I'm trying to find the most general bounds and functions work for this integral:

Question

empty · Answer

$$\int\limits_a^b \sqrt{ \sum_{n=0}^\infty \left( f(x) \right)^n } dx$$

empty · Answer

I guess this is kind of silly, f(x) could be anything as long as the value inside the square root is greater than or equal to zero.

Haha hmm.

anonymous · Answer

One thing you can say is that, at least over the interval $[a,b]$, you have to have $|f(x)|<1$.

empty · Answer

(changed this function to g(x) to avoid confusion)Really this comes from trying to solve a differential equation of this form:

$$y^{(n)} = \tfrac{1}{2} g'(y^{(n-2)})$$

anonymous · Answer

This looks like something that could be handled or at the very least examined with calculus of variations ... assuming the question can be posed appropriately for it.

empty · Answer

That's a good idea I don't know how I'd do that, I really am just playing around with this idea in a more general form:

$$y'' = \frac{dy'}{dx} = \frac{dy'}{dy}\frac{dy}{dx} = \frac{dy'}{dy}y' $$ 

So if you have something of this form:

$$y''=f'(y)$$ you can make it separable:

$$y' \frac{dy'}{dy} = f'(y)$$

anonymous · Answer

Provided that $|f(x)|<1$ for $a\le x\le b$,
$$\int_a^b\sqrt{\sum_{n\ge0}f(x)^n}\,\mathrm{d}x=\int_a^b \sqrt{\frac{1}{1-f(x)}}\,\mathrm{d}x$$(along with the usual conditions needed for the integral to exist). Not sure what else can be done with this in this general form.

empty · Answer

hmmm I wonder if I approximate the sum as for $f(x)<1$

$$\sum_{n=0}^\infty f(x)^n  \approx \int_0^\infty f(x)^ndn = -\ln f(x)$$

$$\int_a^b \sqrt{-\ln f(x) } dx$$

Basically I'm just trying to find a closed form even if I gotta fudge it now haha.