0
$\begingroup$

Problem: Let

$$f(x)=\frac{1}{1+\frac{x}{1+\frac{x}{\cdots}}}$$

Then does

$$\int_{0}^{\infty}e^{-x}f(x)dx=:I$$

have a closed form ?

The first step in the Continued Fraction involves not surprisingly the exponential integral and the golden ratio.

Showing the integral converges is also not hard .

Attempt :

We can use the generating serie of the Catalan's number after @MartinR comment and the classical serie for exponential .

I will update if I go further.

Question :

Using a method involving only real numbers can we find a closed form for $I$?

Context: previous thought:

I was reading this What is the combinatoric significance of an integral related to the exponential generating function?

And got as example the integral above

Further remark :

The result is interesting because we have two representation for the generating function so I think the result is already established elsewhere

As $_2F_2$ appears in the closed form see as tank hypergeometric function.Some special case are discussed .

Update 1:

Omg see On the cubic counterpart of Ramanujan's $\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots$? it becomes crazy .

Update 2:

The link @TitoPiezas can be completed with a CF of $2F2$ representation see https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F2/10/

Update 3 :

One can find a power series with :

$$x^{1/4}(x^{1/4}-\frac{x}{(x+1)(x+2)}-\frac{x}{(x+1)(x+2)(x+3)})\simeq \frac{(1+4x^2)^{1/2}-1}{2x^{1/2}},x>0$$ Second question :

Update 4+5:

Using Mittag-Leffler summation +Catalan generating series we got an expression +Abel summation

If there no way with "real method" how to prove it (see comment section)?

$\endgroup$
0

1 Answer 1

3
+50
$\begingroup$

Well, strictly speaking, solving for the closed form of the continued fraction isn't that hard.

The closed form is given by $$\frac1{1+\overset{\infty}{\underset{n=1}{\large{\text{K}}}} \frac x1} = \frac1{1+\frac{2x}{\sqrt{4x+1}+1}} = \frac{\sqrt{4x+1}-1}{2x}$$

This can be found by solving for $y$ in the equation $y = \frac xy - 1$ and choosing to take the positive version by inspection (if anyone has a better method, please tell lol)

enter image description here

So, the integral just collapses into finding $$\int_0^\infty \frac{\sqrt{1+4x}-1}{2xe^x}\text{ d}x$$

Here, both Rubi and Mathematica cannot give the antiderivative, and I've not tinkered around with it much to be able to solve it analytically for now. Regardless, the definite integral is given by Mathematica to be $$\frac1{12}{}_2F_2\left(1, 1; 2, \frac52; \frac14\right) - \frac\pi2\operatorname{erfi}\left(\frac12\right) - 1 + \frac\gamma2 + e^{\frac14} \pi^{\frac12} \approx 0.68623$$ which matches the result you give.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .