Solving a 2nd-order linear recurrence with variable coefficients

Question

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Question: I am seeking a general closed-form evaluation for the following definite integral.

For nonnegative real parameters $a,b,z\in\mathbb{R}^{+}\land a<z\le b$ and integer index $n\in\mathbb{Z}$, define the auxiliary function $f_{n}{\left(a,b;z\right)}$ via the integral representation $$f_{n}{\left(a,b;z\right)}:=\int_{a}^{z}\frac{t^{n}}{\sqrt{\left(t-a\right)\left(b-t\right)}}\,\mathrm{d}t.\tag{1}$$

Specifically, I am requesting a closed-form that's purely in terms of elementary functions (the problem would be practically trivial if hypergeometric type functions were permitted here). We know in principle that every integral of the form $\int R{\left(x,\sqrt{px^{2}+qx+c}\right)}\,\mathrm{d}x$, where $R{\left(x,y\right)}$ is a rational function of two variables, possesses an elementary antiderivative as a consequence of Euler's substitutions and the partial fraction decomposition algorithm.

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For $n\in\mathbb{N}\land n\ge2\land a,b,z\in\mathbb{R}^{+}\land a<z\le b$, we can derive a second-order linear non-homogeneous recurrence relation with variable coefficients for the terms $f_{n}{\left(a,b;z\right)}$:

$$\begin{cases} &\small{f_{0}{\left(a,b;z\right)}=2\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)},}\\ &\small{f_{1}{\left(a,b;z\right)}=-\sqrt{\left(z-a\right)\left(b-z\right)}+\left(a+b\right)\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)},}\\ &\small{n\,f_{n}=\left(\frac{a+b}{2}\right)\left(2n-1\right)F_{n-1}-ab\left(n-1\right)F_{n-2}-z^{n-1}\sqrt{\left(z-a\right)\left(b-z\right)}.}\tag{2}\\ \end{cases}$$

A general solution to recurrence relation $(2)$ would in principle solve the integral in $(1)$, but my research into general methods for solving second-order recurrence relations didn't turn up anything of much help.

How would I go about evaluating either $(1)$ or $(2)$?

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Answer 1

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Answer:

We first consider the case for negative integer index. Assuming $a,b,z\in\mathbb{R}^{+}\land a<z\le b\land n\in\mathbb{N}$, we have

$$\begin{align} f_{-n}{\left(a,b;z\right)} &=\int_{a}^{z}\frac{t^{-n}}{\sqrt{\left(t-a\right)\left(b-t\right)}}\,\mathrm{d}t\\ &=\int_{a^{-1}}^{z^{-1}}\frac{-\left(u^{-1}\right)^{-n}}{u^{2}\sqrt{\left(u^{-1}-a\right)\left(b-u^{-1}\right)}}\,\mathrm{d}u;~~~\small{\left[t^{-1}=u\right]}\\ &=\int_{z^{-1}}^{a^{-1}}\frac{u^{n}}{u\sqrt{\left(1-au\right)\left(bu-1\right)}}\,\mathrm{d}u\\ &=\frac{1}{\sqrt{ab}}\int_{z^{-1}}^{a^{-1}}\frac{u^{n-1}}{\sqrt{\left(u-b^{-1}\right)\left(a^{-1}-u\right)}}\,\mathrm{d}u\\ &=\frac{1}{\sqrt{ab}}\int_{b^{-1}}^{a^{-1}}\frac{u^{n-1}}{\sqrt{\left(u-b^{-1}\right)\left(a^{-1}-u\right)}}\,\mathrm{d}u\\ &~~~~~-\frac{1}{\sqrt{ab}}\int_{b^{-1}}^{z^{-1}}\frac{u^{n-1}}{\sqrt{\left(u-b^{-1}\right)\left(a^{-1}-u\right)}}\,\mathrm{d}u.\tag{3}\\ \end{align}$$

Then for $z=b$,

$$f_{-n}{\left(a,b;b\right)}=\frac{1}{\sqrt{ab}}f_{n-1}{\left(b^{-1},a^{-1};a^{-1}\right)},\tag{4a}$$

and for $z<b$

$$f_{-n}{\left(a,b;z\right)}=\frac{f_{n-1}{\left(b^{-1},a^{-1};a^{-1}\right)}-f_{n-1}{\left(b^{-1},a^{-1};z^{-1}\right)}}{\sqrt{ab}}.\tag{4b}$$

Thus, the negative index case reduces to the nonnegative index case, so it suffices to evaluate the latter.

