 2, 2018 at 15:02
• BTW, the proposed integral does not converge Commented Dec 3, 2018 at 9:19

(Not an answer, just a comment that was too long).

You can prove that the value of the integral is $$\frac{c^2}{2\sqrt{6}} + O(c)$$ with the following algebraic simplifications. First note that the integral can be written as $$I = \frac{1}{\sqrt{6}}\int_0^c \sqrt{ (y-1)^2 + \frac{6c-1}{2y+1}} \ dy.$$ It follows that $$I > \frac{1}{\sqrt{6}} \int_0^c (y-1) \ dy = \frac{c(c-2)}{2\sqrt{6}}.$$ Similarly, using the fact that $$\sqrt{a+b} < \sqrt{a} + \sqrt{b}$$ (which does not quite hold in some regions of the domain but seems to be insignificant for large $$c$$), we have $$I < \frac{1}{\sqrt{6}} \int_0^c (y-1) \ dy + \frac{1}{\sqrt{6}}\int_0^c \sqrt{\frac{6c-1}{2y+1}} \ dy = \frac{c^2}{2\sqrt{6}} + O(c).$$ Thus we can conclude that $$I = \frac{c^2}{2\sqrt{6}} + O(c).$$ I think what could possibly help you is the following:

• If you want just a numerical answer, the function you are integrating is very smooth and convex so getting high precision values is doable.
• If you want a more accurate answer, you need to specify what regions of $$c$$ you are interested in. My answer holds for $$c \rightarrow \infty$$ but there are certainly more accurate answers for other cases such as $$c << 1$$.

$$\color{brown}{\textbf{Edition of 02.12.2018}}$$

HINT

In particular, for $$C=0.135$$ the graph is

for $$C=\frac16$$ the graph is

and for $$C=3.84$$ the graph is

These figures show, that the proposed integral can't calculate the area correctly, and it can be suitable to calculate the area in polar coordinates.

Let $$q_1=r\cos \varphi, \quad q_2=r \sin\varphi,$$ then $$U(r,\varphi) = \dfrac13r^3\sin 3\varphi+\dfrac12r^2.\tag1$$ Taking in account the properties of the sine function, it is sufficiently to consider $$U(r,\varphi)$$ at the interval $$\varphi\in\left(\frac\pi6,\frac\pi2\right).$$ The bounds determines by the system of inequalities $$\begin{cases} \dfrac13r^3\sin 3\varphi+\dfrac12r^2 > 0\\ \dfrac13r^3\sin 3\varphi+\dfrac12r^2 < C.\tag2 \end{cases}$$ The first inequality has the solution $$\begin{cases} r\in(0,\infty),\quad \text{if}\quad \varphi\in\left(\dfrac\pi6,\dfrac\pi3\right)\\ r\in\left(0,-\dfrac3{2\sin3\varphi}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi3,\dfrac\pi2\right) \end{cases}\tag3$$ Factor $$\dfrac4{Cr^3}$$ allows to present the second inequality in the form of $$\dfrac{4}{r^3} - \dfrac2{Cr} > \dfrac{4\sin3\varphi}{3C},$$ or $$4\left(\dfrac ar\right)^3-3\dfrac{a}r > p,\tag4$$ where $$a=\sqrt{\dfrac{3C}2},\quad p=2a\sin3\varphi.\tag5$$ $$\textbf{If p < 1,}$$ then can be used representation $$\cos\left(3\arccos\left(\dfrac ar\right)\right) > p.$$ Then $$\dfrac ar\in \begin{cases} [0,1],\quad\text{if}\quad p\in[-\infty,-1)\\ \left[\cos\left(\dfrac13\arccos p\right),\infty \right]\bigcup\left[0,\cos\left(\dfrac{2\pi}3-\dfrac13\arccos p\right)\right],\quad\text{if}\quad p\in[-1,1] \end{cases}$$ (see also Wolfram Alpha)

$$r\in \begin{cases} [a,\infty],\text{ if }p\in[-\infty,-1)\\ \left[0,\dfrac a{\cos\left(\dfrac13\arccos p\right)}\right] \bigcup\left[\dfrac a{\cos\left(\dfrac{2\pi}3-\dfrac13\arccos p\right)},\infty\right],\text{ if } p\in[-1,1], \end{cases}\tag6$$ $$\textbf{If p > 1,}$$ then can be used representation $$\cosh\left(3\cosh^{-1}\left(\dfrac ar\right)\right) > p,$$ $$r < \dfrac a{\cosh\left(\dfrac13\cosh^{-1}p\right)},$$ wherein $$\cosh^{-1}x = \log(x+\sqrt{x^2-1}),$$ $$\cosh\left(\dfrac13\cosh^{-1}x\right)=\dfrac12\left(\sqrt[3]{x+\sqrt{x^2-1}}+\dfrac1{\sqrt[3]{x+\sqrt{x^2-1}}}\right).$$ So $$r < \dfrac {2a}{\sqrt[3]{p+\sqrt{p^2-1}}+\sqrt[3]{p-\sqrt{p^2-1}}}.\tag7$$ Besides, $$r\ge0.$$ Let us consider two examples.

