Asymptotic Estimate of Singular Behaviour of an Integral

Question

I am considering a function $f(x)$ defined by an integral as follows:

$$ f(x) = \int_0^1 dt \frac{\sqrt{t}}{(1 -xt^2)^{\beta}} ,$$

which depends on real $x< 1$ and real $\beta > 1$. First note that $f(0) = 2/3$ for all $\beta$. In terms of an asymptotic estimate of the singular behaviour for $x$ close to $1$, it can be shown that one has

$$ f(x) \sim A (1-x)^{1 - \beta} , $$

for some constant $A$.

Is there some way to evaluate or estimate what $A$ is? It seems from some numerics I did (which might be mistaken) that

$$A = (2 \beta - 2)^{-1}.$$

Edit: It can be show that this is correct by using the hypergeometric function and its known asymptotics to derive $A$, but this kind of seems like cheating so I am looking for a more elegant way to show that my guess for $A$ is correct.

Answer 1

Using asymptotic properties of the Gaussian hypergeometric functions.

If $\beta$ is a positive number, the asymptotic of $$f(x) = \int_0^1 \frac{\sqrt{t}}{(1 -x\,t^2)^{\beta}} \,dt=\frac{2}{3} \,\, _2F_1\left(\frac{3}{4},\beta ;\frac{7}{4};x\right)$$ is in general

$$f(x) \sim \frac{(1-x)^{\beta-1}}{2(\beta-1) }$$

For example, for $\beta=\frac 32$, $\frac{3-x}{2 \sqrt{1-x}}$ is much better than $\sqrt{1-x}$

Answer 2

Too long for a comment $$f(x) = \int_0^1 \frac{\sqrt{t}}{(1 -xt^2)^{\beta}}dt\overset{xt^2=s}{=}\frac{x^{-3/4}}2\int_0^x\frac{s^{-1/4}}{(1-s)^\beta}ds$$ Denoting $\epsilon=1-x$ $$f(x)=\frac{(1-\epsilon)^{-3/4}}2\int_\epsilon^1t^{-\beta}(1-t)^{-1/4}dt$$ Using the decomposition $$(1-t)^{-1/4}=1+\frac t4+\frac{5t^2}{32}+...$$ we can choose as many terms as we need - depending on $\beta$.

For example, for $\beta\in(1;2)$ we can write $$f(x)=\frac{(1-\epsilon)^{-3/4}}2\left(\int_\epsilon^1t^{-\beta}\Big((1-t)^{-1/4}-1\Big)dt+\int_\epsilon^1t^{-\beta}dt\right)$$ As the first integral is regular at $\epsilon=0$ $$=\frac{(1-\epsilon)^{-3/4}}2\left(\int_0^1t^{-\beta}\Big((1-t)^{-1/4}-1\Big)dt+o(1)+\int_\epsilon^1t^{-\beta}dt\right)$$ $$f(x)=\frac{(1-x)^{1-\beta}}{2(\beta-1)}+O(1);\,\,\beta\in(1;2)$$ For $\beta\in(2;3)$ we keep two terms in the first integral; namely, present $f(x)$ in the form $$f(x)=\frac{(1-\epsilon)^{-3/4}}2\left(\int_\epsilon^1t^{-\beta}\Big((1-t)^{-1/4}-1-\frac t4\Big)dt+\int_\epsilon^1t^{-\beta}\Big(1+\frac t4\Big)dt\right)$$ Again, the first integral is regular at $\epsilon=0$, and we have for two leading asymptotic terms ($\epsilon =1-x$) $$f(x)\sim\frac{(1-\epsilon)^{-3/4}}2\int_\epsilon^1t^{-\beta}\Big(1+\frac t4\Big)dt\sim\frac{(1-\epsilon)^{-3/4}}2\left(\frac{\epsilon^{1-\beta}}{\beta-1}+\frac{\epsilon^{2-\beta}}{4(\beta-2)}\right)$$ Decomposing also $(1-\epsilon)^{-3/4}=1+\frac34\epsilon+O(\epsilon^2)$ $$f(x)\sim\frac{(1-x)^{1-\beta}}{2(\beta-1)}+\left(\frac38\frac1{\beta-1}+\frac18\frac1{\beta-2}\right)(1-x)^{2-\beta}+O(1);\,\,\beta\in(2;3)$$ etc.

Answer 3

$\require{cancel}$

Note. The following might just be an extremely long comment and not be what OP is looking for. However, this has several interesting ideas, identities, and another way of finding a solution to $A$.

