Prove that the integral is equal to ##\pi^2/8##

[Meden Agan]

Homework Statement

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$

Relevant Equations

Fubini's Theorem, arcsin representation integral.

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$

Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$

The representation integral of $\arcsin$ is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$
Plugging identity above into $(1)$ with $u = \sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}$, we obtain
$$I = \int\limits_0^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{x - x^2}} \int_0^{\sqrt{\frac{\left(x - 1\right) \left( x - 1 + \sqrt{9 - 16x} \right)}{1-2x}}} \frac{\mathrm dt}{\sqrt{1 - t^2}} \, \mathrm dx.$$
Since the integrand is non-negative and continuous over the rectangular domain $D := \left\{(x, t) : 0 \leq x \leq \frac{\sqrt{2}}{4}, \,\, 0 \leq t \leq f(x)\right\}$ ($f(x)$ is the root of the numerator), Fubini's Theorem allows us to interchange the order:
$$I = \int\limits_0^1 \frac{\mathrm dt}{\sqrt{1 - t^2}} \int\limits_{x_*(t)}^{x^*(t)} \frac{\mathrm dx}{\sqrt{x- x^2}},$$
where $x_*(t)$ and $x^*(t)$ are the closed solutions of the equation
$$t^2 = \frac{(x - 1)\left(x - 1 + x \sqrt{9 - 16x}\right)}{1-2x}. \tag{*}$$
Now, computing the closed-form solutions of Equation $(*)$ looks like a lot of work. And even WolframAlpha returns a tremendous expression.

I'd like your help to get on with that integral.

 

[Meden Agan]

Homework Statement: Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$
Relevant Equations: Fubini's Theorem, arcsin representation integral.

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$

Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$

The representation integral of $\arcsin$ is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$
Plugging identity above into $(1)$ with $u = \sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}$, we obtain
$$I = \int\limits_0^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{x - x^2}} \int_0^{\sqrt{\frac{\left(x - 1\right) \left( x - 1 + \sqrt{9 - 16x} \right)}{1-2x}}} \frac{\mathrm dt}{\sqrt{1 - t^2}} \, \mathrm dx.$$
Since the integrand is non-negative and continuous over the rectangular domain $D := \left\{(x, t) : 0 \leq x \leq \frac{\sqrt{2}}{4}, \,\, 0 \leq t \leq f(x)\right\}$ ($f(x)$ is the root of the numerator), Fubini's Theorem allows us to interchange the order:
$$I = \int\limits_0^1 \frac{\mathrm dt}{\sqrt{1 - t^2}} \int\limits_{x_*(t)}^{x^*(t)} \frac{\mathrm dx}{\sqrt{x- x^2}},$$
where $x_*(t)$ and $x^*(t)$ are the closed solutions of the equation
$$t^2 = \frac{(x - 1)\left(x - 1 + x \sqrt{9 - 16x}\right)}{1-2x}. \tag{*}$$
Now, computing the closed-form solutions of Equation $(*)$ looks like a lot of work. And even WolframAlpha returns a tremendous expression.

I'd like your help to get on with that integral.

Typo. The representation integral of $\arcsin$ is $$\arcsin u = \int\limits_{0}^{u} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$

 

[fresh_42]

I have tried to solve this by my favorite method: "remove what disturbs the most" and set $y^2=9-16x ,$ which gave me some symmetries in the integrand:
$$ \begin{align*}&\dfrac{1}{\sqrt{x-x^2}}\operatorname{arcsin}\sqrt{\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}\\
&=\sqrt{\dfrac{1}{4}-\dfrac{1}{256}(y^2-1)^2}\operatorname{arcsin}\sqrt{\dfrac{1}{32}\left(((y^2-1)+8)(y+1)-\dfrac{8}{y+1}\right)}\end{align*} $$
or
$$
\dfrac{16}{\sqrt{(8-uv)(8+uv)}}\operatorname{arcsin}\sqrt{\dfrac{(8+uv)v}{32}-\dfrac{1}{4v}}
$$
with $u=y-1\, , \,v=y+1.$ But I have no idea how to use it, or how to get rid of the single $v$ terms.

Edit: I haven't adjusted $dx$ yet, so maybe this is a chance to simplify the expression further.

 

[Meden Agan]

I have tried to solve this by my favorite method: "remove what disturbs the most" and set $y^2=9-16x ,$ which gave me some symmetries in the integrand:
$$ \begin{align*}&\dfrac{1}{\sqrt{x-x^2}}\operatorname{arcsin}\sqrt{\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}\\
&=\sqrt{\dfrac{1}{4}-\dfrac{1}{256}(y^2-1)^2}\operatorname{arcsin}\sqrt{\dfrac{1}{32}\left(((y^2-1)+8)(y+1)-\dfrac{8}{y+1}\right)}\end{align*} $$
or
$$
\dfrac{16}{\sqrt{(8-uv)(8+uv)}}\operatorname{arcsin}\sqrt{\dfrac{(8+uv)v}{32}-\dfrac{1}{4v}}
$$
with $u=y-1\, , \,v=y+1.$ But I have no idea how to use it, or how to get rid of the single $v$ terms.

Mhm... I've discovered something which puzzles me, but I have no idea how to prove it.
From post #1, we have the double integral $$I = \int\limits_0^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{x - x^2}} \int_0^{\sqrt{\frac{\left(x - 1\right) \left( x - 1 + \sqrt{9 - 16x} \right)}{1-2x}}} \frac{\mathrm dt}{\sqrt{1 - t^2}} \, \mathrm dx.$$
If we change $x$-bounds to $0$ and $1$ (then, $0 \leqslant x \leqslant 1$) and $t$-bounds to $\frac{1}{2} \left(1- \sqrt{1-t^2}\right)$ and $\frac{1}{2} \left(1+ \sqrt{1-t^2}\right)$ (then, $\frac{1}{2} \left(1- \sqrt{1-t^2}\right) \leqslant t \leqslant \frac{1}{2} \left(1+ \sqrt{1-t^2}\right)$) we actually obtain $\dfrac{\pi^2}{8}$ as final result (interchanging order, eventually).
The integrand function must remain intact.

How do you explain this? Is it possible to perform such a change of bounds?

 

[fresh_42]

I think I have a mistake in the second line. Should probably be $(y^2-1)^{-2}$ but I haven't checked.

I don't understand your question. A better way to write an integral is $\displaystyle{\int_{x=a}^{x=b}f(x)\,dx}$ so any substitution leads to new bounds. The $x$-bounds from $0$ to $\sqrt{2}/4$ are just nice to have all roots positive. My substitution showed me that we have only powers of $2$ everywhere, so this looks right. What do you mean by changing the bounds to $[0,1]$? Shouldn't this result in another value? If you want that, you should substitute $z=\dfrac{4}{\sqrt{2}} x.$ I tried this first, but it didn't help a lot. Scaling is the least problem.

