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

[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.

 

[Meden Agan]

$$\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)$$

How did you obtain that?

 

[Meden Agan]

I did WA make the integral.

\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_{\sqrt{9-2\sqrt{8}}}^3 p(y)\operatorname{arsinh}(q(y))\,dy
\end{align*}
\begin{align*}
\int_a^b uv'&= uv|_a^b -\int_a^b u'v\, , \,
z=q(y)\, , \,\dfrac{dz}{dy}=q'\, , \,dy=\dfrac{dz}{q'}
\end{align*}
\begin{align*}
v&=\int_{\sqrt{9-2\sqrt{8}}}^3\dfrac{p(y)}{q'(y)}\,dz=
\int_{\sqrt{9-2\sqrt{8}}}^3\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}\,dy\\
v&=-\dfrac{1}{2}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\\
u'&=\dfrac{d}{dz}\operatorname{arsinh}(z)=\dfrac{1}{\sqrt{1+q(y)^2}} =4\sqrt{2}\sqrt{ \dfrac{y+1}{-y^4-2y^3+18y+81}}\\
\end{align*}
\begin{align*}
I&=2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\underbrace{\operatorname{arsinh}(z)}_{=u}\underbrace{\dfrac{p(y)}{q'(y)}}_{=v'}\,dz\\
&=-\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\\
&+\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\dfrac{\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)}{\dfrac{1}{\sqrt{1+q(y)^2}}}\,dy\\
&=-\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*}

All correct, as far as I'm concerned.
Previously, I got confused because I had the variables $y$ and $z$ interchanged.

Now, do you have any ideas on how to tackle the remaining integral?

 

[fresh_42]

How did you obtain that?

The logarithm was obtained from WA. The entire calculation will follow. I think I made an error. Let me check this first.

 

[fresh_42]

Here is my correction. It isn't quite so nice with the corrected factor for
$$
u'=\dfrac{du}{dz}=\dfrac{d}{dz}\operatorname{arsinh}(z)=\dfrac{1}{\sqrt{1+z^2}}=\dfrac{1}{\sqrt{1+q(y)}}
$$
and therefore
$$
u'v\,dz = \dfrac{q'(y)}{\sqrt{1+q(y)^2}}\cdot v \,dy
$$
All in all:

\begin{align*}
\int_a^b uv'&= uv|_a^b -\int_a^b u'v\, , \,
z=q(y)\, , \,\dfrac{dz}{dy}=q'\, , \,dy=\dfrac{dz}{q'}
\end{align*}
\begin{align*}
v&=\int_{\sqrt{9-2\sqrt{8}}}^3\dfrac{p(y)}{q'(y)}\,dz=
\int_{\sqrt{9-2\sqrt{8}}}^3\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}\,dy\\
v&=-\dfrac{1}{2}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\\
u'&=\dfrac{d}{dz}\operatorname{arsinh}(z)=\dfrac{1}{\sqrt{1+q(y)^2}} =4\sqrt{2}\sqrt{ \dfrac{y+1}{-y^4-2y^3+18y+81}}\\
\end{align*}
\begin{align*}
q'(y)&=-\dfrac{1}{8\sqrt{2}}\sqrt{\dfrac{(3y^4+8y^3+6y^2+63)^2}{-(y^2+7)(y^2+2y-7)(y+1)^3}} \\
\sqrt{1+q(y)^2}&=\dfrac{1}{4\sqrt{2}} \sqrt{\dfrac{-y^4-2y^3+18y+81}{y+1}}\\
\dfrac{q'(y)}{\sqrt{1+q(y)^2}}&=-\dfrac{1}{2}\sqrt{
\dfrac{(3y^4+8y^3+6y^2+63)^2}{(y^2+7)(y^2+2y-7)(y+1)^2(y^4+2y^3-18y-81)}}
\end{align*}
\begin{align*}
I&=2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\underbrace{\operatorname{arsinh}(z)}_{=u}\underbrace{\dfrac{p(y)}{q'(y)}}_{=v'}\,dz\\
&=-\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\\
&+\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\dfrac{q'(y)}{\sqrt{1+q(y)^2}}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\,dy
\end{align*}

It is in principle solvable by integration by parts, although definitely very inconvenient, especially during the heat wave we currently have here.