Observations

First, observe

$$f(x)=\int_0^1 \frac{\sqrt{t}}{(1-xt^2)^{\beta}}dt$$

at least resembles or is literally a special case of the Hypergeometric function:

$${}_2F_1(a,b;c;z)=\frac{1}{\text{B}(b,c-b)}\int_0^1 \frac{t^{b-1}(1-z)^{c-b-1}}{(1-zt)^a}dt$$

Where $\text{B}(b,c-b)$ is the beta function.

How would you recognize this relation? I didn't–Wolfram did :) Working with such an integral I would assume a knowledge of such functions, especially in a course. However, I will try my best to make this answer as accessible as possible.

Second, the Eulerian transformations:

\begin{align}{}_2F_1(a,b;c;z)&=(1-z)^{-a}{}_2F_1(a,c-b;c;\frac{z}{z-1})\\&=(1-z)^{-b}{}_2F_1(c-a,b;c;\frac{z}{z-1})\\&=(1-z)^{c-a-b}{}_2F_1(c-a,c-b;c;z)\end{align}

These transformations have a factor of $(1-z)$, which can help in finding $A$:

$$f(x)\sim A(1-x)^{1-\beta}$$

---

Working backward

A quick WolframAlpha search gives us, where ${}_2F_1$ is a hypergeometric function: \begin{align}f(x) &= \int_0^1 \frac{\sqrt{t}}{(1 -xt^2)^{\beta}}dt\\&=\frac23 \,{}_2F_1\left(\frac34,\beta;\frac74;x\right)\end{align}

For a special case where $\beta=\frac74$, $$f(x)=\frac23 (1-x)^{-3/4}$$

We have such special cases when $b=c$ for a function in the form ${}_2F_1(a,b;c;z)$.

I have tried substitutions and other tricks, but I couldn't figure out how WolframAlpha found the exact representation of $f(x)$.

Edit. I posted this as a question and there you can see how to arrive to this Hypergeometric function.

Series Representations

Note that the Hypergeometric function is named so because it is a generalization of the geometric series. Namely, when $b=c$,

$${}_2F_1(a,b;b;z)=1+x+x^2+x^3+\dots$$

Then, by definition,

$$f(x)=\sum_{n=0}^{\infty}\frac{(3/4)_n(\beta)_n}{(7/4)_n}\frac{x^n}{n!}=3\sum_{n=0}^{\infty}\frac{(\beta)_nx^n}{(4n+3)n!}\tag{1}$$

We can do this because

$$\frac{(a)_n}{(a+1)_n}=\frac{a}{a+n}$$

A few identities by playing around: $$\frac{\Gamma(a+n)\Gamma(a+1)}{\Gamma(a)\Gamma(a+n+1)}=\frac{a}{a+n}$$ $$(a)_n\Gamma(a)=\Gamma(a+n)$$

We'll get back to Equation $(1)$ soon.

Eulerian transformation: solving for $A$

Applying the Eulerian transformation,

\begin{align}f(x)={}_2F_1\left(\frac34,\beta;\frac74;x\right)&=(1-x)^{1-\beta}{}_2F_1\left(1,\frac74-\beta;\frac74;x\right)\\&=A(1-x)^{1-\beta}\end{align}
$$\therefore A={}_2F_1\left(1,\frac74-\beta;\frac74;x\right)$$

Turn this into a series by definition:

$$A=\sum_{n=0}^{\infty}\frac{\cancel{(1)_n}(\frac74-\beta)_n}{(\frac74)_n}\frac{x^n}{\cancel{n!}}\tag{2}$$

And we are done here.

Peculiar results

What we find is truly beautiful. With equations $(1)$ and $(2)$,

$$f(x)=3\sum_{n=0}^{\infty}\frac{(\beta)_nx^n}{(4n+3)n!}=A(1-x)^{1-\beta}=(1-x)^{1-\beta}\sum_{n=0}^{\infty}\frac{x^n(\frac74-\beta)_n}{(\frac74)_n}$$

Rewriting in a more dramatic way:

$$(1-x)^{1-\beta}\sum_{n=0}^{\infty}\frac{x^n(\frac74-\beta)_n}{(\frac74)_n}=3\sum_{n=0}^{\infty}\frac{x^n(\beta)_n}{(4n+3)n!}$$

Which means, with @ClaudeLeibovici's asymptotics,

$$\sum_{n=0}^{\infty}\frac{(1-x)^{1-\beta}x^n(\frac74-\beta)_n}{(\frac74)_n}=\frac{1}{2\beta-2}$$

I apologize if any these results are wrong. I am still in school, so it's very difficult to do math and other work.