 

[pines-demon]

Homework Statement: Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$
Relevant Equations: Fubini's Theorem, arcsin representation integral.

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$

Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$

The representation integral of $\arcsin$ is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$
Plugging identity above into $(1)$ with $u = \sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}$, we obtain
$$I = \int\limits_0^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{x - x^2}} \int_0^{\sqrt{\frac{\left(x - 1\right) \left( x - 1 + \sqrt{9 - 16x} \right)}{1-2x}}} \frac{\mathrm dt}{\sqrt{1 - t^2}} \, \mathrm dx.$$
Since the integrand is non-negative and continuous over the rectangular domain $D := \left\{(x, t) : 0 \leq x \leq \frac{\sqrt{2}}{4}, \,\, 0 \leq t \leq f(x)\right\}$ ($f(x)$ is the root of the numerator), Fubini's Theorem allows us to interchange the order:
$$I = \int\limits_0^1 \frac{\mathrm dt}{\sqrt{1 - t^2}} \int\limits_{x_*(t)}^{x^*(t)} \frac{\mathrm dx}{\sqrt{x- x^2}},$$
where $x_*(t)$ and $x^*(t)$ are the closed solutions of the equation
$$t^2 = \frac{(x - 1)\left(x - 1 + x \sqrt{9 - 16x}\right)}{1-2x}. \tag{*}$$
Now, computing the closed-form solutions of Equation $(*)$ looks like a lot of work. And even WolframAlpha returns a tremendous expression.

I'd like your help to get on with that integral.

First get rid of the prefactor by going $x=\sin^2(\theta)$, the rest should be good trigonometry.

 

[Meden Agan]

First get rid of the prefactor by going $x=\sin^2(\theta)$, the rest should be good trigonometry.

Do you mean I should substitute $x = \sin^2 (\theta)$ into $\dfrac{(x - 1)\left(x - 1 + x \sqrt{9 - 16x}\right)}{1-2x}$? If yes, I fail to see how this helps. If not, I can't see what you mean.

 

[Meden Agan]

What do you mean by changing the bounds to $[0,1]$? Shouldn't this result in another value?

I mean in the double integral, if we set $t= \sqrt{\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}$, bounds of the outer integral change to $[0,1]$. As a standard procedure for double integrals, simply substitute $0$ and $\dfrac{\sqrt 2}{4}$ into $t= \sqrt{\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}$. Yes?

 

[fresh_42]

Let's see. We have
$$
\displaystyle{I=\int_{x=0}^{x=\sqrt{2}/4} f(x)\cdot \operatorname{arcsin} g(x) \,dx }
$$
and substitute $g(x)=u$ which I omit for clarity reasons and
$$
\operatorname{arcsin}g(x)=\int_{t=0}^{t=g(x)} \dfrac{dt}{\sqrt{1-t^2}}.
$$
This yields
\begin{align*}I&=\int_{x=0}^{x=\sqrt{2}/4} f(x)\cdot \int_{t=0}^{t=g(x)} \dfrac{dt}{\sqrt{1-t^2}} \,dx \end{align*}

Here is where it starts to become problematic. Since the inner limits depend on the outer, we cannot simply exchange the integration order. You shifted the complicated argument of the arcus sine into the upper integral limit. If you now exchange the order, then this limit hangs, so to speak, in the air.

Your solution to this mystery seems to be a rescaling of the variable $t,$ such that $t'=g(x)=1.$ But that would transfer the ugly upper bound back into the integrand. Let's see if we can rescale the other variable. Say we set $x'=\alpha x.$ Our equation now becomes
\begin{align*}
I&=\dfrac{1}{\alpha}\int_{x'=0}^{x'=\alpha\sqrt{2}/4} f\left(\dfrac{x'}{\alpha}\right)\cdot \int_{t=0}^{t=g(x'/\alpha)} \dfrac{dt}{\sqrt{1-t^2}} \,dx'
\end{align*}
and we want to choose $\alpha$ such that $g\left(\dfrac{x'}{\alpha}\right)=1.$

This would solve your problem if you could identify the value of $\alpha ,$ but it looks awkward.

You should give @pines-demon 's idea at least a try.

 

[Meden Agan]

You should give @pines-demon 's idea at least a try.

I get nothing.

We have
$$I = \int_0^{\sqrt{2}/4} \frac{\arcsin\sqrt{\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}}{\sqrt{x-x^2}}\,\mathrm{d}x.$$ Under trigonometric substitution $x=\sin^2 \theta$, with $0 \leqslant \theta \leqslant \theta_1$ and $\theta_1 = \arcsin\sqrt{\dfrac{\sqrt{2}}{4}}$, we have $\mathrm{d}x = 2\sin \theta \cos \theta$ and $\sqrt{x-x^2} = \sin \theta \cos \theta$. The integral becomes
$$I = 2 \int_0^{\theta_1} \arcsin \left(\sqrt{R(\theta)}\right)\,\mathrm{d}\theta, \qquad R(\theta) = \frac{\left(\sin^2\theta-1 \right)\left(\sin^2\theta-1+\sin^2\theta\sqrt{9-16\sin^2\theta}\right)}{1-2\sin^2\theta}.$$
To simplify computation, set $A=\sin^2 \theta$ and $B=1-A=\cos^2\theta$. Then, $\sin^2\theta-1 = -B$ and $1-2\sin^2\theta = B-A = \cos 2\theta$, while $9-16\sin^2\theta = 1+8\cos \left(2\theta\right)$. Reassembling factors gives
$$R(\theta) = \frac{B \left(B-A\sqrt{1+8(B-A)}\right)}{B-A}.$$
Let's write $B-A\sqrt{1+8 \left(B-A \right)}$ as $(B-A)+A \left(1-\sqrt{1+8 \, \left(B-A \right)}\right)$.
Thus:
$$\begin{aligned}
R(\theta) &= \frac{B \left[\left(B-A \right) + A \left(1-\sqrt{1+8(B-A)}\right)\right]}{B-A} \\
&= \frac{B \left(B-A \right) + AB \left(1-\sqrt{1+8(B-A)}\right)}{B-A} \\
&=\frac{B \left(B-A \right)}{\left(B-A\right)} + \frac{AB \left(1-\sqrt{1+8(B-A)}\right)}{B-A} \\
&= \frac{B \cancel{\left(B-A \right)}}{\bcancel{\left(B-A\right)}} + \frac{AB \left(1-\sqrt{1+8(B-A)}\right)}{B-A}\\
&= B + \frac{AB \left(1-\sqrt{1+8(B-A)}\right)}{B-A} \\
&= \cos^2 \theta + \frac{\sin^2 \theta \cos^2 \theta \left(1-\sqrt{1+8 \cos(2\theta)}\right)}{\cos(2 \theta)}.
\end{aligned}$$
In order for the final result to be $\pi^2/8$, we need that $\dfrac{\sin^2 \theta \cos^2 \theta \left(1-\sqrt{1+8 \cos(2\theta)}\right)}{\cos(2 \theta)} = 0$. And that is not true in general :-(

