# Computing $\int_{0 }^{1} (1-x^{\frac{1 } {a}})^{-\frac{1 } {2 }} dx$

I was wondering how to compute directly this integral without using beta/gamma functions: $$\int_{0 }^{1} (1-x^{\frac{1 } {a}})^{-\frac{1 } {2 }} dx$$,

$$a\to 0$$

Wolfram Alpha said its equal to 1, but I can't get it by direct calculation.

I tried the substitution :

$$t=x^{\frac{1} {a}}$$ so $$t^a=x$$ then $$dx=at^{a-1}dt$$

But failed to calculate the new integral.

Thank you

• Change of variable $y=x^{1/a}$ and then split the integral $\int_0^1$ into $\int_0^\epsilon+\int_\epsilon^1$. Show the first one goes to 1 and the second one goes to 0. Mar 21, 2020 at 8:59
• @user58955 that's exactly my problem, to show, after substitution, that the integral equals to 1 Mar 21, 2020 at 9:30
• I think you need to split it up into two parts. The second part is well-defined and it's $a$ times some bounded integral, and as $a\to 0$, it goes to $0$. In the first part, you can first ignore $1/\sqrt{1-y}$ because it is close to 1 on $[0,\epsilon]$. Pretending it to be $1$ gives you the answer easily. Then you need to choose $\epsilon$ to show that the loss of replacing $1/\sqrt{1-y}$ with $1$ is small. Mar 21, 2020 at 9:40
• Duplicate with answer here. Mar 21, 2020 at 14:05

I think the limit does not exist. only for $$a\rightarrow 0^+$$ the limit exists.

In first step With out using beta function I show

$$lim_{a\rightarrow 0^+} \int_{0 }^{1} (1-x^{\frac{1 } {a}})^{-\frac{1 } {2 }} dx=1$$

$$a$$ must be positive since $$(1-x^{\frac{1 } {a}})^{-\frac{1 } {2 }}=\frac{1}{(1-x^{\frac{1}{a}})^{\frac{1}{2}}}$$ so $$1-x^{\frac{1}{a}}>0$$ so $$a>0$$. so this limit is valid only for $$a\rightarrow 0^+$$

x<-.5
a<--.0001
(1-x^(1/a))^(-.5) #NaN


if $$0 so $$0 for $$a> 0$$

$$lim_{a\rightarrow 0^{+}} \int_{0 }^{1} (1-x^{\frac{1 } {a}})^{-\frac{1 } {2 }} dx$$

$$=lim_{n\rightarrow +\infty} \int_{0 }^{1} (1-x^n)^{-\frac{1 } {2 }} dx$$

define $$0\leq f_n=(1-x^n)^{-\frac{1 } {2 }}=\frac{1}{(1-x^n)^{\frac{1 } {2 }}}$$ so $$f_n$$ are decreasing and non-negative so by monotone-convergence-theorem $$lim \int f_n=\int lim f_n$$ $$\overset{MCT}{=} \int_{0 }^{1} lim_{n\rightarrow +\infty} (1-x^n)^{-\frac{1 } {2 }} dx$$

since $$0 so $$x^{n} \rightarrow 0$$ $$=\int_{0 }^{1} 1 dx=1$$ .

With beta function let $$a> 0$$

$$t=x^{\frac{1} {a}}$$ so $$t^a=x$$ then $$dx=at^{a-1}dt$$

$$=\int_{0 }^{1} (1-t)^{-\frac{1 } {2 }} at^{a-1}dt$$ $$=a\int_{0 }^{1} (1-t)^{\frac{1 } {2 }-1} t^{a-1}dt$$ so $$a> 0$$ Beta_distribution (this is why $$a\rightarrow 0^+$$) $$=a B(a,\frac{1}{2})=a\frac{\Gamma(a)\Gamma(\frac{1}{2})}{\Gamma(a+\frac{1}{2})}$$

$$=\frac{\Gamma(a+1)\Gamma(\frac{1}{2})}{\Gamma(a+\frac{1}{2})} \rightarrow 1 \hspace{1cm} when \, \, a\rightarrow 0^+$$

Rcode
by a -->0+
a<<-.00001
fun<-function(x){
(1-x^(1/a))^(-.5)
}
integrate(fun,lower=0,upper=1)  ### 1 with absolute error < 1.1e-14
by a -->0-
a<<--.00001
integrate(fun,lower=0,upper=1)
#Error in integrate(fun, lower = 0, upper = 1) : non-finite function value

• Thanks but I asked without beta function. That's the way I solved it before. Mar 21, 2020 at 11:14
• I did not see that Mar 21, 2020 at 11:14
• Thank you, I understand your solution Mar 21, 2020 at 14:11

Let me write a solution following my comments at the top.