 

[Meden Agan]

Here is my correction. It isn't quite so nice with the corrected factor for
$$
u'=\dfrac{du}{dz}=\dfrac{d}{dz}\operatorname{arsinh}(z)=\dfrac{1}{\sqrt{1+z^2}}=\dfrac{1}{\sqrt{1+q(y)}}
$$
and therefore
$$
u'v\,dz = \dfrac{q'(y)}{\sqrt{1+q(y)^2}}\cdot v \,dy
$$
All in all:

\begin{align*}
\int_a^b uv'&= uv|_a^b -\int_a^b u'v\, , \,
z=q(y)\, , \,\dfrac{dz}{dy}=q'\, , \,dy=\dfrac{dz}{q'}
\end{align*}
\begin{align*}
v&=\int_{\sqrt{9-2\sqrt{8}}}^3\dfrac{p(y)}{q'(y)}\,dz=
\int_{\sqrt{9-2\sqrt{8}}}^3\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}\,dy\\
v&=-\dfrac{1}{2}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\\
u'&=\dfrac{d}{dz}\operatorname{arsinh}(z)=\dfrac{1}{\sqrt{1+q(y)^2}} =4\sqrt{2}\sqrt{ \dfrac{y+1}{-y^4-2y^3+18y+81}}\\
\end{align*}
\begin{align*}
q'(y)&=-\dfrac{1}{8\sqrt{2}}\sqrt{\dfrac{(3y^4+8y^3+6y^2+63)^2}{-(y^2+7)(y^2+2y-7)(y+1)^3}} \\
\sqrt{1+q(y)^2}&=\dfrac{1}{4\sqrt{2}} \sqrt{\dfrac{-y^4-2y^3+18y+81}{y+1}}\\
\dfrac{q'(y)}{\sqrt{1+q(y)^2}}&=-\dfrac{1}{2}\sqrt{
\dfrac{(3y^4+8y^3+6y^2+63)^2}{(y^2+7)(y^2+2y-7)(y+1)^2(y^4+2y^3-18y-81)}}
\end{align*}
\begin{align*}
I&=2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\underbrace{\operatorname{arsinh}(z)}_{=u}\underbrace{\dfrac{p(y)}{q'(y)}}_{=v'}\,dz\\
&=-\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\\
&+\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\dfrac{q'(y)}{\sqrt{1+q(y)^2}}\log\left(1-y^2+\sqrt{y^4-2y^2-63}\right)\,dy
\end{align*}

It is in principle solvable by integration by parts, although definitely very inconvenient, especially during the heat wave we currently have here.

Mhm. This looks awful.
What shall we do?

 

[fresh_42]

Mhm. This looks awful.

I'm afraid I have to agree. Especially, as all the negative roots and singularities haven't even been addressed.

I don't think I'll have the patience to actually solve that last integral, possible or not. The integral of the log doesn't have a closed form, so it has to be differentiated in the integration by parts. But that means that the first factor has to be integrated. Maybe it's getting easier to divide that whole thing by powers of y and introduce the new variable 1/y, but I'm not very confident.

What shall we do?

Go back to Start.

$I=\dfrac{\pi^2}{8}$ is a simple expression. And $I=\displaystyle{\int_{-a}^{+a}\operatorname{arcsin}\sqrt{f(\alpha)f(-\alpha)}\,d\alpha}$ is a simple integral. It's crying out for a connection to find.

 

[Meden Agan]

Go back to Start.

$I=\dfrac{\pi^2}{8}$ is a simple expression. And $I=\displaystyle{\int_{-a}^{+a}\operatorname{arcsin}\sqrt{f(\alpha)f(-\alpha)}\,d\alpha}$ is a simple integral. It's crying out for a connection to find.

I totally agree. That is actually what I was going to say to you.

 

[Meden Agan]

@fresh_42 Anything new on this beast of an integral?
Yesterday I tried to come up with something, but it's only a remark and can't be enough to evaluate the integral.

 

[fresh_42]

@fresh_42 Anything new on this beast of an integral?
Yesterday I tried to come up with something, but it's only a remark and can't be enough to evaluate the integral.

Not really. I am at
$$
I=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot (z^2)'}{\sqrt{z^2-z^4}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha}{\sqrt{t(1-t)}}\,dt
$$
with
$$
f(\alpha)f(-\alpha)=\dfrac{\cos^2(\alpha)-4\cos^2(\beta)\sin^2(\alpha)}{1-\tan^2(\alpha)}=z^2=t
$$
and
$$
y(\alpha)^2=9-16\sin^2(\alpha)=\sqrt{1+8\cos(2\alpha)}=16\cos^4(\beta)
$$
I hope to use the intermediate variable $\beta$ to avoid differentiation of the square root as long as possible. $z'$ is really inconvenient and results in a long polynomial expression. Same happens with changing the variables, then $\alpha$ carries all the burden.

