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The improper integral is

$$\int^1_0\frac{e^x-1-x}{x^2\sqrt{x}}dx.$$

It is from an old calculus exam. It should be convergent. I've looked around on how to solve it but I've had no success at all.

Edit: $0\leq\frac{e^x-1-x}{x^2\sqrt{x}}\leq\frac{e^x}{x^3}$ when $0\leq x\leq 1$ and since $\lim_{x\to0^+}x^3\frac{e^x}{x^3}=1$ the improper integral is convergent. Is this right?

Gary
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per persson
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    Use the Maclaurin series for $e^x$ (you may need to justify using it, but it’s pretty straightforward). – Clayton Dec 20 '23 at 06:42
  • The integral of $1/x^3$ at $0$ is divergent, i.e., your conclusion is false. – Gary Dec 20 '23 at 07:10

2 Answers2

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Since $\frac{e^x-1-x}{x^2}\rightarrow\frac{1}{2}$ as $x\rightarrow 0^+$ we can find $\delta>0$ so that $\Big|\frac{e^x-x-1}{x^2}-\frac{1}{2}\Big|<\frac{1}{4}$ for any $x\in (0,\delta)$.

But then $$0<\frac{1}{4\sqrt{x}}< \frac{e^x-x-1}{x^2}\frac{1}{\sqrt{x}}<\frac{3}{4\sqrt{x}}$$ on the domain $x\in (0,\delta)$.

By direct comparison $\int_0^\delta \frac{e^x-x-1}{x^2\sqrt{x}}dx$ converges and hence so must your integral.

Matthew H.
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Near $x=0$, using the expansion of the exponential function you have: $$\frac{e^x-1-x}{x^2\sqrt x }\sim \frac{x^2/2}{x^2\sqrt x}=\frac12\frac1{\sqrt x} $$ Can you take it from here?

Sine of the Time
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  • Can I get a reference for this method? I've never used expansions to estimate integrals – per persson Dec 20 '23 at 21:07
  • @perpersson are you familiar with Taylor expansion? – Sine of the Time Dec 20 '23 at 21:12
  • yes both taylor and maclaurin – per persson Dec 20 '23 at 21:12
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    @perpersson the method is called asymptotic test (or comparison test) and you can use it for series and integrals. Basically, you study the asymptotic behaviour of the function using its Taylor expansion – Sine of the Time Dec 20 '23 at 21:15
  • can u possibly show me same method but with more rigor? I know $e^x=1+x+x^2/2+x^3B(x)$ where $B(x)$ is bounded near 0. But how do I insert this expression into the integral rigorously? $\int^1_0\frac{1+x+x^2/2 +x^3B(x)-1-x}{x^2\sqrt{x}} $doesn't make sense. – per persson Dec 21 '23 at 02:40