 

[fresh_42]

My result is complicated, too. I wrote $\theta=\alpha$ since it is easier to type with my keyboard settings. I got for the entire integral $I$
$$
I=2\int_0^{\operatorname{arcsin}\left(1/\sqrt[4]{8}\right)} \operatorname{arcsin}\left(
\sqrt{\dfrac{\cos^2(\alpha)-y \sin^2(\alpha)}{1-\tan^2(\alpha)}}
\right)\,d\alpha
$$
where $y=\sqrt{(3-4\sin(\alpha))(3+4\sin(\alpha))}.$

My next step would be to use the Weierstraß substitution and see whether a polynomial expression simplifies that nasty trig expression, but I assume that we cannot get rid of the crucial term $y.$

 

[Meden Agan]

Let's see. We have
$$
\displaystyle{I=\int_{x=0}^{x=\sqrt{2}/4} f(x)\cdot \operatorname{arcsin} g(x) \,dx }
$$
and substitute $g(x)=u$ which I omit for clarity reasons and
$$
\operatorname{arcsin}g(x)=\int_{t=0}^{t=g(x)} \dfrac{dt}{\sqrt{1-t^2}}.
$$
This yields
\begin{align*}I&=\int_{x=0}^{x=\sqrt{2}/4} f(x)\cdot \int_{t=0}^{t=g(x)} \dfrac{dt}{\sqrt{1-t^2}} \,dx \end{align*}

Here is where it starts to become problematic. Since the inner limits depend on the outer, we cannot simply exchange the integration order. You shifted the complicated argument of the arcus sine into the upper integral limit. If you now exchange the order, then this limit hangs, so to speak, in the air.

Your solution to this mystery seems to be a rescaling of the variable $t,$ such that $t'=g(x)=1.$ But that would transfer the ugly upper bound back into the integrand. Let's see if we can rescale the other variable. Say we set $x'=\alpha x.$ Our equation now becomes
\begin{align*}
I&=\dfrac{1}{\alpha}\int_{x'=0}^{x'=\alpha\sqrt{2}/4} f\left(\dfrac{x'}{\alpha}\right)\cdot \int_{t=0}^{t=g(x'/\alpha)} \dfrac{dt}{\sqrt{1-t^2}} \,dx'
\end{align*}
and we want to choose $\alpha$ such that $g\left(\dfrac{x'}{\alpha}\right)=1.$

This would solve your problem if you could identify the value of $\alpha ,$ but it looks awkward.

Mhm, good point.

But what I'm trying to say is something different.
I am saying I discovered that the two double integrals $$\int\limits_0^{\frac{\sqrt{2}}{4}} \frac{1}{\sqrt{x - x^2}} \int_0^{\sqrt{\frac{\left(x - 1\right) \left( x - 1 + \sqrt{9 - 16x} \right)}{1-2x}}} \frac{\mathrm dt}{\sqrt{1 - t^2}} \, \mathrm dx \tag{1}$$
and $$\int\limits_0^{1} \frac{1}{\sqrt{x - x^2}} \int_{\frac{1}{2}\left(1- \sqrt{1-t^2}\right)}^{\frac{1}{2}\left(1+ \sqrt{1-t^2}\right)} \frac{\mathrm dt}{\sqrt{1 - t^2}} \, \mathrm dx \tag{2}$$ are equivalent.

Surprisingly no change of integrand function, just a change of bounds. And evaluating the integral $(2)$ is very easy; the result is $\dfrac{\pi^2}{8}$, as desired.

How do we explain that?

 

[fresh_42]

How did you show equivalence without rescaling $x$ which would be necessary to change the upper bound of the outer integral? A numerical coincidence isn't proof.

And I don't understand the inner integral. How can you have the same variable in the limits as for the integration variable? This is what you should avoid by all means! Always.

 

[Meden Agan]

My result is complicated, too. I wrote $\theta=\alpha$ since it is easier to type with my keyboard settings. I got for the entire integral $I$
$$
I=2\int_0^{\operatorname{arcsin}\left(1/\sqrt[4]{8}\right)} \operatorname{arcsin}\left(
\sqrt{\dfrac{\cos^2(\alpha)-y \sin^2(\alpha)}{1-\tan^2(\alpha)}}
\right)\,d\alpha
$$
where $y=\sqrt{(3-4\sin(\alpha))(3+4\sin(\alpha))}.$

My next step would be to use the Weierstraß substitution and see whether a polynomial expression simplifies that nasty trig expression, but I assume that we cannot get rid of the crucial term $y.$

I have no idea how to handle that...

 

[fresh_42]

I have no idea how to handle that...

Yes, I'm stuck, too.

We can write it as - where $y^2=9-16x$ and $x=\sin^2(\alpha)$ -
$$
I=2\int_0^{\operatorname{arcsin}(1/\sqrt[4]{8})} \operatorname{\arcsin}\left(\cos(\alpha)\sqrt{\dfrac{1-y\tan^2(\alpha)}{1-\tan^2(\alpha)}}\right)\,d\alpha
$$
I tried to set
$$
\dfrac{1-y\tan^2(\alpha)}{1-\tan^2(\alpha)}=\dfrac{\sin^2(\beta)}{\cos^2(\alpha)}
$$
but then I need the derivative according to $\alpha$ to get $d\beta$ which is a nightmare.

The expression has two difficulties: the asymmetry due to $y=\sqrt{9-16x}$ and the fact that the argument isn't a sine value, so that we can't get rid of the $\operatorname{arcsin}.$ We have a fourth root in the argument of the arcus sine and I have no idea how to handle this.

I think there is a trick for the asymmetry in $y,$ but that nasty term depends on the variable, which complicates it. The examples I have seen so far used the Taylor series of $\operatorname{arcsin},$ but that doesn't seem to be an option here. The Weierstraß substitution wasn't of help either.

 

[fresh_42]

I think I made a step forward. Say we abbreviate the integral by
$$
I=\int_0^{\sqrt{2}/4}\dfrac{1}{\sqrt{x-x^2}}\operatorname{arcsin}\sqrt{\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}}\,dx
$$
and make the substitutions $x=\sin^2(\alpha)\, , \,y^2=9-16x\, , \,a=\operatorname{arcsin}(1/\sqrt[4]{8})$ and call the function
$$
f(\alpha)=\dfrac{\cos(\alpha)+\sqrt{y(\alpha)}\sin(\alpha)}{1+\tan(\alpha)}
$$
then I get a symmetric expression
$$
I=I(a)=\int_{-a}^a \operatorname{arcsin} \sqrt{f(-\alpha)f(\alpha)}\,d\alpha
$$
This is still not easy since $f(\alpha)$ is not symmetric, but at least the entire integrand and the boundaries are symmetric now. Maybe we can use this. $f(\alpha)$ alone looks like

 

[pines-demon]

Relevant Equations: Fubini's Theorem, arcsin representation integral.

