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

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.

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  • $\begingroup$ Your mistake was considering that the numerator is equivalent to $x^{\frac{1}{6}}$ when $x$ approaches $0$. $\endgroup$
    – Bowei Tang
    Commented May 17 at 12:25

2 Answers 2

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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 $\alpha\in \left(\dfrac{7}{9},\dfrac{4}{3}\right)$.

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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 $4/3$.

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