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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.

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    If $y$ is the expression inside, then $y = x+ \frac{x}{y}$, so that $y^2 = xy+x$. Solve for $y$, and then solve the integral. – 2017-01-20
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    "Solve" is the wrong word here. "Evaluate" is appropriate. One solves problems; one solves equations; one _evaluates_ expressions. – 2017-01-20
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    @MichaelHardy More precisely "solve" is the **right** word to use in "Solve for $y$", but the **wrong** word to use is "solve the integral". – 2017-01-20
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    @mweiss : Except that nothing called $y$ appears here. – 2017-01-20
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    @MichaelHardy I assumed you were responding to the comment directly above yours, which uses the word "solve" twice, rather than to the OP, which does not contain the word at all. (But I see now that it *did* contain the word in its original form, which was edited by the time I got to it.) – 2017-01-20

3 Answers 3

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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?

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    THX ! I got it. – 2017-01-20
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    @AlexGuru Your welcome. – 2017-01-20
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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.

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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.