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

[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.

 

[Meden Agan]

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.

@fresh_42 I found a solution to this integral here. Does it convince you?
I can't understand how they derived equation $(3)$ from equation $(2)$.

 

[fresh_42]

@fresh_42 I found a solution to this integral here. Does it convince you?
I can't understand how they derived equation $(3)$ from equation $(2)$.

Looks nice at first sight. They say how to do it. Looks like a bit of work to check it. I'll go through it if I have some time. $\theta'$ has been my problem, too.

 

[Meden Agan]

Looks nice at first sight. They say how to do it. Looks like a bit of work to check it. I'll go through it if I have some time. $\theta'$ has been my problem, too.

Sure. But if you differentiate $\Theta(t) = \arcsin \left(\sqrt{E(t)}\right)$ with respect to $t$, you don't obtain the expression for $\Theta'(t)$ given by $(3)$.

 

[robphy]

@fresh_42 I found a solution to this integral here. Does it convince you?
I can't understand how they derived equation $(3)$ from equation $(2)$.

Sadly, there's no context.
Where did this integral arise?
What motivated the specific change of variables?
What characterizes the class of similar integrals solved by this method?
From what's presented, it seems it's "guess and check".

 

[pines-demon]

Sadly, there's no context.
Where did this integral arise?
What motivated the specific change of variables?
What characterizes the class of similar integrals solved by this method?
From what's presented, it seems it's "guess and check".

See post #26.

 

[fresh_42]

Sure. But if you differentiate $\Theta(t) = \arcsin \left(\sqrt{E(t)}\right)$ with respect to $t$, you don't obtain the expression for $\Theta'(t)$ given by $(3)$.

And if you do what he wrote, you won't obtain it either. $(2)$ results in a monster polynomial in $t,$ $\sin\theta$ and $\sin 2\theta.$ To claim "one arrives at" seems a bit too optimistic to me. Where does the root come from? It looks as if he put all the trouble I had, and that prevented me from solving it, into these three words. Quite as if he calculated from behind and claimed what he needed to arrive at the correct number.

You are right. What would really be the clue we were looking for is still hidden. I have had all those formulas, too, at one or another stage of my attempts and failed between $(2)$ and $(3).$ To summarize it as "one arrives at" is perky.

 

[fresh_42]

I had something similar:
$$
I=2\int_0^a\left(\int_0^{\sqrt{f(\alpha)f(-\alpha)}} \dfrac{1}{\sqrt{1-t^2}}\,dt\right)\,d\alpha\, , \,a=\operatorname{arcsin}\left(1/\sqrt[4]{8}\right)
$$
The problem is the inner integral limit.