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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

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  • $\begingroup$ 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. $\endgroup$
    – user58955
    Mar 21, 2020 at 8:59
  • $\begingroup$ @user58955 that's exactly my problem, to show, after substitution, that the integral equals to 1 $\endgroup$
    – Deb. U
    Mar 21, 2020 at 9:30
  • $\begingroup$ 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. $\endgroup$
    – user58955
    Mar 21, 2020 at 9:40
  • $\begingroup$ Duplicate with answer here. $\endgroup$ Mar 21, 2020 at 14:05

3 Answers 3

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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 $0<x<1$ so $0<x^{\frac{1}{a}}<1$ 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 $0<x<1$ 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
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  • $\begingroup$ Thanks but I asked without beta function. That's the way I solved it before. $\endgroup$
    – Deb. U
    Mar 21, 2020 at 11:14
  • $\begingroup$ I did not see that $\endgroup$
    – Masoud
    Mar 21, 2020 at 11:14
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    $\begingroup$ Thank you, I understand your solution $\endgroup$
    – Deb. U
    Mar 21, 2020 at 14:11
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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 $\lim_{a\to 0^+} I(a)\leq 1$.

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  • $\begingroup$ I'll try to follow your steps. What is the range of epsilon? Is it between 0 to 1? $\endgroup$
    – Deb. U
    Mar 21, 2020 at 11:31
  • $\begingroup$ How do you know what steps to carry out? Does it come with practice? $\endgroup$
    – Ryder Rude
    Mar 21, 2020 at 11:40
  • $\begingroup$ @Deb.U Yes, imagine that $\epsilon$ is a very small number. $\endgroup$
    – user58955
    Mar 21, 2020 at 12:23
  • $\begingroup$ @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. $\endgroup$
    – user58955
    Mar 21, 2020 at 12:36
  • $\begingroup$ @Deb.U If it looks all right to you, would you please accept it as an answer? $\endgroup$
    – user58955
    Mar 22, 2020 at 1:59
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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 $p$.

In your example,

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

$m = 0$, $n = \frac{1}{a}$, $p = -\frac{1}{a}$, $a = 1$, $b = -1$.

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  • $\begingroup$ Thank you for your effort but that's only gives me a suggestion for the substitution which I tried somehow, I detailed in my question. My problem is to compute it after substitution. $\endgroup$
    – Deb. U
    Mar 21, 2020 at 9:21

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