What is the actual statement or context of this integral? Does it explictly say that it can be calculated using Fubini's theorem? Or is this experimental math?

 

[Meden Agan]

Does it explictly say that it can be calculated using Fubini's theorem?

No, but I just felt it was the easiest way to evaluate it.

 

[pines-demon]

No, but I just felt it was the easiest way to evaluate it.

So what is the context? Where you playing with numerics?

 

[Meden Agan]

So what is the context? Where you playing with numerics?

No. My instructor proposed that integral.

 

[PeroK]

I was worried you might all be wasting your time, but a quick numerical estimate (on Excel) indicates that the result might well be true.

 

[fresh_42]

I gave WA the integrand, and here is what you want to calculate: the area below ...

I also fed WA with the integral

https://www.wolframalpha.com/input?i=y(x)=integral+(from+0+to+arcsin(1/root(root(8)))+)+arcsin+(+(((cos(x)+root(root(9-16*sin^2(x)))*sin(x))/(1+tan(x)))*((cos(-x)+root(root(9-16*sin^2(-x)))*sin(-x))/(1+tan(-x))))^(1/2)+)+dx+

and I received $I/2=0.61685$ numerically versus $\dfrac{\pi^2}{16}= 0.61685027506808491367715568749226 \ldots$

At least, I didn't make a mistake with all my substitutions and manipulations of the $\operatorname{arcsin}$ argument.

The question is really how to solve $\displaystyle{I=\int_0^a \operatorname{arcsin}\sqrt{f(\alpha)f(-\alpha)} \,d\alpha.}$

One thing, I haven't tried so far is to use the identity
$$
\operatorname{arcsin}(xy)=-i \log\left(\sqrt{1-x^2 y^2}+ i xy\right)\,.
$$

Note: The deviation of $\sqrt{f(\alpha)f(-\alpha)}$ from a suitably scaled circle yields, with arcus sine and integral, the value $0.636772,$ only $1\%$ off.

 

[Meden Agan]

One thing, I haven't tried so far is to use the identity
$$
\operatorname{arcsin}(xy)=-i \log\left(\sqrt{1-x^2 y^2}+ i xy\right)\,.
$$

Mhm... Do you think that identity can make anything easier? IMO, probably complicates the integrand even more.

 

[fresh_42]

That's why I haven't tried it so far. $\sqrt{f(\alpha)f(-\alpha)}$ looked like a perfect circle,

but this wasn't the case, although the deviation is really small:

This nasty little bulk makes the function so complicated. Our curve is a circle with a flat spot around $1/2.$

I'm currently back at
$$
f(\alpha)f(-\alpha)=\dfrac{1}{32}\dfrac{(y^2+7)(y^2+2y-7)}{y+1}
$$
and look for a parameterization that is suited to draw the root and apply the arcus sine. The fact that the result can be noted exactly is the curse here. It makes you think that there is a hidden trick somewhere.

 

[pines-demon]

No. My instructor proposed that integral.

Is an instructor for an undergrad course? is the integral physics related? Is it a course of math, physics or something in particular? what techniques would they expect you to know? Any additional context would help

 

[renormalize]

It turns out that this very integral was posted 5 days ago on Mathematics Stack Exchange by "Dan" (is that you @Meden Agan ?): https://math.stackexchange.com/ques...4-frac1-sqrtx-x2-arcsin-sqrt-fracx-1x-1x-sqrt
The first answer there by "user170231" provides an impressive proof in terms of dilogarithms.

 

[Meden Agan]

It turns out that this very integral was posted 5 days ago on Mathematics Stack Exchange by "Dan" (is that you @Meden Agan ?): https://math.stackexchange.com/ques...4-frac1-sqrtx-x2-arcsin-sqrt-fracx-1x-1x-sqrt
The first answer there by "user170231" provides an impressive proof in terms of dilogarithms.

Mhm... A delightful surprise! Most likely my instructor took that integral from there.
However, I would be interested to hear what we get with @fresh_42's method in post #24 :-)

 

[Meden Agan]

The question is really how to solve $\displaystyle{I=\int_0^a \operatorname{arcsin}\sqrt{f(\alpha)f(-\alpha)} \,d\alpha.}$

How about differentiation under integral sign or integral representation of $\arcsin$? What do you think?

 

[fresh_42]

How about differentiation under integral sign or integral representation of $\arcsin$? What do you think?

I think that there are nice possibilities to get rid of the arcus sine.

I used
\begin{align*}
I&=2\int_{0}^a \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=2\int_{0}^a\int_0^1 \sqrt{\dfrac{f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}\,dt\,d\alpha \\
&=2\int_{0}^a\int_0^1 \dfrac{1}{\sqrt{\left(f(\alpha)f(-\alpha)\right)^{-1}-t^2}}\,dt\,d\alpha\\
&=2\int_{0}^a\int_0^1\dfrac{1}{\sqrt{\dfrac{32(y+1)}{(y^2+2y-7)(y^2+7)}-t^2}}\,dt\,d\alpha
\end{align*}
where $a=\operatorname{arcsin}(1/\sqrt{8})\, , \,y^2=9-16 x\, , \,x=\sin^2(\alpha)$ and
$$
f(\alpha)f(-\alpha)=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}
$$
However, Fubini or not, it leads me directly into the same complicated - now polynomial - integrations as dealt with on MSE.

I have found another funny formula on Wikipedia that at least provide the additional $\pi$ and the logarithm which is apparently part of the deal:
$$
\operatorname{arcsin}R(x) = \int_0^1 \dfrac{1}{\pi \,t}\log\left(\dfrac{t^2+2R(x)t+1}{t^2-2R(x)t+1}\right) \;dt
$$

Looking at the math on MSE shows that there is no easy way to deal with the integrations, even without the trig function.

 

[Meden Agan]

I think that there are nice possibilities to get rid of the arcus sine.

I used
\begin{align*}
I&=2\int_{0}^a \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=2\int_{0}^a\int_0^1 \sqrt{\dfrac{f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}\,dt\,d\alpha \\
&=2\int_{0}^a\int_0^1 \dfrac{1}{\sqrt{\left(f(\alpha)f(-\alpha)\right)^{-1}-t^2}}\,dt\,d\alpha\\
&=2\int_{0}^a\int_0^1\dfrac{1}{\sqrt{\dfrac{32(y+1)}{(y^2+2y-7)(y^2+7)}-t^2}}\,dt\,d\alpha
\end{align*}
where $a=\operatorname{arcsin}(1/\sqrt{8})\, , \,y^2=9-16 x\, , \,x=\sin^2(\alpha)$ and
$$
f(\alpha)f(-\alpha)=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}
$$
However, Fubini or not, it leads me directly into the same complicated - now polynomial - integrations as dealt with on MSE.

