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$$\int_0^x \cfrac{1}{1+\int_0^t \cfrac{1}{2+\int_0^{t_1} \cfrac{1}{3+\int_0^{t_2} \cfrac{1}{\cdots} dt_3} dt_2} dt_1} dt =f(x)$$

$$\int_{0}^{x} \frac{1}{n+h_{n+1}(t)}{d} t=h_n(x)$$

$$h'_n(x)(n+h_{n+1}(x))=1$$ $$h'_{n+1}(x)(n+1+h_{n+2}(x))=1$$

I need to find $ h_1(x)=f(x)$

Please help me how to express $f(x)$ as known functions or power series?

Thanks a lot for answers

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As a general rule, it is bad to write $\int_0^x f(x) dx$, because there are two different usages of the variable $x$. As a rule, it should be written as $\int_0^x f(t)dt$. It's a little harder to do that with your chained "continued-fraction-like" integration, perhaps... – Thomas Andrews Sep 11 '12 at 14:54
@ThomasAndrews: I have editted my question. Thank you very much for advice. – Mathlover Sep 11 '12 at 14:59
I changed \frac to \cfrac in your continued fraction. The difference between the appearances of the fraction before and after is a nice illustration of the utility of \cfrac. It's easier to read now. – Michael Hardy Sep 11 '12 at 17:58
@MichaelHardy : Really it is so. Thanks a lot for format change. – Mathlover Sep 11 '12 at 19:04
up vote 4 down vote accepted

It seems that additionally you require $h_n(0)=0$ for all $n$ (because $\int_0^0=0$). This allows us to compute the power series of $h_1$ to arbitrary precision. For example working down from $n=4$: $$h_4(x)=O(x)$$ $$h_3'(x)=\frac1{3+O(x)}=\frac13+O(x)$$ $$h_3(x)=\frac13x+O(x^2)$$ $$h_2'(x) = \frac1{2+\frac13x+O(x^2)}=\frac12-\frac1{12}x+O(x^2)$$ $$h_2(x)=\frac12x-\frac1{24}x^2+O(x^3)$$ $$h_1'(x) = \frac1{1+\frac12x-\frac1{24}x^2+O(x^3)}=1-\frac12x+\frac7{24}x^2+O(x^3)$$ $$h_1(x) = x-\frac14x^2+\frac7{72}x^3+O(x^4)$$ Starting with $n=10$, I get $$h_1(x) = x - \frac{1}{4} x^2 + \frac{7}{72} x^3 - \frac{149}{3456} x^4 + \frac{21193}{1036800} x^5 - \frac{235619}{23328000} x^6 + \frac{1408454377}{274337280000} x^7 - \frac{1227854784917}{460886630400000} x^8 + \frac{6524483827239647}{4645737234432000000} x^9+ O(x^{10}). $$ I don't know if the sequence of coefficients should be well-known.

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thanks for answer. Can we find $\lim x-->\infty$ $f(x)$? Any suggestion? – Mathlover Sep 12 '12 at 9:27
I calculated the coefficients up to $x^{1000}$. Apparently they are alternating and their absolute value grows slowly (reaching about 0.59 there, if I remember yesterdays output correctly). It might still tend to $\phi$, $\frac23$, $1$ or $\infty$ - who knows? – Hagen von Eitzen Sep 12 '12 at 9:36

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