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

[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.

 

[fresh_42]

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

Here is my plan:

\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\\
&=2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3} \underbrace{\left[\dfrac{p(y)}{q'(y)}\right]}_{=u}\underbrace{\log\left|\sqrt{1+z^2}+z\right|}_{=v'}\,dz\\
&=2\left[\dfrac{p(y)}{q'(y)}\cdot\left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\qquad\qquad \longrightarrow C\\
&- 2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left( \left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\dfrac{d}{dz}\dfrac{p(y)}{q'(y)}\right)\,dz\\
&=C+2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left(\sqrt{z^2+1}\dfrac{d}{dz}\dfrac{p(y)}{q'(y)}\,dz\right)-2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}z\log\left(\sqrt{z^2+1}+z\right)\,dz
\end{align*}

The second integral has a closed form; $C$ can be calculated, so we are left with the first integral. I carried the variable within the boundaries to avoid the mistake of not adjusting them. $q'=dq(y)/dy=dz/dy.$

But I admit that this might again result in an inconvenient polynomial.

 

[Meden Agan]

Here is my plan:

\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\\
&=2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3} \underbrace{\left[\dfrac{p(y)}{q'(y)}\right]}_{=u}\underbrace{\log\left|\sqrt{1+z^2}+z\right|}_{=v'}\,dz\\
&=2\left[\dfrac{p(y)}{q'(y)}\cdot\left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\qquad\qquad \longrightarrow C\\
&- 2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left( \left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\dfrac{d}{dz}\dfrac{p(y)}{q'(y)}\right)\,dz\\
&=C+2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left(\sqrt{z^2+1}\dfrac{d}{dz}\dfrac{p(y)}{q'(y)}\,dz\right)-2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}z\log\left(\sqrt{z^2+1}+z\right)\,dz
\end{align*}

The second integral has a closed form; $C$ can be calculated, so we are left with the first integral. I carried the variable within the boundaries to avoid the mistake of not adjusting them. $q'=dq(y)/dy=dz/dy.$

But I admit that this might again result in an inconvenient polynomial.

Mhm, there are some major problems.

1) Do we care that $p(y)$ is complex when $y \in (-3,3)$?
2) When $y=3$, $z= i$. So we'd have a complex term as bound.
3) How do we evaluate $\dfrac{\mathrm d}{\mathrm dz}\dfrac{p(y)}{q'(y)}$?

 

[fresh_42]

Mhm, there are some major problems.

1) Do we care that $p(y)$ is complex when $y \in (-3,3)$?
2) When $y=3$, $z= i$. So we'd have a complex term as bound.
3) How do we evaluate $\dfrac{\mathrm d}{\mathrm dz}\dfrac{p(y)}{q'(y)}$?

1) It was your integration that led to the minus sign. That also happens when we substitute the arctangent with a logarithm. I was hoping that the imaginary parts would cancel if we evaluated the integral. E.g., $p(y)/q'(y)$ is defined on the small interval we need, $y\in \left[\sqrt{9-2\sqrt{8}},3\right].$
2) Same. Let's first see where we end up.
3) Differentiation is easy. Although the polynomials I get are terrible. I calculated
$$
\dfrac{p(y)}{(q'(y))^2}=-128\sqrt{\dfrac{y^2(y+1)^6 (y^2+7)(y^2+2y-7)^2}{(3y^4+8y^3+6y^2+63)^4(y^2-9)}}
$$
Happy differentiation! And the factor $q'(y)\sqrt{1+q(y)^2}$ doesn't look much better. As usual, you start hoping that the awkward terms cancel out somewhere, but instead, they get more and more.

 

[Meden Agan]

3) Differentiation is easy. Although the polynomials I get are terrible. I calculated
$$
\dfrac{p(y)}{(q'(y))^2}=-128\sqrt{\dfrac{y^2(y+1)^6 (y^2+7)(y^2+2y-7)^2}{(3y^4+8y^3+6y^2+63)^4(y^2-9)}}
$$
Happy differentiation! And the factor $q'(y)\sqrt{1+q(y)^2}$ doesn't look much better. As usual, you start hoping that the awkward terms cancel out somewhere, but instead, they get more and more.

Shouldn't we express this as a function of $z$? How do we differentiate an expression in $y$ with respect to the variable $z$?

 

[fresh_42]

Shouldn't we express this as a function of $z$? How do we differentiate an expression in $y$ with respect to the variable $z$?

I only needed $z$ as an intermediate variable. Expressing terms like $p(y)/q'(y)$ by $z=q(y)$ is almost impossible. So I integrated by $z$ but kept the main variable $y.$ We have
$$q'(y)=\dfrac{d}{dy}q(y)=\dfrac{dz}{dy}$$
so we can switch back and forth as long as we keep track of the variables. The downside is that we get additional factors $q'(y)$, however, still better than pressing $z$ into $p(y).$ (I think.)