This whole thing is frankly unmanageable. Perhaps changing the upper bound to the inner integral (by giving it as a function of $\alpha$)? I mean, maybe a different integral representation of $\arcsin$.
What do you think?

 

[fresh_42]

What do you think?

I think this:

This whole thing is frankly unmanageable.

The whole part is typical of trig integration problems. We have $(x-a)(x+a)$ or $x^2-2ax+c$ and $c$ is not $a^2,$ where we needed $(x-b)^2$ instead. Here we have this $x\sqrt{9-16x}$ term that spoils the party. There is a book Gradshteyn/Ryzhik "TABLE OF INTEGRALS, SERIES, AND PRODUCTS" with more than1200 pages. Maybe there could be found a trick to get rid of that inconvenient middle term.

I thought that perhaps the fact that $f(0)=1$ and $f(-\operatorname{arcsin}(1/\sqrt{8}))=0$ could help find a nice parameterization of the boundary curve that is nearly a circle.

 

[Meden Agan]

The whole part is typical of trig integration problems. We have $(x-a)(x+a)$ or $x^2-2ax+c$ and $c$ is not $a^2,$ where we needed $(x-b)^2$ instead. Here we have this $x\sqrt{9-16x}$ term that spoils the party. There is a book Gradshteyn/Ryzhik "TABLE OF INTEGRALS, SERIES, AND PRODUCTS" with more than1200 pages. Maybe there could be found a trick to get rid of that inconvenient middle term.

I thought that perhaps the fact that $f(0)=1$ and $f(-\operatorname{arcsin}(1/\sqrt{8}))=0$ could help find a nice parameterization of the boundary curve that is nearly a circle.

Totally agree. I have also tried to approach the integral with complex analysis, but all useless and monstrously meaningless algebra.

Trying to write $\arcsin \sqrt{f(\alpha)f(-\alpha)}$ in order to evaluate an integral, we can write
$${\huge \int}\limits_{\displaystyle 0}^{\sqrt{\dfrac{y^2+2y-7}{y+1}}} \frac{\sqrt{\dfrac{y^2+7}{32}}}{\sqrt{1-\dfrac{y^2+7}{32}t^2}} \, \mathrm dt.$$
Replacing $t \mapsto \sqrt{\dfrac{t^2+2t-7}{t+1}}$:
$$\frac{1}{2}{\huge \int}\limits_{\displaystyle 2\sqrt 2 -1}^{\displaystyle y} \frac{\sqrt{y^2+7}}{\sqrt{32 \,(t+1)-(y^2+7)(t^2+2t+9)}} \cdot {\color{red}{\frac{(t^2+2t+9)}{\sqrt{t^2+2t-7} \, (t+1)}}} \, \mathrm dt.$$ Then we can apply Fubini and interchange the order of integration. Thus, the red part can be carried out of the integral.
This whole thing is problematic for me. I have no idea how to tackle it. By converting $\arcsin$ into $\arctan$, the problem seems to be at least reduced.
Why is that, in your opinion?

 

[fresh_42]

I'm not sure. There are countless formulas involving both. They transform by an expression $x\to \dfrac{x}{\sqrt{1-x^2}}$ and $\operatorname{arctan} x = 2\operatorname{arctan} \dfrac{x}{1+\sqrt{1+x^2}},$ or $\operatorname{arcsin}\left(\dfrac{2x}{1+x^2}\right)=2\operatorname{arctan}(x).$ The book I referred to has 270 occurrences of $\operatorname{arcsin}.$ These transformations can help because they have a similar structure to the original polynomial. But if so, then $2\operatorname{arcsin}(x)=\operatorname{arcsin}\left(2x\sqrt{1-x^2}\right)$ could possibly help, too. It is all about getting rid of the root under the root. That's why I introduced $y^2=9-16x.$ At least, it resulted in a polynomial expression with only one big root left.

The second picture in post #24 shows the difference $A-C$ where $A$ is the argument $\sqrt{f(\alpha)f(-\alpha)}$ of the $\operatorname{arcsin}$ and $C$ a circle, an ellipse to be exact since I rescaled the $x$-axis to make it a circle. Anyway, it shows that the difference itself is already a weird curve, so any attempts to write $f(\alpha)f(-\alpha)$ as a square are doomed.

I hesitated to study the solution on MSE since it looked like a lot of additional scribbling to do if I wanted to fully understand it, and trying things myself is more fun. Did anyone but me notice that the area under the curve we are looking for (image in post #22) looks like the tower of a Russian church?

Here is my official monster polynomial:
$$
I=2\int_0^{-a}\int_0^1 \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\,dt\,dy
$$
The minus sign came from switching from $d\alpha$ to $dy.$

 

[Meden Agan]

Here is my official monster polynomial:
$$
I=2\int_0^{-a}\int_0^1 \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\,dt\,dy
$$
The minus sign came from switching from $d\alpha$ to $dy.$

Mhm, unfortunately it doesn't work.
I fed WolframAlpha with that integral here, and returned a completely different value than expected (complex valued).
Are you sure the integrand and the bounds of outer integral are correct?

 

[fresh_42]

Mhm, unfortunately it doesn't work.
I fed WolframAlpha with that integral here, and returned a completely different value than expected (complex valued).
Are you sure the integrand and the bounds of outer integral are correct?

You are right. I made the standard mistake and forgot to adjust the limits when changing from $d\alpha$ to $dy.$ Here are the correct bounds:
$$
I=2\int_{ \sqrt{ 9-2\sqrt{8} } }^{3}\int_0^1 \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\,dt\,dy
$$
$y=3$ leads to a zero in the denomiator which we don't have in the former expression
$$
I=2\int_{0}^a\int_0^1 \sqrt{\dfrac{f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}\,dt\,d\alpha
$$
so I'm not sure whether this can be handled. I used $\cos(\operatorname{arcsin}(u))=\sqrt{1-u^2}$ which produced that zero.

 

[Meden Agan]

$$
I=2\int_{ \sqrt{ 9-2\sqrt{8} } }^{3}\int_0^1 \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\,dt\,dy
$$

Still not working, see here. The numerical value should be $\dfrac{\pi^2}{16} \approx 0.6168{\color{red}{50}}...$, different from $0.6168{\color{red}{68}}$.
Something must be wrong with the integrand.

 

[fresh_42]

Still not working, see here. The numerical value should be $\dfrac{\pi^2}{16} \approx 0.6168{\color{red}50}...$, different from $0.6168{\color{red}68}$.
Something must be wrong with the integrand.