But I cannot get this picture out of my mind

which is so nice. There must be some chance to use this beauty. The only problem is that $z$ is not differentiable at $\pm a$ with its vertical tangents.

 

[robphy]

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?

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

No. My instructor proposed that integral.

With some context (e.g, a plot or some geometric figure),
there may be symmetries that could be exploited to simplify the calculation.
(For example, what is the role of $9-16x$?)
This may be just an obfuscated identity.

 

[fresh_42]

With some context (e.g, a plot or some geometric figure),
there may be symmetries that could be exploited to simplify the calculation.

See my post #69. And I already posted it a couple of posts before.

(For example, what is the role of $9-16x$?)
This may be just an obfuscated identity.

The expression under the root is $1+8\cos(2x)$ which, together with $a=\operatorname{arcsin}(1/\sqrt[4]{8})$, explains the central role of $8$, including the solution, but that doesn't help much.

What makes the problem interesting is the asymmetry caused by that root. This is a standard problem in the integration of trig functions: a linear term in a quadratic expression. That's why it is interesting.

 

[Meden Agan]

Not really. I am at
$$
I=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot (z^2)'}{\sqrt{z^2-z^4}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha}{\sqrt{t(1-t)}}\,dt
$$
with
$$
f(\alpha)f(-\alpha)=\dfrac{\cos^2(\alpha)-4\cos^2(\beta)\sin^2(\alpha)}{1-\tan^2(\alpha)}=z^2=t
$$
and
$$
y(\alpha)^2=9-16\sin^2(\alpha)=\sqrt{1+8\cos(2\alpha)}=16\cos^4(\beta)
$$
I hope to use the intermediate variable $\beta$ to avoid differentiation of the square root as long as possible. $z'$ is really inconvenient and results in a long polynomial expression. Same happens with changing the variables, then $\alpha$ carries all the burden.

But I cannot get this picture out of my mind

View attachment 362901
which is so nice. There must be some chance to use this beauty. The only problem is that $z$ is not differentiable at $\pm a$ with its vertical tangents.

The most beautiful form into which we can convert the original integral, according to the picture, is:
$$2\int\limits_{0}^{\arcsin \left(1/\sqrt[4]{8}\right)}\arcsin\left(\sqrt{\frac{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)\sqrt{9-16\ \sin^{2}x}}{1-\tan^{2}x}}\right) \, \mathrm dx.$$
Now, maybe we can do this by applying Feynman's technique differentiating with respect to a real parameter $t?$ Or by converting $\arcsin$ into $\arccos$ to get rid of the outer square root?
What do you think?

 

[fresh_42]

I thought about Feynman, but haven't looked deeper into it. My integral is $\displaystyle{\int_0^a \dfrac{\alpha z'}{\sqrt{1-z^2}}\,d\alpha}.$ So either I have this nasty $z'(\alpha)$ or a nasty $\alpha(z)$ if I switch to $dz.$ At least I got rid of the inverse sine. The problem is thus, that I cannot separate $\alpha$ and $z.$ I reduced the problem to $\displaystyle{\int_{z=0}^{z=1} \dfrac{1-\alpha}{\sqrt{1-z^2}}\,dz}$ or likewise $\displaystyle{\int_{z=0}^{z=1} \dfrac{\alpha}{\sqrt{1-z^2}}\,dz} .$ If we cannot separate $\alpha$ and $z$ by inverting $f(\alpha),$ maybe we can merge the two in a new variable.

