# Closed form for $\int_0^{\pi/2}\frac{\sqrt{1+\sin\phi}}{\sqrt{\sin2\phi}\,\sqrt{\sin\phi+\cos\phi}}d\phi$

Is it possible to evaluate this integral in a closed form? $$I=\int_0^{\pi/2}\frac{\sqrt{1+\sin\phi}}{\sqrt{\sin2\phi} \,\sqrt{\sin\phi+\cos\phi}}d\phi$$ Its approximate numeric value is $$I\approx3.0184739081025456585366323697639091178278225680831890465473990823...$$

• Have you tried the Weirstrass substitution? – Steven Stadnicki Apr 2 '14 at 23:18
• @Jeff: this is obviously not homework, so the interrogation is not necessary. Laila posts some challenging integrals and would like to see someone evaluate them analytically, using human power rather than just plugging into Maple or Mathematica. – Ron Gordon Apr 2 '14 at 23:32
• @RonGordon This may very well be homework/research problem, I have seen much more challenging problems as math homework given before. Are you assuming it is not? And if it is not homework,and the integral is for fun, well that is even more of a reason to make an effort as to what you have tried, rather than just posting a numerical result. – Jeff Faraci Apr 3 '14 at 0:02
• @Jeff: really? What class at what school? I am having a hard time believing that, but I'm curious to find out what sort of teacher assigns integrals that take many hours of effort, without it being obvious that they are able to be evaluated. I simply encourage Laila and others to post these - I find them fun to try out. In any case, she did not label it homework, so give her the benefit of the doubt. – Ron Gordon Apr 3 '14 at 0:06
• High School of Mathematics- Plovdiv, Bulgaria. As a student preparing for the IMO many years ago, they would give us plenty of integrals that would take MANY hours of effort. (Many integrals I post are from those days, the 1 that I put up for the bounty specifically). Sofia University (bulgaria)-any analysis class there or special functions course. Utrecht University (netherlands). I am not familiar with the US curriculum however though, so I cannot speak for this @RonGordon. I do encourage the posting of these too, of course they are fun however I do expect the OP to show work though. – Jeff Faraci Apr 3 '14 at 0:20

The integral equals $\sqrt{2+\sqrt{8}} \cdot \Omega$, where $\Omega$ is the real half-period $\Omega = 1.3736768699491\ldots$ of the elliptic curve $$E : y^2 = x^3 - 4 x^2 - 4 x,$$ i.e. the complete elliptic integral $$\Omega = \int_{2-\sqrt{8}}^0 \frac{dx}{\sqrt{x^3-4x^2-4x}} = \int_{2+\sqrt{8}}^\infty \frac{dx}{\sqrt{x^3-4x^2-4x}}$$ (the integrand can also be brought to the classical form ${\bf K}(k) = \int_0^1 dz \, / \sqrt{(1-z^2) (1-k^2 z^2)}$, but with a more complicated $k$ and probably also an elementary factor more complicated than our $\sqrt{2+\sqrt{8}}$).

Here's gp code for this formula:

sqrt(2+sqrt(8)) * ellinit([0,-4,0,-4,0])[15]


The curve $E$ is reasonably nice, with conductor $128=2^7$ and $j$-invariant $10976 = 2^5 7^3 = 1728 + 2^5 17^2$; but $E$ does not have complex multiplication (CM), so we do not expect to get a simpler form as would be possible for a CM curve [e.g. $\int_1^\infty dx/\sqrt{x^3-1}$ is a Beta integral, and $\int_0^\infty dx/\sqrt{x^3+4x^2+2x} = \Gamma(1/8) \Gamma(3/8) / (4\sqrt{\pi})$].

Harry Peter already used the trigonometric substitution $$(\cos \phi, \sin \phi, d\phi) = \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}, \frac{2 \, dt}{1+t^2} \right)$$ (which I guess is the "Weierstrass substitution" suggested in the comment of Steven Stadnicki) to write $I$ as $$\int_0^1 \frac{(1+t)dt}{\sqrt{(1+2t-t^2) (t-t^3)}},$$ which is a half-period of the holomorphic differential $(1+t) dt/u$ on the hyperelliptic curve $C: u^2 = (1+2t-t^2) (t-t^3)$ of genus $2$. Most such periods cannot be simplified further, but this one is special because the curve has more symmetry than just the "hyperelliptic involution" $(t,u) \leftrightarrow (t,-u)$. In particular $C$ has an involution $$\iota: (t,u) \leftrightarrow \left( \frac{1-t}{1+t}, \frac{2^{3/2}}{(1+t)^3} u \right)$$ which also sends the interval $(0,1)$ to itself, reversing the orientation. This suggests splitting the integral at the midpoint $t_0 := \sqrt{2} - 1$ and applying the change of variable $(t,dt) \leftarrow ((1-t)/(1+t), -2\,dt/(1+t)^2)$ to the integral over $(t_0,1)$ to obtain $\sqrt{2} \int_0^{t_0} dt/u$. Hence $$I = \int_0^{t_0} \frac{(\sqrt{2}+1+t)dt}{\sqrt{(1+2t-t^2) (t-t^3)}}$$ and now the change of variable $X = t + (1-t)/(1+t)$ transforms $I$ to an elliptic integral corresponding to the quotient curve $C\,/\langle\iota\rangle$. While $C\,/\langle\iota\rangle$ has irrational coefficients involving $\sqrt{2}$, it has rational $j$-invariant, so we can find coordinates that identify $C\,/\langle\iota\rangle$ with our curve $E$ with rational coefficients, though at the cost of introducing the factor $\sqrt{2+\sqrt{8}}$ into the formula for $I$ given at the start of this answer.

• You have solutions that surpass what I know. THank you for this work, this is the best kind of stuff I've seen on here. – Jeff Faraci May 2 '14 at 4:54
• Thank you very much for your answer! A good lesson to learn for me. – user 1591719 May 2 '14 at 15:05

\begin{align} \int_0^\frac{\pi}{2}\dfrac{\sqrt{1+\sin\phi}}{\sqrt{\sin2\phi}\sqrt{\sin\phi+\cos\phi}}d\phi &=\int_0^1\dfrac{2}{1+t^2}\dfrac{\sqrt{1+\dfrac{2t}{1+t^2}}}{\sqrt{\dfrac{4t(1-t^2)}{(1+t^2)^2}}\sqrt{\dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}}}dt\\ &=\int_0^1\dfrac{1+t}{\sqrt t\sqrt{1-t^2}\sqrt{1+2t-t^2}}dt\\ &=\int_0^1\dfrac{\sqrt{1+t}}{\sqrt t\sqrt{1-t}\sqrt{1+(\sqrt2+1)t}\sqrt{1-(\sqrt2-1)t}}dt\\ &=\pi F_D^{(3)}\left(\dfrac{1}{2},-\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},1;-1,-\sqrt2-1,\sqrt2-1\right). \end{align}

According to Lauricella hypergeometric series.

• Use double dollar signs instead of single dollar signs when writing your LaTex in. If you use single dollar sign, the integral sign is way too small relative to the integrand. – Jeff Faraci Apr 3 '14 at 15:17
• This is trivial. – user85798 Apr 30 '14 at 20:57
• @Oliver The solution below provides a trivial answer for us:) – Jeff Faraci May 2 '14 at 4:55