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Define the auxiliary function $g_{k}{\left(z\right)}$ for nonnegative integer $k\in\mathbb{N}\cup\{0\}$ and real $z\in\mathbb{R}\land\left|z\right|\le1$ via

$$g_{k}{\left(z\right)}:=\int_{0}^{z}\frac{t^{2k}}{\sqrt{1-t^{2}}}\,\mathrm{d}t.\tag{5}$$

Assume $a,b,z\in\mathbb{R}^{+}\land a<z\le b\land n\in\mathbb{N}\cup\{0\}$. Then, with definition $(5)$ above, we have

$$\begin{align} f_{n}{\left(a,b;z\right)} &=\int_{a}^{z}\frac{t^{n}}{\sqrt{\left(t-a\right)\left(b-t\right)}}\,\mathrm{d}t\\ &=\int_{0}^{z-a}\frac{\left(a+u\right)^{n}}{\sqrt{u\left(b-a-u\right)}}\,\mathrm{d}u;~~~\small{\left[t-a=u\right]}\\ &=\int_{0}^{\frac{z-a}{b-a}}\frac{\left[a+\left(b-a\right)v\right]^{n}}{\sqrt{v\left(1-v\right)}}\,\mathrm{d}v;~~~\small{\left[u=\left(b-a\right)v\right]}\\ &=2\int_{0}^{\sqrt{\frac{z-a}{b-a}}}\frac{\left[a+\left(b-a\right)w^{2}\right]^{n}}{\sqrt{1-w^{2}}}\,\mathrm{d}w;~~~\small{\left[\sqrt{v}=w\right]}\\ &=2\int_{0}^{\sqrt{\frac{z-a}{b-a}}}\frac{\sum_{k=0}^{n}\binom{n}{k}a^{n-k}\left(b-a\right)^{k}w^{2k}}{\sqrt{1-w^{2}}}\,\mathrm{d}w\\ &=2\sum_{k=0}^{n}\binom{n}{k}a^{n-k}\left(b-a\right)^{k}\int_{0}^{\sqrt{\frac{z-a}{b-a}}}\frac{w^{2k}}{\sqrt{1-w^{2}}}\,\mathrm{d}w\\ &=2\sum_{k=0}^{n}\binom{n}{k}a^{n-k}\left(b-a\right)^{k}g_{k}{\left(\sqrt{\frac{z-a}{b-a}}\right)}.\tag{6}\\ \end{align}$$

We next turn to the intermediate problem of finding a closed-form for the auxiliary function $g_{k}{\left(z\right)}$. One finds that,

$$g_{k}{\left(z\right)}=\frac{\binom{2k}{k}\arcsin{\left(z\right)}}{4^{k}}-\frac{\binom{2k}{k}}{2^{2k+1}}\sqrt{1-z^{2}}\sum_{j=0}^{k-1}\frac{\left(2z\right)^{2j+1}}{\left(2j+1\right)\binom{2j}{j}}.\tag{7}$$

For proof of identity $(7)$, see the appendix below.

Then,

$$\small{g_{k}{\left(\sqrt{\frac{z-a}{b-a}}\right)}=\frac{\binom{2k}{k}\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)}}{4^{k}}-\sqrt{\frac{b-z}{z-a}}\sum_{j=0}^{k-1}\frac{\binom{2k}{k}\left(\frac{z-a}{b-a}\right)^{j+1}}{\left(2j+1\right)\binom{2j}{j}4^{k-j}}}.\tag{8}$$

