Evaluate $\int_{0}^{1}\left ( x+\frac{x}{x+\frac{x}{x+\frac{x}{x+\cdots }}} \right )dx$

Question

How to evaluate the following integral $$\int_0^1 \left ( x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{x+\cdots }}} \right )dx$$ I have no idea how to deal with the continued fraction.

Answer 1

HINT$$f(x)=x+\frac{x}{x+\dfrac{x}{x+\dfrac{x}{x+\cdots }}} \implies f(x)=x+\frac{x}{f(x)} \implies f(x)^2-xf(x)-x=0$$ Solve for $f(x)$. Use the fact that $f(x)>0$ to get $$f(x)=\frac{x+\sqrt{x^2+4x}}{2}$$ So we have $$\int_{0}^{1} \frac{x+\sqrt{x^2+4x}}{2} dx$$ Can you take it from here?

Answer 2

Let the expression given under integral be A. Hnece $$x+\frac{x}{A}=A$$ So $$A=\frac{x+\sqrt{\left ( x+2 \right )^{2}-2^{2}}}{2}$$ Now the integral become $$\frac{1}{2}\int_0^1 x+\sqrt{\left ( x+2 \right )^{2}-2^{2}}\, \mathrm{d}x$$ hope you can take it from here.

Answer 3

If

$$y=x+\cfrac x{x+\cfrac x{x+\cfrac x\ddots}}$$

then it follows that

$$y=x+\frac xy$$

Which upon solving (with $y\ge0$),

$$y=\frac{x+\sqrt{x^2+4x}}2$$

And thus, we have

$$\int_0^1\frac{x+\sqrt{x^2+4x}}2\ dx=\frac{1+3\sqrt5}4-2\operatorname{csch}^{-1}(2)$$

where we used the inverse hyperbolic cosecant.