State of the art:
\begin{align*}
I&=2\left[\dfrac{p(y)}{q'(y)}\cdot\left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3} - \ldots\\
&\phantom{=}\dfrac{1}{2}\left[\left(-z\sqrt{z^2+1}+2z^2\log\left(\sqrt{z^2+1}+z\right)\right)+\operatorname{arcsinh}(z)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3}+ \ldots\\
&\phantom{=}2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\sqrt{q(y)^2+1}\left(\dfrac{d}{dy}\dfrac{p(y)}{q'(y)}\right)\,dy
\end{align*}
where
$$
\sqrt{q(y)^2+1}=\dfrac{1}{4\sqrt{2}}\sqrt{ \dfrac{-y^4-2y^3+18y+81}{y+1}}
$$
and
$$
\dfrac{d}{dy}\dfrac{p(y)}{q'(y)}==-8\sqrt{2}\dfrac{d}{dy}\sqrt{\dfrac{y^2(y+1)^3 (y^2+2y-7)}{(3y^4+8y^3+6y^2+63)^2(9-y^2)}}
$$

So basically solved ... uhm ... in principle ... uhm

Edit (in order to calculate the constants):
$$
z=q(y)=\dfrac{1}{4\sqrt{2}}\sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{y+1}}
$$
and
$$
\dfrac{p(y)}{q'(y)}=-8\sqrt{2}\sqrt{\dfrac{y^2(y+1)^3 (y^2+2y-7)}{(3y^4+8y^3+6y^2+63)^2(9-y^2)}}
$$

Edit: Calculation error corrected.

 

[Meden Agan]

I only needed $z$ as an intermediate variable. Expressing terms like $p(y)/q'(y)$ by $z=q(y)$ is almost impossible. So I integrated by $z$ but kept the main variable $y.$ We have
$$q'(y)=\dfrac{d}{dy}q(y)=\dfrac{dz}{dy}$$
so we can switch back and forth as long as we keep track of the variables. The downside is that we get additional factors $q'(y)$, however, still better than pressing $z$ into $p(y).$ (I think.)

State of the art:
\begin{align*}
I&=2\left[\dfrac{p(y)}{q'(y)}\cdot\left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3} - \ldots\\
&\phantom{=}\dfrac{1}{2}\left[\left(-z\sqrt{z^2+1}+2z^2\log\left(\sqrt{z^2+1}+z\right)\right)+\operatorname{arcsinh}(z)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3}+ \ldots\\
&\phantom{=}2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\sqrt{q(y)^2+1}q'(y)\left(\dfrac{d}{dy}\dfrac{p(y)}{(q'(y))^2}\right)\,dy
\end{align*}
where
$$
\sqrt{q(y)^2+1}q'(y)=-\dfrac{1}{64}\sqrt{\dfrac{(y^4+2y^3+46y-17)(3y^4+8y^3+6y^2+63)^2}{-(y+1)^4(y^2+7)(y^2+2y-7)}}
$$
and
$$
\dfrac{d}{dy}\dfrac{p(y)}{(q'(y))^2}=-128\dfrac{d}{dy} \left(\sqrt{\dfrac{y^2(y+1)^6 (y^2+7)(y^2+2y-7)^2}{(3y^4+8y^3+6y^2+63)^4(y^2-9)}}\right)
$$

So basically solved ... uhm ... in principle ... uhm

All correct, but it's almost impossible to integrate all that stuff. Take a look here.

How about using the function $\sinh^{-1}$ instead of the logarithm? Maybe something gets simplified?

 

[fresh_42]

All correct, but it's almost impossible to integrate all that stuff. Take a look here.

I think it can be simplified in a better way by sorting the factor polynomials. But I agree, it doesn't look good.

We have a circle with a flat negative and irregular bump. That's the problem: to find a nice parameterization of that buckle.

How about using the function $\sinh^{-1}$ instead of the logarithm? Maybe something gets simplified?

Well, simpler than $I=\displaystyle{\int_{-a}^a \operatorname{arcsin}\sqrt{f(\alpha)f(-\alpha)}\,d\alpha}$ is hardly possible. This leads into the world of trig functions instead of polynomials. The function was
$$
f(\alpha)=\dfrac{\cos(\alpha)+\sqrt{y(\alpha)}\sin(\alpha)}{1+\tan(\alpha)}
$$

$\sinh^{-1}$ helps calculating the constants in my expression. The integration should be the same.

Maybe we should give it a try! $\sinh^{-1}$ plus integration by parts.

 

[Meden Agan]

Maybe we should give it a try! $\sinh^{-1}$ plus integration by parts.

Does that simplify anything? I have the feeling we are left with the same old disgusting polynomials.

 

[fresh_42]

Does that simplify anything? I have the feeling we are left with the same old disgusting polynomials.

I made a mistake and corrected it. That simplifies the formulas a bit. Integration by parts with $\operatorname{arsinh}$ has always two possibilities, depending on which factor is which in the formula. One led to an $\operatorname{arsinh}$ with an even nastier factor to integrate, so I changed the roles.

I have three integrations pending. That will take a while. But
$$
I=2\int_{\sqrt{9-2\sqrt{8}}}^3\underbrace{\operatorname{arsinh}(z)}_{=u}\underbrace{\dfrac{p(y)}{q'(y)}}_{=v'}\,dz =[uv|_a^b -\int_a^b u'v
$$
could work. At least the polynomials aren't quite as nasty.

 

[fresh_42]

Ignoring the singularities and without a good idea to solve the remaining integral, I arrived at:
\begin{align*}
I&=-\left[\operatorname{arsinh}\sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)}}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\right]_{\sqrt{9-2\sqrt{8}}}^3+\ldots\\
&+\dfrac{1}{4\sqrt{2}}\int_{\sqrt{9-2\sqrt{8}}}^{3}\sqrt{\dfrac{-y^4-2y^3+18y+81}{y+1}}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\,dy
\end{align*}
with your idea of using $\operatorname{arsinh}$ and a suitable integration by parts. At least an improvement compared with those polynomials before.