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

  • Thread starter Thread starter Meden Agan
  • Start date Start date
  • Tags Tags
    Definite integral
Click For Summary
The integral $$I = \int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx$$ is proposed to equal $$\frac{\pi^2}{8}.$$ The discussion involves using the representation of the arcsine function and applying Fubini's Theorem to interchange the order of integration, allowing for a transformation of the integral into a double integral. Participants explore various substitutions, including trigonometric substitutions, to simplify the integrand and compute the integral more effectively. The complexity of the integral and the challenge of finding closed-form solutions for the limits of integration are highlighted, with suggestions for further simplification and exploration of the integral's properties.
  • #101
Meden Agan said:
@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.
 
Physics news on Phys.org
  • #102
fresh_42 said:
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)##.
 
  • #103
Meden Agan said:
@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".
 
  • #104
robphy said:
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.
 
  • #105
Meden Agan said:
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.
 
  • #106
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.