$I = \int_{c^{-a}}^{\infty} \left(1-\left(1-\frac{b}{1+x^{1/a}}\right)^{t}\right)\mathrm{d}x$, as $t \to \infty$ How to to find a closed-form expression for the following integral as $t \to \infty$:
$$I = \int_{c^{-a}}^{\infty} \left(1-\left(1-\frac{b}{1+x^{1/a}}\right)^{t}\right)\mathrm{d}x,$$
where $a \in (0, 1)$, $b \in (0, 1]$, $c > 0$, and $t > 0$? 
My attempt: I tried using binomial expansion as $\left(1-\frac{b}{1+x^{1/a}}\right)^{t} = \sum_{k = 0}^{\infty}\binom{t}{k}\left(-\frac{b}{1+x^{1/a}}\right)^{k}$. Then
$$I = \int_{c^{-a}}^{\infty} \sum_{k = 1}^{\infty}\binom{t}{k}\left(-\frac{b}{1+x^{1/a}}\right)^{k}\mathrm{d}x.$$
I could not proceed from here. 
 A: $$
I(t) = \int_{c^{-a}}^{\infty} \left(1-\left(1-\frac{b}{1+x^{1/a}}\right)^{t}\right)\mathrm{d}x
 = a\int_{c^{-1}}^{\infty}y^{a-1} \left(1-\left(1-\frac{b}{1+y}\right)^{t}\right)\mathrm{d}y \\
= y^{a} \left(1-\left(1-\frac{b}{1+y}\right)^{t}\right)\Bigg|_{y=c^{-1}}^\infty  + {tb}\int_{c^{-1}}^{\infty}\frac{y^a}{(1+y)^2} \left(1-\frac{b}{1+y}\right)^{t-1}\mathrm{d}y\\  = K(t) + tb \int_0^{(1+c^{-1})^{-1}} \left(\frac{1-z}z\right)^a (1-bz)^{t-1} dz =: K(t) + J(t).
$$
Then, obviously, $K(t) = -c^{-a}+o(t^{-1})$, $t\to\infty$. The last integral can be analysed in a standard manner. Denote $d = (1+c^{-1})^{-1}$, and let $C$ be a large number. Rewrite the integral as
$$
\int_0^{C/(t-1)} \left(\frac{1-z}z\right)^a e^{(t-1)\log(1-bz)} dz + \int_{C/(t-1)}^d \left(\frac{1-z}z\right)^a e^{(t-1)\log(1-bz)} dz.
$$
Observe that $$\int_{C/(t-1)}^d \left(\frac{1-z}z\right)^a e^{(t-1)\log(1-bz)} dz\le \int_0^{d} z^{-a} e^{-Cb}dz = e^{-Cb}d^{1-a}/(1-a).$$
Further,
$$
\int_0^{C/(t-1)} \left(\frac{1-z}z\right)^a e^{(t-1)\log(1-bz)} dz\\
 = (t-1)^{a-1}\int_0^C u^{-a} \left(1-\frac{u}{t-1}\right)\exp\left\{(t-1)\log\left(1-\frac{bu}{t-1}\right)\right\}du \\
\sim t^{a-1}\int_0^C u^{-a}e^{-bu}du = (tb)^{a-1} \Gamma(1-a,C), t\to \infty.
$$
By letting $C\to\infty$ (I leave technicalities for you), we get 
$$
J(t) \sim (tb)^a\Gamma(1-a), t\to\infty;
$$
the error seems to be $O(t^{a-1})$. As a result, 
$I(t) = (tb)^a\Gamma(1-a) - c^{-a} + O(t^{a-1})$, $t\to\infty$.