I'd rather assume that this tiny error is due to the singularity. If you want to check what I've done, here is it:

The polynomial:
\begin{align*}
y^2&=y(\alpha)^2=9-16\sin^2(\alpha)=9-16x\\
x-1&=-\dfrac{7+y^2}{16}\\
1-2x&=\dfrac{y^2-1}{8}
\end{align*}
\begin{align*}
f(\alpha)f(-\alpha)&=\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}\\
&=\dfrac{1}{32}\dfrac{(-7-y^2)((-7-y^2)+(9-y^2)y)}{y^2-1}\\
&=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y-1)(y^2+7)}{y^2-1}\\
&=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}
\end{align*}

The integral:
$a=\operatorname{arcsin}(1/\sqrt[4]{8})$
\begin{align*}
I&=2\int_{0}^a \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=2\int_{0}^a\int_0^1 \sqrt{\dfrac{f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}\,dt\,d\alpha
\end{align*}
\begin{align*}
\sqrt{\dfrac{ f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}&=\sqrt{
\dfrac{ 2^{-5}\dfrac{(y^2+2y-7)(y^2+7) }{y+1} }{ \dfrac{y+1}{y+1} - 2^{-5}\dfrac{(y^2+2y-7)(y^2+7)t^2}{y+1} }}\\
&=\sqrt{\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)-(y^2+2y-7)(y^2+7)t^2}}\\
&=\dfrac{1}{\sqrt{\dfrac{32(y+1)}{(y^2+2y-7)(y^2+7)}-t^2}}\\
&=\sqrt{\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)-t^2(y^2+2y-7)(y^2+7)}}\\
I&=2\int_0^a\int_0^1\sqrt{\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)-t^2(y^2+2y-7)(y^2+7)}}\,dt\,d\alpha
\end{align*}
\begin{align*}
y^2&=9-16\sin^2(\alpha)\\
\sqrt{\dfrac{9-y^2}{16}}&=\sin(\alpha)\,,\,\alpha=\operatorname{arcsin}\left(\sqrt{\dfrac{9-y^2}{16}}\right)\\
\alpha&=0 \longrightarrow y=3\, , \,\alpha=a\longrightarrow y=\sqrt{9-\dfrac{16}{\sqrt{8}}}=\sqrt{9-2\sqrt{8}}\\
\dfrac{dy}{d\alpha}&=-\dfrac{16\sin(\alpha)\cos(\alpha)}{y}
=-\dfrac{\sqrt{144-16y^2}}{y}\cos\left(\operatorname{arcsin}\left(\sqrt{\dfrac{9-y^2}{16}}\right)\right)\\
\dfrac{dy}{d\alpha}&=-\dfrac{\sqrt{144-16y^2}}{y}\sqrt{1-\dfrac{9-y^2}{16}}=-\dfrac{\sqrt{144-16y^2}}{y}\sqrt{\dfrac{y^2+7}{16}}\\
\dfrac{dy}{d\alpha}&=-\dfrac{\sqrt{(9-y^2)(y^2+7)}}{y}\\
d\alpha&=-\dfrac{y\,dy}{\sqrt{(9-y^2)(y^2+7)}}
=-\sqrt{\dfrac{y^2}{(9-y^2)(y^2+7)}}\,dy
\end{align*}
$$
I=2\int_{\sqrt{9-2\sqrt{8}}}^{3}\int_0^1 \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\,dt\,dy
$$

 

[Meden Agan]

I'd rather assume that this tiny error is due to the singularity. If you want to check what I've done, here is it:

The polynomial:
\begin{align*}
y^2&=y(\alpha)^2=9-16\sin^2(\alpha)=9-16x\\
x-1&=-\dfrac{7+y^2}{16}\\
1-2x&=\dfrac{y^2-1}{8}
\end{align*}
\begin{align*}
f(\alpha)f(-\alpha)&=\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}\\
&=\dfrac{1}{32}\dfrac{(-7-y^2)((-7-y^2)+(9-y^2)y)}{y^2-1}\\
&=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y-1)(y^2+7)}{y^2-1}\\
&=\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}
\end{align*}

The integral:
$a=\operatorname{arcsin}(1/\sqrt[4]{8})$
\begin{align*}
I&=2\int_{0}^a \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=2\int_{0}^a\int_0^1 \sqrt{\dfrac{f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}\,dt\,d\alpha
\end{align*}
\begin{align*}
\sqrt{\dfrac{ f(\alpha)f(-\alpha)}{1-f(\alpha)f(-\alpha)t^2}}&=\sqrt{
\dfrac{ 2^{-5}\dfrac{(y^2+2y-7)(y^2+7) }{y+1} }{ \dfrac{y+1}{y+1} - 2^{-5}\dfrac{(y^2+2y-7)(y^2+7)t^2}{y+1} }}\\
&=\sqrt{\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)-(y^2+2y-7)(y^2+7)t^2}}\\
&=\dfrac{1}{\sqrt{\dfrac{32(y+1)}{(y^2+2y-7)(y^2+7)}-t^2}}\\
&=\sqrt{\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)-t^2(y^2+2y-7)(y^2+7)}}\\
I&=2\int_0^a\int_0^1\sqrt{\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)-t^2(y^2+2y-7)(y^2+7)}}\,dt\,d\alpha
\end{align*}
\begin{align*}
y^2&=9-16\sin^2(\alpha)\\
\sqrt{\dfrac{9-y^2}{16}}&=\sin(\alpha)\,,\,\alpha=\operatorname{arcsin}\left(\sqrt{\dfrac{9-y^2}{16}}\right)\\
\alpha&=0 \longrightarrow y=3\, , \,\alpha=a\longrightarrow y=\sqrt{9-\dfrac{16}{\sqrt{8}}}=\sqrt{9-2\sqrt{8}}\\
\dfrac{dy}{d\alpha}&=-\dfrac{16\sin(\alpha)\cos(\alpha)}{y}
=-\dfrac{\sqrt{144-16y^2}}{y}\cos\left(\operatorname{arcsin}\left(\sqrt{\dfrac{9-y^2}{16}}\right)\right)\\
\dfrac{dy}{d\alpha}&=-\dfrac{\sqrt{144-16y^2}}{y}\sqrt{1-\dfrac{9-y^2}{16}}=-\dfrac{\sqrt{144-16y^2}}{y}\sqrt{\dfrac{y^2+7}{16}}\\
\dfrac{dy}{d\alpha}&=-\dfrac{\sqrt{(9-y^2)(y^2+7)}}{y}\\
d\alpha&=-\dfrac{y\,dy}{\sqrt{(9-y^2)(y^2+7)}}
=-\sqrt{\dfrac{y^2}{(9-y^2)(y^2+7)}}\,dy
\end{align*}
$$
I=2\int_{\sqrt{9-2\sqrt{8}}}^{3}\int_0^1 \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\,dt\,dy
$$

Seems all fine to me.
The inner integral can be evaluated easily, can't it? It comes out to be a logarithmic form, if I'm right.
As per indefinite integral, I obtain:
$${\huge \int} \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\, \mathrm dt = \frac{|y|}{\sqrt{(y^2-9)(y^2+7)}} \ln \left(\left|\sqrt{-t^2\frac{(y^2+7)(y^2+2y-7))}{32 \, (y+1)}+1} + t \sqrt{-\frac{(y^2+7)(y^2+2y-7))}{32 \, (y+1)}}\right|\right) + C \qquad C \in \mathbb R.$$
Do you confirm?

