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How do I show that the following improper integral $$\int^1_0\frac{e^x-1-x}{x^2\sqrt{x}}dx$$ is convergent by using Taylor expansion rigorously?

So $e^x=1+x+x^2/2+x^3B(x)$ where $B(x)$ bounded close to zero. It doesn't make sense to just insert this in the integral directly. But maybe the following works? There exists a $\delta>0$ such that $\int^\delta_0\frac{1+x+x^2/2+x^3B(x)-1-x}{x^2\sqrt{x}}dx=\int^\delta_0\frac{1/2}{\sqrt{x}}+\sqrt{x}B(x)dx$ which is convergent since $B(x)$ is bounded.

per persson
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  • The problem arises near the $0$ bound, since there is there a division by zero. You have to find an equivalent of the integrand, and then you will know whether the integral is convergent. What you have done works too. – Jean-Claude Arbaut Dec 22 '23 at 12:24
  • You could use the following on $[0,,1]$:$$\frac12\le\frac12+\frac16x+\cdots=\frac{e^x-1-x}{x^2}\le e-2,$$since the Taylor series is increasing as it has positive coefficients. Now bound your integral between multiples of $\int_0^1x^{-1/2}dx$. – J.G. Dec 22 '23 at 12:39

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