 

[robphy]

The most beautiful form into which we can convert the original integral, according to the picture, is:
$$2\int\limits_{0}^{\arcsin \left(1/\sqrt[4]{8}\right)}\arcsin\left(\sqrt{\frac{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)\sqrt{9-16\ \sin^{2}x}}{1-\tan^{2}x}}\right) \, \mathrm dx.$$

Possibly helpful:
$$\frac{\left(\left(\cos x\right)^{2}-\left(\sin x\right)^{2}\sqrt{9-16\left(\sin x\right)^{2}}\right)}{1-\left(\tan x\right)^{2}}$$
can be factored to read
$$\cos\left(x\right)^{2}\left(\frac{\left(\cos\left(x\right)-\sin\left(x\right)\left(9-16\sin\left(x\right)^{2}\right)^{\frac{1}{4}}\right)\left(\cos\left(x\right)+\sin\left(x\right)\left(9-16\sin\left(x\right)^{2}\right)^{\frac{1}{4}}\right)}{\left(\cos\left(x\right)-\sin\left(x\right)\right)\left(\cos\left(x\right)+\sin\left(x\right)\right)}\right)$$

 

[fresh_42]

Yes, that's basically my formula for $f(\alpha)f(-\alpha).$ It reveals the symmetry of the curve. If we ignore the inverse sine, then we have almost a circle. But that circle has a flat spot, which is exactly the problem here.

 

[Meden Agan]

I reduced the problem to $\displaystyle{\int_{z=0}^{z=1} \dfrac{1-\alpha}{\sqrt{1-z^2}}\,dz}$ or likewise $\displaystyle{\int_{z=0}^{z=1} \dfrac{\alpha}{\sqrt{1-z^2}}\,dz} .$ If we cannot separate $\alpha$ and $z$ by inverting $f(\alpha),$ maybe we can merge the two in a new variable.

Mhm... Could you show that?

 

[fresh_42]

Mhm... Could you show that?

Integration by parts:
$$
\int_a^b u'v=[uv]_a^b-\int_a^b uv'
$$
\begin{align*}
z&=\sqrt{f(\alpha)f(-\alpha)}\, , \,z'=\dfrac{dz}{d\alpha}\, , \,z'\,d\alpha=dz\, , \,d\alpha=\dfrac{dz}{z'}\\
\int\dfrac{dz}{z'}&=\int d\alpha= \alpha\, , \,z(0)=1\, , \,z(a)=0
\end{align*}

\begin{align*}
I&=\int_{-a}^{+a} \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=\int_{\alpha=-a}^{\alpha=a}\dfrac{1}{z'}\operatorname{arcsin}(z)\,dz\\
&=\underbrace{\left[\alpha \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\right]_{\alpha=-a}^{\alpha=a}}_{=0\text{ since }f(-a)=0}-\int_{\alpha=-a}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha
\end{align*}
and
\begin{align*}
I&=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot (z^2)'}{\sqrt{z^2-z^4}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha}{\sqrt{t-t^2}}\,dt\\
&=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha-1}{\sqrt{(1-z)(1+z)}}\,dz -2\int_{\alpha=0}^{\alpha=a}\dfrac{1}{\sqrt{(1-z)(1+z)}}\,dz\\
&=2\int_{\alpha=0}^{\alpha=a}\dfrac{1-\alpha}{\sqrt{(1-z)(1+z)}}\,dz+2\int_{z=0}^{z=1}\dfrac{1}{\sqrt{(1-z)(1+z)}}\,dz\\
&=2\int_{\alpha=0}^{\alpha=a}\dfrac{(1-\sqrt{\alpha})(1+\sqrt{\alpha})}{\sqrt{(1-z)(1+z)}}\,dz+\pi =\pi -2\int_{z=0}^{z=1}\dfrac{1-\alpha}{\sqrt{1-z^2}}\,dz
\end{align*}

 

[Meden Agan]

Integration by parts:
$$
\int_a^b u'v=[uv]_a^b-\int_a^b uv'
$$
\begin{align*}
z&=\sqrt{f(\alpha)f(-\alpha)}\, , \,z'=\dfrac{dz}{d\alpha}\, , \,z'\,d\alpha=dz\, , \,d\alpha=\dfrac{dz}{z'}\\
\int\dfrac{dz}{z'}&=\int d\alpha= \alpha\, , \,z(0)=1\, , \,z(a)=0
\end{align*}

