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 
 A: 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

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