So we need to find $$I(a) = \int_0^1 \frac{a}{t^{1-a}\sqrt{1-t}} dt = \int_0^{\epsilon} \frac{a}{t^{1-a}\sqrt{1-t}} dt + \int_{\epsilon}^1 \frac{a}{t^{1-a}\sqrt{1-t}} dt =: I_1(a) + I_2(a)$$

We control $$I_2(a)$$ first. $$0\leq I_2(a) \leq \int_{\epsilon}^1 \frac{a}{\epsilon^{1-a}\sqrt{1-t}}dt = \frac{a}{\epsilon^{1-a}}\cdot 2\sqrt{1-\epsilon}.$$ Now let's examine $$I_1(a)$$. Compare it with $$\hat{I}_1(a) = \int_0^\epsilon \frac{a}{t^{1-a}\sqrt{1-\epsilon}}dt = \frac{\epsilon^a}{\sqrt{1-\epsilon}}.$$ The difference $$\hat I_1(a) - I_1(a)$$ is \begin{align*} \int_0^\epsilon \frac{a}{t^{1-a}}\left(\frac{1}{\sqrt{1-\epsilon}}-\frac{1}{\sqrt{1-t}}\right)dt &\leq \int_0^\epsilon \frac{a}{t^{1-a}} \frac{\epsilon-t}{\sqrt{(1-\epsilon)(1-t)}(\sqrt{1-t}+\sqrt{1-\epsilon})} dt \\ &\leq \int_0^\epsilon \frac{a}{t^{1-a}}\cdot \frac{\epsilon}{2(1-\epsilon)^{3/2}}dt \\ &= \frac{\epsilon^{a+1}}{2(1-\epsilon)^2}. \end{align*} This implies that $$I_1(a) \leq \hat I_1(a)\leq I_1(a) + \frac{\epsilon^{a+1}}{2(1-\epsilon)^{3/2}},$$ or $$\frac{\epsilon^a}{\sqrt{1-\epsilon}} - \frac{\epsilon^{a+1}}{2(1-\epsilon)^{3/2}} \leq I_1(a) \leq \frac{\epsilon^a}{\sqrt{1-\epsilon}}.$$ Combining with our estimate of $$I_2$$, $$\frac{\epsilon^a}{\sqrt{1-\epsilon}} - \frac{\epsilon^{a+1}}{2(1-\epsilon)^{3/2}} \leq I(a) \leq \frac{\epsilon^a}{\sqrt{1-\epsilon}} + \frac{a}{\epsilon^{1-a}}\cdot 2\sqrt{1-\epsilon}.$$ This holds for arbitrary small $$\epsilon > 0$$.

It is clear that $$I(a)$$ is monotone w.r.t. $$a$$, so taking the limit on both sides of the left inequality above, $$\lim_{a\to 0^+} I(a) \geq \frac{1}{\sqrt{1-\epsilon}} - \frac{\epsilon}{2(1-\epsilon)^{3/2}}$$ for any small $$\epsilon > 0$$, which implies that $$\lim_{a\to 0^+} I(a)\geq 1$$. Similarly using the upper bound one obtains that $$\lim_{a\to 0^+} I(a)\leq 1$$.

• I'll try to follow your steps. What is the range of epsilon? Is it between 0 to 1? Mar 21, 2020 at 11:31
• How do you know what steps to carry out? Does it come with practice? Mar 21, 2020 at 11:40
• @Deb.U Yes, imagine that $\epsilon$ is a very small number. Mar 21, 2020 at 12:23
• @RyderRude Splitting an integral into pieces and handling each piece separately is a typical argument in analysis. How you split the integral usually depends on the singularity points. Mar 21, 2020 at 12:36
• @Deb.U If it looks all right to you, would you please accept it as an answer? Mar 22, 2020 at 1:59

This is an example of the integration of differential binom, i.e.

$$\int x^m(a + bx^n)^p dx,$$ where $$a, b \in \mathbb{R}$$ (irrational numbers), $$m, n, p \in \mathbb{Q}$$ (rational numbers).

It can be expressed in elementary functions in the three following cases only:

1) $$p$$ is integer. Then, the following change of variable is used: $$x = t^k$$, $$k$$ is a common denominator of $$m$$ and $$n$$.

2) $$\frac{m+1}{n}$$ is integer. Then, the following change of variable is used: $$a + bx^n = t^s$$, where $$s$$ is a denominator of $$p$$.

3) $$p+\frac{m+1}{n}$$ is integer. Then, the following change of variable is used: $$ax^{-n} + b = t^s$$, where $$s$$ is a denominator of $$p$$.

$$x^m(a+bx^n)^p = (1-x^\frac{1}{a})^{-\frac{1}{2}} \Leftrightarrow$$
$$m = 0$$, $$n = \frac{1}{a}$$, $$p = -\frac{1}{a}$$, $$a = 1$$, $$b = -1$$.