\begin{align*}
I&=\int_{-a}^{+a} \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\,d\alpha=\int_{\alpha=-a}^{\alpha=a}\dfrac{1}{z'}\operatorname{arcsin}(z)\,dz\\
&=\underbrace{\left[\alpha \operatorname{arcsin} \sqrt{f(\alpha)f(-\alpha)}\right]_{\alpha=-a}^{\alpha=a}}_{=0\text{ since }f(-a)=0}-\int_{\alpha=-a}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha
\end{align*}
and
\begin{align*}
I&=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot (z^2)'}{\sqrt{z^2-z^4}}\,d\alpha=-\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha}{\sqrt{t-t^2}}\,dt\\
&=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha-1}{\sqrt{(1-z)(1+z)}}\,dz -2\int_{\alpha=0}^{\alpha=a}\dfrac{1}{\sqrt{(1-z)(1+z)}}\,dz\\
&=2\int_{\alpha=0}^{\alpha=a}\dfrac{1-\alpha}{\sqrt{(1-z)(1+z)}}\,dz+2\int_{z=0}^{z=1}\dfrac{1}{\sqrt{(1-z)(1+z)}}\,dz\\
&=2\int_{\alpha=0}^{\alpha=a}\dfrac{(1-\sqrt{\alpha})(1+\sqrt{\alpha})}{\sqrt{(1-z)(1+z)}}\,dz+\pi =\pi -2\int_{z=0}^{z=1}\dfrac{1-\alpha}{\sqrt{1-z^2}}\,dz
\end{align*}

Fantastic work. The integral reduces to $2 \displaystyle \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz$. We have to prove it is equal to $\pi^2/8$.
Now, there is no chance of separating $\alpha$ and $z$ by inverting $f(\alpha)$.
So, how can we merge the two in a new variable?

 

[fresh_42]

Fantastic work. The integral reduces to $2 \displaystyle \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz$. We have to prove it is equal to $\pi^2/8$.
Now, there is no chance of separating $\alpha$ and $z$ by inverting $f(\alpha)$.
So, how can we merge the two in a new variable?

You can see from the many versions that I listed for my integral that I'm still searching for a good key. At least the integral limits can easily be replaced by $[0,1].$ And the minus sign you asked about is due to the negative value of $z',$ the right part of the curve.

 

[robphy]

Possibly useful:
https://www.doubtnut.com/qna/121560332
https://www.doubtnut.com/qna/642670805


$$\int_{0}^{1}\frac{\arcsin x}{\sqrt{1-x^{2}}}dx = \frac{\pi^2}{8}$$

 

[Meden Agan]

You can see from the many versions that I listed for my integral that I'm still searching for a good key. At least the integral limits can easily be replaced by $[0,1].$ And the minus sign you asked about is due to the negative value of $z',$ the right part of the curve.

Possibly useful:
https://www.doubtnut.com/qna/121560332
https://www.doubtnut.com/qna/642670805


$$\int_{0}^{1}\frac{\arcsin x}{\sqrt{1-x^{2}}}dx = \frac{\pi^2}{8}$$

All of that is puzzling me.
We have to prove that $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz = \frac{\pi^2}{8}.$$
Now, some well-known integrals of this type are:
$$2 \int\limits_{0}^{1}\frac{\frac{1}{2} \, \arcsin z}{\sqrt{1-z^{2}}} \, \mathrm dz = \frac{\pi^2}{8}, \qquad 2 \int\limits_{0}^{1}\frac{\frac{1}{2} \, \arcsin \sqrt{1-z^2}}{\sqrt{1-z^{2}}} \, \mathrm dz = \frac{\pi^2}{8}.$$
We have $\sqrt{f(\alpha) f(-\alpha)} =\sqrt{\dfrac{\cos^{2}\left(\alpha\right)-\sin^{2}\left(\alpha\right)\sqrt{9-16\ \sin^{2} \left(\alpha\right)}}{1-\tan^{2}\left(\alpha\right)}}$. If we can prove that $\alpha = \dfrac{1}{2} \, \arcsin z$ or $\alpha = \dfrac{1}{2} \, \arcsin \sqrt{1-z^2}$ returns $z = \sqrt{f(\alpha) f(-\alpha)}$, we are done.
But plugging those values of $\alpha$ into the expression $z = \sqrt{f(\alpha) f(-\alpha)}$, that doesn't happen.

Instead, we have the following plots.

Red plot for $\alpha = \dfrac{1}{2} \, \arcsin z$, blue plot for $\alpha = \dfrac{1}{2} \, \arcsin \sqrt{1-z^2}$.

How can we explain that?

