# "Simple" integral equation

Let $$H(z)$$ be a continuous solution of the problem $$H(z)=\frac1{1-z}\int_z^1 \frac{2\zeta}{1+\zeta} H(\zeta^2)\,d\zeta,\ \ \ z\in[0,1);\ \ \ H(1)=1.$$ Is it true that $$H(0)=1-\ln2$$? The question is directly related to this one https://math.stackexchange.com/questions/4748129/asymptotics-of-sequence-of-rational-numbers .

Small update: I did not check all the details, but, it seems that the existence of the solution of this equation can be proved by applying the standard arguments - it is a contraction mapping acting on continuous functions satisfying $$H(1)=1$$. Hence, the corresponding fixed point (the solution $$H(z)$$) exists. I already calculated this by using the Picard-Lindelöf scheme $$H_{n+1}(z)=\frac1{1-z}\int_z^1 \frac{2\zeta}{1+\zeta} H_n(\zeta^2) \, d\zeta,\ \ \ H_0(z)\equiv1.$$ In fact, the iterations work well even for discontinuous functions, but the continuity at $$z=1$$ is required.

• I think it is worth mentioning that there can be only one such function (and it can be obtained by the iterative procedure starting with $H(z) = 1$). Nov 22, 2023 at 15:36
– AAK
Nov 22, 2023 at 15:56

That is a rather tough puzzle (took me two full days) with a rather short solution.

The first step is the differential equation Fred already mentioned: $$(1-z^2)H'(z)-(1+z)H(z)+2zH(z^2)=0\,.$$ Now define $$F(\alpha)=\int_0^1 z^\alpha H(z)\,dz$$, $$\alpha>-1$$. Then, by the condition $$H(1)=1$$, we have $$F(\alpha)\approx 1/\alpha$$ for large $$\alpha$$.

Integration of the differential equation against $$z^\alpha$$, $$\alpha>0$$ yields $$-\alpha F(\alpha-1)+(\alpha+2)F(\alpha+1)-F(\alpha)-F(\alpha+1)+F(\alpha/2)=0.$$ Summing over odd $$\alpha$$ from $$1$$ to $$A$$, we get $$-\sum_{0\le \alpha\le A+1}F(\alpha)+(A+2)F(A+1)+\sum_{1\le \alpha\le A, \alpha\text{ odd}}F(\alpha/2)=0$$ where the first sum is over all integer $$\alpha$$ in the range (each positive integer is either odd, or odd plus $$1$$), or, equivalently, $$\sum_{0\le\alpha\le A}(-1)^\alpha F(\alpha/2)+\sum_{\frac A2<\alpha\le A+1}F(\alpha)=(A+2)F(A+1)\,.$$

Now notice that $$H(0)=\int_0^1\frac 1{1+\sqrt z}H(z)\,dz=\sum_{\alpha\ge 0}(-1)^{\alpha}F(\alpha/2)$$ (I leave it to you to justify the convergence). Thus, passing to the limit as $$A\to+\infty$$ in the last displayed equation and using the large $$\alpha$$ asymptotics for $$F(\alpha)$$ with $$\alpha>A/2$$, we get $$H(0)+\log 2=1\,,$$ as desired.

• Super, this solution is quite nice!
– AAK
Nov 24, 2023 at 2:45
• @AAK Thanks! I'm glad I could be of some help :-) Nov 24, 2023 at 2:56

Alternative simple proof - integration by parts: $$\int_0^{1-a}\frac{H(z)}{1-z}dz=\int_0^{1-a}\frac1{(1-z)^2}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta=$$ $$\frac1{1-z}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta\bigg\rvert_{z=0}^{z=1-a}+\int_0^{1-a}\frac{2\zeta}{1-\zeta^2}H(\zeta^2)d\zeta,$$ which leads to $$\int_{1-2a+a^2}^{1-a}\frac{H(z)}{1-z}dz= H(1-a)-H(0)$$ Since $$H(1)=1$$ and $$\int_{1-2a}^{1-a}\frac{dz}{1-z}\to\ln2\ \ \ for\ \ \ a\to0,$$ we obtain the result.

• I cannot believe that I have not seen this before, very strange in my case.
– AAK
Nov 25, 2023 at 11:19
• Beautiful! It is curious that the integration against $\frac 1{1-z}$ up to $a\approx 1$ you used and adding the moments up to $A\approx\infty$ (i.e., integrating against $1+z+\dots+z^A$) I used are quite close to each other :-) Nov 25, 2023 at 19:57
• Yes, thanks! And another goal was to obtain this new integral functional equation useful for other expansions.
– AAK
Nov 25, 2023 at 21:13

This is an incomplete answer, the last step is missing (yet).

We can differentiate the OPs equation to get \begin{align}\tag{1}\label{eq:1} (1-z^2) H'(z)-(z+1) H(z)+2 z H(z^2)=0. \end{align} The series expansion of $$H(z)$$ around $$z=0$$, \begin{align}\tag{2}\label{eq:2} H(z)=\sum_{n=0}^\infty h_n z^n, \end{align} fulfills the recurrence relation (assuming $$H(0)=1$$) \begin{align} h_0&=h_1=1\\ h_n&=\frac 1 n\left[ (n-1)h_{n-2}+h_{n-1} + ((-1)^n+1)h_{n/2-1}\right], \tag{3}\label{eq:3} \end{align} such that \begin{align}\tag{4}\label{eq:4} H(z)=1+z+\frac{2 z^3}{3}-\frac{z^4}{3}+\frac{7 z^5}{15}-\frac{z^6}{5}+\frac{13 z^7}{35}-\frac{31 z^8}{105}+\frac{281 z^9}{945}+O\left(z^{10}\right) \end{align} Some Mathematica code:

Clear[hn]; hn[0]=1; hn[1]=1;
hn[n_Integer]:=hn[n] = ((n-1) hn[n-2] + hn[n-1] - If[EvenQ[n], 2 hn[n/2-1], 0])/n
Hs[z_]=Sum[hn[n] z^n, {n, 0, 10}]
(1-z^2) Hs'[z] + 2 z Hs[z^2] - (z+1) Hs[z] + O[z]^10

So we have to show that \begin{align}\tag{5}\label{eq:5} H(1)=\sum_{n=0}^\infty h_n = \frac 1 {1-\ln 2}. \end{align} Numerically, this is the case, the partial sums of \eqref{eq:5} converge with $$O(1/n)$$.

• Yes, and there is a little bit more simple sequence (not a series) that is equivalent to the problem, see the original question in math.stackexchange.com. In both, the difficult component $h_{n/2}$ appears. We have a few equivalent formulations of this problem - discrete and continuous, but I do not know which one is simpler to solve.
– AAK
Nov 23, 2023 at 9:42