 

[fresh_42]

Seems all fine to me.
The inner integral can be evaluated easily, can't it? It comes out to be a logarithmic form, if I'm right.
As per indefinite integral, I obtain:
$${\huge \int} \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\, \mathrm dt = \frac{|y|}{\sqrt{(y^2-9)(y^2+7)}} \ln \left(\left|\sqrt{-t^2\frac{(y^2+7)(y^2+2y-7))}{32 \, (y+1)}+1} + t \sqrt{-\frac{(y^2+7)(y^2+2y-7))}{32 \, (y+1)}}\right|\right) + C \qquad C \in \mathbb R.$$
Do you confirm?

Unfortunately, WA spits out an $\operatorname{arctan}$ or, likewise, a complex logarithm, so this will take me a while to differentiate it. Did you use substitutions to shorten that procedure?

I think I'll start with the arcus tangent and see where it leads me.

 

[Meden Agan]

Did you use substitutions to shorten that procedure?

I fed this site with the integral. I put $$\sqrt{\frac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}$$ into the bar and changed the integration variable from $x$ to $t$.
You can see substitutions by clicking on the “Show steps” button.

 

[fresh_42]

My result was way more complicated. I didn't differentiate yours but used WA for the easy integration and got
$$I=\int \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\, dt=\int \sqrt{\dfrac{A}{B-Ct^2}}\,dt $$
WA calculated this to
$$
I=\sqrt{\dfrac{A}{C}}\operatorname{arctan} \left( \dfrac{t\sqrt{C}}{\sqrt{B-Ct^2}} \right)
$$
which I verified by differentiation, also on WA. Then I used
$$
\operatorname{arctan}(w)=\dfrac{1}{2i}\log \dfrac{1+iw}{1-iw}
$$
to turn it into a logarithm. That gave me
\begin{align*}
I&=
\int \sqrt{\dfrac{A}{B-Ct^2}}\,dt = \sqrt{\dfrac{A}{C}}\operatorname{arctan} \left( \dfrac{t\sqrt{C}}{\sqrt{B-Ct^2}} \right)\\
&=\dfrac{1}{2i}\sqrt{\dfrac{A}{C}}\log\left(\dfrac{1+i\dfrac{t\sqrt{C}}{\sqrt{B-Ct^2}}}{1-i\dfrac{t\sqrt{C}}{\sqrt{B-Ct^2}}}\right)\\[6pt]
&=\sqrt{\dfrac{A}{-4C}}\log\left(\dfrac{\sqrt{B-Ct^2}+t\sqrt{-C}}{\sqrt{B-Ct^2}-t\sqrt{-C}}\right)=\sqrt{\dfrac{A}{-4C}}\log\left(\dfrac{(\sqrt{B-Ct^2}+t\sqrt{-C})^2}{B}\right)
\end{align*}
The leading factor is
$$
\sqrt{\dfrac{A}{-4C}}=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}
$$
the same as yours. The factor $1/2$ can be turned into a root of the logarithm argument, but I preferred to have the easy factor instead of the root.

Now came the ugly part, and I'm not sure I haven't made any copy+paste errors.
\begin{align*}
\dfrac{(\sqrt{B-Ct^2}+t\sqrt{-C})^2}{B}&=\dfrac{B-Ct^2+2t(-BC+C^2t^2)-Ct^2}{B}=1-2t^2 C+2t\dfrac{C^2t^2-BC}{B}
\end{align*}
Re-substitution $B=32(y+1)(9-y^2)$ and $C=(y^2+2y-7)(y^2+7)(9-y^2)$ gave me
\begin{align*}
I&=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\quad\cdot\\
&\cdot\log(1-2t^2\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)}+\ldots\\
&\ldots + 2t\dfrac{\sqrt{(y^2+2y-7)(y^2+7)} }{32(y+1)}\sqrt{t^2(y^2+2y-7)(y^2+7)-32(y+1)})
\end{align*}

Given that I made no mistakes, esp. copy+paste or sign errors, I really hope that the integration bounds $[0,1]$ for $t$ simplify it before the other integration starts.

 

[fresh_42]

I think there is an error in my calculation since the logarithm isn't defined on the domain we need for $y.$

 

[Meden Agan]

I think there is an error in my calculation since the logarithm isn't defined on the domain we need for $y.$

Yes, double-check this step:

\begin{align*}
\dfrac{(\sqrt{B-Ct^2}+t\sqrt{-C})^2}{B}&=\dfrac{B-Ct^2+2t(-BC+C^2t^2)-Ct^2}{B}=1-2t^2 C+2t\dfrac{C^2t^2-BC}{B}
\end{align*}

 

[fresh_42]

Yes, double-check this step:

Yes, I forgot the root, but only here. My calculation has it. However, I ended up with
\begin{align*}
J&=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\cdot\log(1-2t^2\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)}+\ldots\\
&\ldots + 2t\dfrac{\sqrt{(y^2+2y-7)(y^2+7)} }{32(y+1)}\sqrt{t^2(y^2+2y-7)(y^2+7)-32(y+1)})
\end{align*}
for the new integrand. That is
\begin{align*}
J(1)-J(0)&=J(1)=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\cdot \log(1-\dfrac{(y^2+2y-7)(y^2+7)}{16(y+1)}+\ldots\\
&\ldots \dfrac{1}{16(y+1)}\sqrt{(y^2+2y-7)^2(y^2+7)^2-32(y+1)(y^2+2y-7)(y^2+7)}\\
&=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\cdot \left[\log(D)-\log(16(y+1))\right]
\end{align*}
with
$$
D=16(y+1)-(y^2+2y-7)(y^2+7)+\sqrt{(y^2+2y-7)^2(y^2+7)^2-32(y+1)(y^2+2y-7)(y^2+7)}
$$
This is horrible to integrate. And I think $\log(D)$ isn't defined on $[\sqrt{9-2\sqrt{8}},3].$ This means we cannot pretend as if the complex logarithm were real by drawing $i$ under the root.

I wish I could understand where your relatively simple logarithm argument came from. Guess, I'll bite into the sour apple as we say here and differentiate it to show at least that it is correct.