$$\textbf{Example C=0.135, a=0.45}$$ The control points are $$p\left(\dfrac\pi6\right)=0.9,\quad p\left(\dfrac\pi3\right)=0,\quad p\left(\dfrac{2\pi}5\right)=-0.593083,\quad p\left(\dfrac\pi2\right)=-0.9,$$ $$r\left(\dfrac\pi6\right)\in(0,0.455304),\quad r\left(\dfrac\pi3\right)\in(0,0.519615),\quad r\left(\dfrac{2\pi}5\right)\in(0,0.519615)\cup(2.43582,2.55195),\quad r\left(\dfrac\pi2\right)\in((0,0.721023)\cup(1.23406,1.5)).$$

System $$(2)$$ has solutions \left[\begin{align} r\in\left(0,\dfrac {0.45}{\cos\left(\dfrac13\arccos (0.9\sin3\varphi)\right)}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi6,\dfrac\pi2\right)\\ r\in\left(\dfrac {0.45}{\cos\left(\dfrac{2\pi}3-\dfrac13\arccos(0.9\sin3\varphi) \right)},-\dfrac3{2\sin3\varphi}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi3,\dfrac\pi2\right) \end{align}\right.\tag8 The obtained results correspond with the first graph.

$$\textbf{Example C=3.84, a=2.4}$$ The control points are $$p\left(\dfrac\pi6\right)=4.8,\quad p\left(\dfrac\pi3-\dfrac13\arcsin\dfrac5{24}\right)=1,\quad p\left(\dfrac\pi3-\dfrac13\arcsin 0.2\right)=0.96,\quad p\left(\dfrac\pi3\right)=0,\quad p\left(\dfrac\pi3+\dfrac13\arcsin 0.2\right)=-0.96,\quad p\left(\dfrac\pi3+\dfrac13\arcsin\dfrac5{24}\right)=-1,\quad p\left(\dfrac{2\pi}5\right)=-2.82137, \quad p\left(\dfrac\pi2\right)=-4.8,$$ $$r\left(\dfrac\pi6\right)\in(0,1.85345),\quad r\left(\dfrac\pi3-\dfrac13\arcsin\dfrac5{24}\right)=(0,2.4),\quad r\left(\dfrac\pi3-\dfrac13\arcsin0.2\right)=(0,2.41078),\quad r\left(\dfrac\pi3\right)\in(0,2.77128),\quad r\left(\dfrac\pi3+\dfrac13\arcsin0.2\right)=(0,4.14103)\cup(5.76975,7.5),\quad r\left(\dfrac\pi3+\dfrac13\arcsin\dfrac5{24}\right)=(0,7.2),\quad r\left(\dfrac{2\pi}5\right)\in(0,2.55195),\quad r\left(\dfrac\pi2\right)\in(0,1.5).$$

System $$(2)$$ has solutions $$\begin{cases} r\in\left(0,\dfrac {4.8}{\sqrt[3]{4.8\sin3\varphi+\sqrt{(4.8\sin3\varphi)^2-1}}+\sqrt[3]{4.8\sin3\varphi-\sqrt{(4.8\sin3\varphi)^2-1}}}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi6,\dfrac\pi3-\dfrac13\arcsin\dfrac5{24}\right)\\ r\in\left(0,\dfrac {2.4}{\cos\left(\dfrac13\arccos (4.8\sin3\varphi)\right)}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi3-\dfrac13\arcsin\dfrac5{24},\dfrac\pi3\right)\\ r\in\left(0,\dfrac {2.4}{\cos\left(\dfrac13\arccos (4.8\sin3\varphi)\right)}\right)\bigcup\left(\dfrac {2.4}{\cos\left(\dfrac{2\pi}3-\dfrac13\arccos(4.8\sin3\varphi) \right)},-\dfrac3{2\sin3\varphi}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi3,\dfrac\pi3+\dfrac13\arcsin\dfrac5{24}\right)\\ r\in\left(0,-\dfrac3{2\sin3\varphi}\right),\quad \text{if}\quad \varphi\in\left(\dfrac\pi3+\dfrac13\arcsin\dfrac5{24},\dfrac\pi2\right) \end{cases}\tag9$$ The obtained results correspond with the third graph.

