Question

$\displaystyle\rm \red{{ \lim_{ \alpha \to \infty }} \bigg( \int_{0}^1 \frac{1}{1 + {x}^ \alpha } \: dx \bigg)^{ \alpha } }$​

Answer 1

Answer:

Let's check separately convergence at 00 and 11, by splitting the integral at c∈(0,1)c∈(0,1).

If α≥0α≥0 there is no question about convergence at 00. If α<0α<0, you can consider

∫c01−xα1−xdx=∫c011−xdx−∫c0xα1−xdx

∫0c1−xα1−xdx=∫0c11−xdx−∫0cxα1−xdx

The first integral is not a problem, so we tackle the second one with the substitution t=1/xt=1/x to get

∫∞1/ctβ−1t−1dt

∫1/c∞tβ−1t−1dt

where β=−α>0β=−α>0. This is asymptotic to tβ−2tβ−2 and we have convergence if and only if β−2<−1β−2<−1, hence α>−1α>−1.

Hence our integral converges at 00 if and only if α>−1α>−1.

Now note that

limx→11−xα1−x=α

limx→11−xα1−x=α

so there is no real problem with convergence at 11.

Answer 2

We can evaluate this limit using the dominated convergence theorem. First, note that the integrand is positive and bounded by 1 for all x ∈ [0, 1]. Therefore, we can apply the dominated convergence theorem if we can show that the integrand converges pointwise to 0 as α → ∞.

Consider the function $f(x) = \frac{1}{1+x^\alpha}$ for fixed x ∈ [0, 1]. As α → ∞, we have:

$f(x) = \frac{1}{1+x^\alpha} \to \begin{cases} 1 & \text{if } x = 0 \\ 0 & \text{if } x \in (0, 1] \end{cases}$

Therefore, the integrand converges pointwise to the function g(x) = 0 for x ∈ (0, 1] and g(0) = 1. This function is integrable on [0, 1] since it is continuous except at a single point x = 0.

By the dominated convergence theorem, we have:

$\lim_{\alpha \to \infty} \left(\int_0^1 \frac{1}{1+x^\alpha} dx\right)^\alpha = \left(\int_0^1 \lim_{\alpha \to \infty} \frac{1}{1+x^\alpha} dx\right)^\alpha = \left(\int_0^1 g(x) dx\right)^\alpha = 1^\alpha = 1$

Therefore, the limit is equal to 1.