 

[fresh_42]

I made another integration by parts since the numerator $\alpha\cdot z'$ (cp. post #77) was so inviting. That resulted in
($v$ is one part of the integration by parts)
\begin{align*}
I&=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha=2\int_{v=0}^{v=a}z\,dv
\end{align*}
which is nice since all the complications are perfectly hidden. Less beautiful is it if we write it as
\begin{align*}
I&=2\int_{\alpha=0}^{\alpha=a}\sqrt{f(\alpha)f(-\alpha)}\dfrac{d}{d\alpha}\left(\dfrac{\alpha}{\sqrt{1-f(\alpha)f(-\alpha)}}\right)\,d\alpha
\end{align*}
But in that version, we are left with one variable, no inverse sine functions, and only ordinary trig functions, a polynomial, and roots. Plus, we have reasonable integration limits. (If I made no mistakes.)

 

[Meden Agan]

I made another integration by parts since the numerator $\alpha\cdot z'$ (cp. post #77) was so inviting. That resulted in
($v$ is one part of the integration by parts)
\begin{align*}
I&=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot z'}{\sqrt{1-f(\alpha)f(-\alpha)}}\,d\alpha=2\int_{v=0}^{v=a}z\,dv
\end{align*}
which is nice since all the complications are perfectly hidden. Less beautiful is it if we write it as
\begin{align*}
I&=2\int_{\alpha=0}^{\alpha=a}\sqrt{f(\alpha)f(-\alpha)}\dfrac{d}{d\alpha}\left(\dfrac{\alpha}{\sqrt{1-f(\alpha)f(-\alpha)}}\right)\,d\alpha
\end{align*}
But in that version, we are left with one variable, no inverse sine functions, and only ordinary trig functions, a polynomial, and roots. Plus, we have reasonable integration limits. (If I made no mistakes.)

Mhm, but unfortunately that derivative is frankly unmanageable. We seem to have complicated the integral again.

 

[fresh_42]

Mhm, but unfortunately that derivative is frankly unmanageable. We seem to have complicated the integral again.

Sure. Ultimately, there is no way to cheat. The problem has to be addressed somewhere. As long as we operate with $f(\alpha)$ and $f'(\alpha),$ we have a manageable expression. My different integrals are only a variety of possibilities, hoping that some of them would provide a lever to tackle that "flat spot".

 

[robphy]

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

This must be a popular question.

• https://artofproblemsolving.com/community/c7h3603912_prove_that_the_following_integral_equals_pi28
• https://math.stackexchange.com/ques...4-frac1-sqrtx-x2-arcsin-sqrt-fracx-1x-1x-sqrt
  

  Context
  The integral comes from the following probability question: "The vertices of a triangle are uniformly random points on a circle. The side lengths in random order are a,b,c. Simulations suggest that $P(ab^3+a^3b<c^4)=\dfrac12$. Can this be proved?"

---

Possibly useful:
partial summary of results

www.desmos.com/calculator/ivgtbpfvtt

 

[Meden Agan]

Sure. Ultimately, there is no way to cheat. The problem has to be addressed somewhere. As long as we operate with $f(\alpha)$ and $f'(\alpha),$ we have a manageable expression. My different integrals are only a variety of possibilities, hoping that some of them would provide a lever to tackle that "flat spot".

This must be a popular question.

• https://artofproblemsolving.com/community/c7h3603912_prove_that_the_following_integral_equals_pi28
• https://math.stackexchange.com/ques...4-frac1-sqrtx-x2-arcsin-sqrt-fracx-1x-1x-sqrt

---

Possibly useful:
partial summary of results

www.desmos.com/calculator/ivgtbpfvtt

IMO, the expression $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz = \frac{\pi^2}{8}$$ is the best way of representing the integral.
Perhaps it is enough to choose $\alpha(z)$ so that the integrand is symmetric in the interval $(-1, 1)$, like the original function?
The problem is that there would be many $\alpha(z)$ functions satisfying that requirement. And not all of them return $\pi^2/8$ as a result.

 

[fresh_42]

The problem is that we have $\alpha$ and $p(\operatorname{trig}(\alpha))$ in one equation - a dimension conflict. The trick that solves such problems should help us here, too. Guess I have to search in that integral book for such expressions.

 

[Meden Agan]

The problem is that we have $\alpha$ and $p(\operatorname{trig}(\alpha))$ in one equation - a dimension conflict. The trick that solves such problems should help us here, too. Guess I have to search in that integral book for such expressions.

Any significant developments?
I've come up with a heuristic argument, but I'm becoming more and more convinced that is not correct either.

 

[fresh_42]

Any significant developments?
I've come up with a heuristic argument, but I'm becoming more and more convinced that is not correct either.