 

[Meden Agan]

Yes, I forgot the root, but only here. My calculation has it. However, I ended up with
\begin{align*}
J&=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\cdot\log(1-2t^2\dfrac{(y^2+2y-7)(y^2+7)}{32(y+1)}+\ldots\\
&\ldots + 2t\dfrac{\sqrt{(y^2+2y-7)(y^2+7)} }{32(y+1)}\sqrt{t^2(y^2+2y-7)(y^2+7)-32(y+1)})
\end{align*}
for the new integrand. That is
\begin{align*}
J(1)-J(0)&=J(1)=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\cdot \log(1-\dfrac{(y^2+2y-7)(y^2+7)}{16(y+1)}+\ldots\\
&\ldots \dfrac{1}{16(y+1)}\sqrt{(y^2+2y-7)^2(y^2+7)^2-32(y+1)(y^2+2y-7)(y^2+7)}\\
&=\dfrac{1}{2}\sqrt{\dfrac{y^2}{(y^2+7)(y^2-9)}}\cdot \left[\log(D)-\log(16(y+1))\right]
\end{align*}
with
$$
D=16(y+1)-(y^2+2y-7)(y^2+7)+\sqrt{(y^2+2y-7)^2(y^2+7)^2-32(y+1)(y^2+2y-7)(y^2+7)}
$$

Yes, that's what I get.

I fed WolframAlpha with the integral here, and returned the value $\dfrac{\pi^2}{8} \approx 1.2337...$ (even though with a minus sign, why?).
However, how on earth do we integrate this? Is there a human way?

 

[fresh_42]

Seems all fine to me.
The inner integral can be evaluated easily, can't it? It comes out to be a logarithmic form, if I'm right.
As per indefinite integral, I obtain:
∫y2(y2+2y−7)32(y+1)(9−y2)−t2(y2+2y−7)(y2+7)(9−y2)dt=|y|(y2−9)(y2+7)ln⁡(|−t2(y2+7)(y2+2y−7))32(y+1)+1+t−(y2+7)(y2+2y−7))32(y+1)|)+CC∈R.
Do you confirm?

Confirmed!

Spoiler

\begin{align*}
&\int \sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}\, dt \\&= \frac{|y|}{\sqrt{(y^2-9)(y^2+7)}} \log \left(\left|\sqrt{-t^2\frac{(y^2+7)(y^2+2y-7))}{32 \, (y+1)}+1} + t \sqrt{-\frac{(y^2+7)(y^2+2y-7))}{32 \, (y+1)}}\right|\right) + C \qquad C
\end{align*}
Set $p(y)=\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}$ and $q(y)=\sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)} }.$
\begin{align*}
\dfrac{d}{dt}p(y)&\log\left(\left|\sqrt{1+t^2q(y)^2}+tq(y)\right|\right)=p(y)\dfrac{\dfrac{d}{dt}\sqrt{1+t^2q(y)^2}+ \dfrac{d}{dt}tq(y)}{\sqrt{1+t^2q(y)^2}+tq(y)}\\[12pt]
&=\dfrac{q(y)p(y)}{\sqrt{1+t^2q(y)^2}+tq(y)}\left[1+\dfrac{tq(y)}{\sqrt{1+t^2q(y)^2}}\right]\\[12pt]
&=\dfrac{p(y)q(y)}{\sqrt{1+t^2q(y)^2}+tq(y)}\dfrac{\sqrt{1+t^2q(y)^2}+tq(y)}{\sqrt{1+t^2q(y)^2}}=\dfrac{p(y)q(y)}{\sqrt{1+t^2q(y)^2}}\\[12pt]
p(y)q(y)&=\sqrt{\dfrac{y^2}{(9-y^2)}\dfrac{(y^2+2y-7)}{32(y+1)}}=
\sqrt{\dfrac{y^2(y^2+2y-7)}{(9-y^2) 32(y+1)}}
\\[12pt]
\sqrt{1+t^2q(y)^2}&=\sqrt{1-t^2\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)}}=\sqrt{\dfrac{32(y+1)-t^2(y^2+7)(y^2+2y-7)}{32(y+1)}}\\[12pt]
\dfrac{1}{\sqrt{1+t^2q(y)^2}}&=\sqrt{\dfrac{32(y+1)}{32(y+1)-t^2(y^2+7)(y^2+2y-7)}}\\[12pt]
\dfrac{p(y)q(y)}{\sqrt{1+t^2q(y)^2}}&=\sqrt{\dfrac{1}{32(y+1)-t^2(y^2+7)(y^2+2y-7)} \cdot \dfrac{y^2(y^2+2y-7)}{(9-y^2) } }\\[12pt]
&=\sqrt{\dfrac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+7)(y^2+2y-7)(9-y^2)}}
\end{align*}

It is really too hot here to differentiate ...

It means we are now at
\begin{align*}
I&=2\int_{\sqrt{9-2\sqrt{8}}}^3 \left(p(y)\log\left|\sqrt{1+q(y)^2}+q(y)\right|\right) \,dy
\end{align*}
with $p(y)=\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}$ and $q(y)=\sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)} }.$

 

[fresh_42]

Yes, that's what I get.

I fed WolframAlpha with the integral here, and returned the value $\dfrac{\pi^2}{8} \approx 1.2337...$ (even though with a minus sign, why?).

The minus sign comes from the re-substitution of $\alpha$ to $y.$

However, how on earth do we integrate this? Is there a human way?

No idea. Meanwhile, I confirmed your formula, which looks much more promising. I already deleted these monster calculations from my scribblings. It would probably take dozens of substitutions.

 

[Meden Agan]

It means we are now at
\begin{align*}
I&=2\int_{\sqrt{9-2\sqrt{8}}}^3 \left(p(y)\log\left|\sqrt{1+q(y)^2}+q(y)\right|\right) \,dy
\end{align*}
with $p(y)=\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}$ and $q(y)=\sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)} }.$

All I can say here is $\log \left(\sqrt{1+q(y)^2}+q(y)\right) = \sinh^{-1} \left(q(y)\right)$. I'm not even sure it is helpful.
After that, I have no idea.

 

[fresh_42]

All I can say here is $\log \left(\sqrt{1+q(y)^2}+q(y)\right) = \sinh^{-1} \left(q(y)\right)$. I'm not even sure it is helpful.
After that, I have no idea.

I don't think that returning to $\operatorname{arcsin}$ would be an improvement. My formula is awkward if you ask me, even if it should basically be doable.

I like the simplicity in your expression and $\int \log(\sqrt{1+x^2}+x)\,dx$ can be done. So how to deal with $q(y)$ is the question. The rest looks like integration by parts.

 

[Meden Agan]

I like the simplicity in your expression and $\int (\sqrt{1+x^2})+x\,dx$ can be done. So how to deal with $q(y)$ is the question. The rest looks like integration by parts.

Hope you'll be able to outline how to go on. For now I have no idea how to do it.