What is wrong with this method of gauging the convergence of an improper integral?

Question

$$\int_0^{\infty} \frac {(e^{-x} + x - 1)^{\frac 16}} {x^{\alpha}\sqrt {1+x^{\alpha}}} \mathrm{d} x$$

I am convinced that this integral doesn't converge for any $\alpha \leq 0$ . The alpha greater than zero case isn't as immediate however. I was thinking that maybe by considering the asymptotic character of the integrand function and requiring the equivalent function to converge could cover some ground. What I was thinking is $f \sim x^{\frac 16 - \frac 32\alpha} \text{ when } x \rightarrow +\infty$ (with f being the integrand function) and from there deduce that $\alpha > \frac 79.$ Also the same could maybe be applied to the asymptote that forms around zero and deduce that $\alpha < \frac 76$ .

All this was alright in my head but a friend of mine who is better at maths than me said this was somehow wrong but I couldn't get from them exactly why.

Answer 1

Your asymptotics for $x\to+\infty$ is correct, and we indeed deduce that $\alpha>\dfrac{7}{9}$ .

Now, for $x\to 0^+$ , using the Taylor Series of the exponential we find $$e^{-x}+x-1=\dfrac{x^2}{2}+o(x^3)$$

So $$(e^{-x}+x-1)^\frac{1}{6}\sim x^{1/3}$$

Thus, the integrand behaves as $x^{\frac{1} {3}- \alpha}$ when $x\to 0^+$ .

As $\displaystyle\int_{0}^ {1}x ^p\,dx$ converges iff $p>-1$ we need $\dfrac{1} {3}- \alpha>-1\iff \alpha<\dfrac{4}{3}$ .

Therefore, the improper integral from $0$ to $\infty$ will converge iff $\alpha\in \left(\dfrac{7}{9},\dfrac{4}{3}\right)$ .

Answer 2

Your estimate for $\alpha \rightarrow 0$ is indeed wrong, because $e^{-x} + x - 1$ has a double root at $0$ . So the numerator scales a $x^{1/3}$ , and the upper bound for $\alpha$ is $4/3$ .