I want to do the Weierstraß substitution now just to see where I end up. However, it doesn't look as if this would help. E.g.
$$
\sqrt{1+8\cos(2\alpha)}=\dfrac{\sqrt{(3t^2-8t+ 3) (3t^2+8t+3)}}{1+t^2}
$$
and $t(a)$ fulfills $t(a)^4+8t(a)^2-112=0.$ I think it's time to analyze the solution on MSE.

 

[Meden Agan]

I think it's time to analyze the solution on MSE.

Sad. I can't believe that is the only possible solution to the integral.

 

[fresh_42]

Sad. I can't believe that is the only possible solution to the integral.

If we look at the graphic,

then we see an almost linear behavior with slope around $-1$ (or so) when we approach $\alpha=0$ and a very strong curvature toward $\alpha=a$ with an infinite slope. So any substitution $\beta=p(\alpha)$ has to depend on where we are. It's not only non-linear, it also has to switch from almost linear to almost a square root. So what could be a proper scaling of $\alpha,$ making the curve integrable?

The origin of that behavior can be seen if we express $f(\alpha)(f(- \alpha)$ in the original variable $x,$
$$
f(\alpha)f(-\alpha)=\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}
$$
so maybe this expression is more promising than the trigonometric version.

 

[Meden Agan]

If we look at the graphic,

View attachment 362995​

then we see an almost linear behavior with slope around $-1$ (or so) when we approach $\alpha=0$ and a very strong curvature toward $\alpha=a$ with an infinite slope. So any substitution $\beta=p(\alpha)$ has to depend on where we are. It's not only non-linear, it also has to switch from almost linear to almost a square root. So what could be a proper scaling of $\alpha,$ making the curve integrable?

The origin of that behavior can be seen if we express $f(\alpha)(f(- \alpha)$ in the original variable $x,$
$$
f(\alpha)f(-\alpha)=\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}
$$
so maybe this expression is more promising than the trigonometric version.

I agree. Your ideas are always excellent, but unfortunately there's nothing we can do.
The form in post #86 made me hopeful. But unfortunately nothing came of it.

 

[Meden Agan]

@fresh_42 Did you get something new?
If the answer is no, let's go ahead to analyze the solution on MSE.

 

[fresh_42]

@fresh_42 Did you get something new?
If the answer is no, let's go ahead to analyze the solution on MSE.

I found some nice formulas, e.g., with $y^2=9-16x$ we get
$$
\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}=
\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}=\dfrac{1}{32}(y+1)(y^2+7)-\dfrac{1}{4}\dfrac{y^2+7}{y+1}
$$
which makes the polynomial much nicer, but I don't see yet how that helps.

 

[Meden Agan]

I found some nice formulas, e.g., with $y^2=9-16x$ we get
$$
\dfrac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}=
\dfrac{1}{32}\dfrac{(y^2+2y-7)(y^2+7)}{y+1}=\dfrac{1}{32}(y+1)(y^2+7)-\dfrac{1}{4}\dfrac{y^2+7}{y+1}
$$
which makes the polynomial much nicer, but I don't see yet how that helps.

I don't see how that helps either.
Let me know if you get anything useful, I'm stuck.

 

[fresh_42]

The key is Zacky's post here:
https://math.stackexchange.com/ques...s-left-frac3x3-3x4x2-sqrt2-x2/4891224#4891224

I tried a similar substitution $u(y)=\dfrac{y^2+7}{y+1}$ (or the root of it) and got $f(\alpha)f(-\alpha)=2^{-5}u'u(y+1)^2$ which is similar. However, I have not found a nice way to get rid of the root of $f(\alpha)f(-\alpha)$ since our integral is
$$
I=2\int_0^{\operatorname{arcsin}(1/\sqrt[4]{8})}\int_0^1 \sqrt{\dfrac{f(y)f(-y)}{1-t^2f(y)f(-y)}}\,dt\,dy.
$$

 

[elias001]

@Meden Agan can I ask in what class did you get this integral question? I mean what is the course that you are taking? The reason I am asking this id that the integral in question has to do with the topic of integral geometry and geometric probability.

 

[Meden Agan]

@fresh_42 Did you get something new?

 

[fresh_42]

@fresh_42 Did you get something new?

No, I gave it up. I had the feeling that I only shifted the problem from one point of view to another without adding substance. I tried to fight my way through what MSE had to offer, but that wasn't very clear and demanded more work to do, only to understand what they wrote.