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

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

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.

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

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

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 $$4/3$$.