$$\textbf{Finding the area}$$

The area of figure in the polar coordinates equals to

$$S=6\cdot\dfrac12\int\limits_{\pi/6}^{\pi/2}r^2(\varphi)\,\mathrm d\varphi.$$

In particular, for $$C=0.135$$

$$S=6\cdot\dfrac12\int\limits_{\pi/6}^{\pi/2}\dfrac {0.45^2}{\cos^2\left(\dfrac13\arccos (0.9\sin3\varphi)\right)}\,\mathrm d\varphi +6\cdot\dfrac12\int\limits_{\pi/3}^{\pi/2}\left(\dfrac{9}{4\sin^2(3\varphi)} - \dfrac {0.45^2}{\cos^2\left(\dfrac{2\pi}3-\dfrac13\arccos (0.9\sin3\varphi)\right)}\right)\,\mathrm d\varphi \approx 0.968088 + 0.968088 = \mathbf{1.937376}$$ (see also Wolfram Alpha for the first integral and for the second one)

• In (8) do you mean $\pi/6..\pi/3$ ? Commented Dec 2, 2018 at 16:02
• @Diger In $(8)$ the bracket is square. I.e. the first solution exists everywhere. Commented Dec 2, 2018 at 16:32
• Yes, but you write $\varphi \in (\pi/6,\pi/2)$, but isn't this the part where you should separate between $(\pi/6,\pi/3)$ and $(\pi/3,\pi/2)$ as in (3). And what special is the point $2\pi/5$ ? Commented Dec 2, 2018 at 16:40
• @Diger 1. This is another form of presentation. 2. the point $\dfrac{2\pi}5$ is one of the intermediate points, where the formulas and the figures were compared. Commented Dec 2, 2018 at 16:56

The Appell-Lauricella function is defined by the series $$F[\{a,c\};\{b_1,b_2,\dots,b_n\};\{x_1,x_2,\ldots,x_n\}]:=$$ $$=\sum_{i_1,i_2,\ldots,i_n\geq 0}\frac{(a)_{i_1+i_2+\ldots+i_n}(b_1)_{i_1}(b_2)_{i_2}\ldots(b_n)_{i_n}}{(c)_{i_1+i_2+\ldots+i_n}i_1!i_2!\ldots i_n!}x_1^{i_1}x_2^{i_2}\ldots x_n^{i_n},$$ where $$n\geq2$$, $$a,c,b_1,b_2,\ldots,b_n\in\textbf{C}$$ and $$|x_1|<1,|x_2|<1,\ldots,|x_n|<1$$.

Then holds the following

THEOREM. For $$Re(c)>Re(a)>0$$ and $$|x_1|<1,|x_2|<1,\ldots,|x_n|<1$$, we have $$F[\{a,c\};\{b_1,b_2,\dots,b_n\};\{x_1,x_2,\ldots,x_n\}]=$$ $$=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int^{1}_{0}t^{a-1}(1-t)^{c-a-1}(1-x_1t)^{-b_1}(1-x_2t)^{-b_2}\ldots (1-x_nt)^{-b_n}dt.$$

Using the above theorem I will prove that

$$\int^{c}_{0}\sqrt{\frac{c-y^2/2+y^3/3}{1+2y}}dy= \frac{c\sqrt{4-l}}{2\sqrt{6}}|l-1|\times$$ $$\times F\left[\{1,2\};\{\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\};\{-2c,\frac{2c}{l-1},\frac{4c}{4-l-\sqrt{3}\sqrt{(4-l)l}},\frac{4c}{4-l+\sqrt{3}\sqrt{(4-l)l}}\}\right],$$ where $$c=\frac{1}{24}(4-9l+6l^2-l^3)$$.

For to prove the above evaluation make the change of variable $$y\rightarrow -y$$ to get $$\int^{c}_{0}\sqrt{\frac{c-y^2/2+y^3/3}{2y+1}}dy=i\int^{-c}_{0}\sqrt{\frac{y^2/2+y^3/3-c}{-2y+1}}dy,$$ then $$y\rightarrow \frac{1-w}{2}$$ to get $$i\int^{-c}_{0}\sqrt{\frac{y^2/2+y^3/3-c}{-2y+1}}dy=\frac{\sqrt{c}}{4\sqrt{6}}\int^{2c+1}_{1}\sqrt{\frac{24+1/c(w-4)(w-1)^2}{w}}dw.$$ Now if $$c=\frac{1}{24}(4-9l+6l^2-l^3)$$ we can write $$24+(-4+w)(-1+w)^2/c=\frac{24(l-w)(9-6l+l^2-6w+lw+w^2)}{(l-4)(l-1)^2}.$$ Hence we can write the last integral in the form of theorem and use it to get the result, which is the Appell-Lauricella function.