Thus, for $a,b,z\in\mathbb{R}^{+}\land a<z\le b\land n\in\mathbb{N}\cup\{0\}$,

$$\begin{align} f_{n}{\left(a,b;z\right)} &=2\sum_{k=0}^{n}\binom{n}{k}a^{n-k}\left(b-a\right)^{k}g_{k}{\left(\sqrt{\frac{z-a}{b-a}}\right)}\\ &=2\sum_{k=0}^{n}\binom{n}{k}a^{n-k}\left(b-a\right)^{k}\bigg{[}\frac{\binom{2k}{k}\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)}}{4^{k}}\\ &~~~~~-\sqrt{\frac{b-z}{z-a}}\sum_{j=0}^{k-1}\frac{\binom{2k}{k}\left(\frac{z-a}{b-a}\right)^{j+1}}{\left(2j+1\right)\binom{2j}{j}4^{k-j}}\bigg{]}\\ &=2a^{n}\sum_{k=0}^{n}\binom{n}{k}\binom{2k}{k}\left(\frac{b-a}{4a}\right)^{k}\bigg{[}\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)}\\ &~~~~~-\sqrt{\frac{b-z}{z-a}}\sum_{j=0}^{k-1}\frac{4^{j}\left(\frac{z-a}{b-a}\right)^{j+1}}{\left(2j+1\right)\binom{2j}{j}}\bigg{]}\\ &=\small{2a^{n}\arcsin{\left(\sqrt{\frac{z-a}{b-a}}\right)}\sum_{k=0}^{n}\binom{n}{k}\binom{2k}{k}\left(\frac{b-a}{4a}\right)^{k}}\\ &~~~~~\small{-2a^{n}\sqrt{\frac{b-z}{z-a}}\sum_{k=0}^{n}\binom{n}{k}\binom{2k}{k}\left(\frac{b-a}{4a}\right)^{k}\sum_{j=0}^{k-1}\frac{4^{j}\left(\frac{z-a}{b-a}\right)^{j+1}}{\left(2j+1\right)\binom{2j}{j}}}.\blacksquare\tag{9}\\ \end{align}$$

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Appendix:

For each $m\in\mathbb{N}$, the following derivative holds (as may be easily verified through the familiar product rule):

$$\frac{d}{dt}\left[t^{m}\sqrt{1-t^{2}}\right]=\frac{t^{m-1}\left[m-\left(m+1\right)t^{2}\right]}{\sqrt{1-t^{2}}};~~~\small{-1\le t\le1}.$$

Let $z\in[-1,1]$. Then, integrating both sides of the derivative expression above, for each $m\in\mathbb{N}$ we have

$$z^{m}\sqrt{1-z^{2}}=\int_{0}^{z}\frac{t^{m-1}\left[m-\left(m+1\right)t^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t,$$

$$\implies z^{m}\sqrt{1-z^{2}}=m\int_{0}^{z}\frac{t^{m-1}}{\sqrt{1-t^{2}}}\,\mathrm{d}t-\left(m+1\right)\int_{0}^{z}\frac{t^{m+1}}{\sqrt{1-t^{2}}}\,\mathrm{d}t,$$

$$\implies \small{z^{2k+1}\sqrt{1-z^{2}}=\left(2k+1\right)\int_{0}^{z}\frac{t^{2k}}{\sqrt{1-t^{2}}}\,\mathrm{d}t-2\left(k+1\right)\int_{0}^{z}\frac{t^{2\left(k+1\right)}}{\sqrt{1-t^{2}}}\,\mathrm{d}t},$$

where in the last line above we've set $m=2k+1$.

Then, using definition $(5)$, we have for $k\in\mathbb{N}\cup\{0\}\land z\in[-1,1]$,

$$2\left(k+1\right)g_{k+1}{\left(z\right)}-\left(2k+1\right)g_{k}{\left(z\right)}=-z^{2k+1}\sqrt{1-z^{2}},$$

$$\implies g_{k+1}{\left(z\right)}-\frac{\left(2k+1\right)}{2\left(k+1\right)}g_{k}{\left(z\right)}=-\frac{z^{2k+1}\sqrt{1-z^{2}}}{2\left(k+1\right)}.$$

Hence, the terms $g_{k}{\left(z\right)}$ satisfy a first-order linear inhomogeneous recurrence relation with variable coefficients: for $z\in\mathbb{R}\land\left|z\right|\le1$,

$$\begin{cases} &g_{0}{\left(z\right)}=\arcsin{\left(z\right)},\\ &g_{k+1}{\left(z\right)}-\frac{\left(2k+1\right)}{2\left(k+1\right)}g_{k}{\left(z\right)}=-\frac{z^{2k+1}\sqrt{1-z^{2}}}{2\left(k+1\right)};~~~\small{k\in\mathbb{Z}^{\ge0}}.\\ \end{cases}$$

Unlike in the second-order case, there does exist a general method for solving first-order linear recurrences. Dividing both sides of the recurrence relation by $4^{-k-1}\binom{2k+2}{k+1}$, we have

$$\frac{g_{k+1}{\left(z\right)}}{4^{-k-1}\binom{2k+2}{k+1}}-\frac{g_{k}{\left(z\right)}}{4^{-k}\binom{2k}{k}}=-\frac{z^{2k+1}\sqrt{1-z^{2}}}{\left(2k+1\right)4^{-k}\binom{2k}{k}}.$$

Taking the sum of both sides,

$$\frac{g_{k}{\left(z\right)}}{4^{-k}\binom{2k}{k}}-g_{0}{\left(z\right)}=-\frac12\sqrt{1-z^{2}}\sum_{j=0}^{k-1}\frac{\left(2z\right)^{2j+1}}{\left(2j+1\right)\binom{2j}{j}}.$$

Thus,

$$g_{k}{\left(z\right)}=\frac{\binom{2k}{k}\arcsin{\left(z\right)}}{4^{k}}-\frac{\binom{2k}{k}}{2^{2k+1}}\sqrt{1-z^{2}}\sum_{j=0}^{k-1}\frac{\left(2z\right)^{2j+1}}{\left(2j+1\right)\binom{2j}{j}}.\blacksquare$$

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Answer 2

The change of variables $t = x - a$ gives us $$ F_n(a,b;z) = \int_0^{z-a} \dfrac{(a+t)^n}{\sqrt{t (b-a-t)}}\; dt = \sum_{j=0}^n {n \choose j} a^{n-j} F_j(0,b-a; z-a) $$ If you don't mind the sum, we're thus reduced to the case $a=0$. Use the scaling $x = b t$ to get $$ F_n(0,b; z) = b^n \int_0^{z/b} t^{n-1/2} (1-t)^{-1/2}\; dt = b^n F_n(0,1; z/b)$$ Now expand using the binomial series $$ (1-t)^{-1/2} = \sum_{k=0}^\infty \dfrac{(2k)!}{4^k \; k!^2} t^k$$ and integrate term-by-term. The sum can then be expressed in terms of a hypergeometric function $$ F_n(0,1; z) = {\frac {2\;{z}^{n+1/2}\;{\mbox{$_2$F$_1$}(1/2,n+1/2;\,n+3/2;\,z)}}{2\,n+ 1}} $$

Answer 3

I think your recurrence may be solved in terms of a generating function

$$G(x) = \sum_{n=0}^{\infty} F_n x^n$$

so that

$$F_n = \frac1{n!}G^{(n)}(0) $$

Because all of the coefficients are no worse than linear, the recurrence relation turns into a first-order differential equation in $G$ by using the relation

$$\sum_{n=0}^{\infty} n F_n x^n = x G'(x)$$

We multiply the recurrence relation by $x^n$ and sum over $n \ge 2$ to get

$$\sum_{n=2}^{\infty} n F_n x^n - \frac12 (a+b) \sum_{n=2}^{\infty} (2 n-1) F_{n-1} x^n + a b \sum_{n=2}^{\infty} (n-1) F_{n-2} x^n = -\frac{\sqrt{(z-a)(b-z)}}{z} \sum_{n=2}^{\infty} z^n x^n $$

This becomes

$$x G'(x) - x F_1 - (a+b) x^2 G'(x) - \frac12 (a+b) x G(x)+\frac12 (a+b) x F_0+ a b x^3 G'(x)+a b x^2 G(x) = -z \sqrt{(z-a)(b-z)} \frac{x}{1-z x} $$

or

$$(a b x^2- (a+b) x + 1)G'(x) + \left ( a b x - \frac12 (a+b) \right )G(x) = \left (F_1 - \frac12 (a+b) F_0 \right) -z \sqrt{(z-a)(b-z)} \frac{x}{1-z x}$$

The integrating factor for this equation is relatively simple, so we may rewrite this equation as

$$a b \frac{d}{dx} \left [\sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )} G(x) \right ] = \frac{F_1 - \frac12 (a+b) F_0}{\sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )}} - z \sqrt{(z-a)(b-z)} \frac{x}{(1-z x)\sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )} } $$

Integrating, we get

$$a b \sqrt{\left ( \frac1{a}-x\right )\left ( \frac1{b}-x\right )} G(x) +C = \left (F_1 - \frac12 (a+b) F_0 \right ) \log \left(2 a b x+2 \sqrt{a b} \sqrt{1-a x} \sqrt{1-b x}-(a+b)\right) + \sqrt{z-a} \sqrt{b-z} \log \left(2 a b x+2 \sqrt{a b} \sqrt{1-a x} \sqrt{1-b x}-(a+b)\right)+\sqrt{a b} \left[\log (1-x z)-\log \left(x (2 a b+a z+b z)-2 \sqrt{z-a}\sqrt{b-z} \sqrt{1-a x}\sqrt{1-b x} +a+b-2 z\right)\right] $$

where $C$ is a constant of integration determined by $G(0)=F_0$.

While this way looks messy, it does not rely on hypergeometrics.