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I'm trying to show that the integral

$$I:=\int\limits_0^2x^{-1}e^{-x}\mathrm dx$$

diverges.

My attempt: We know that the problem is in the $x \to 0$ part so we have that $x \to 0$ implies $e^{-x} \to 1$ and $x^{-1} \to \infty$. We have that the function is not defined at zero so if we set this integral correctly we have that

$$I= \lim_{\xi \to 0}\int_\xi^2x^{-1}e^{-x}\mathrm dx $$

I think is not possible to interchange the series of $e^x$ with the integral because it is a divergent integral of a series. (Is this correct? Thinking in the answer given HERE).

We can use that $$E\mathrm i(-x) = \int x^{-1}e^{-x}\mathrm dx+\mathrm{Const.}$$

So $$I = E\mathrm i(-2) - \lim_{\xi \to 0}E\mathrm i(-\xi)$$

Can anyone set a hint for what can I do?

Martin Sleziak
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R.W
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    You could compare it to $x^{-1}e^{-2}$. – Jonas Meyer Oct 26 '17 at 22:31
  • Using that $x^{-1}e^{2}> x^{-1}e^{-x}$ and that $ln(x) \to \infty$? @JonasMeyer – R.W Oct 26 '17 at 22:32
  • But I don't expect that the integral converges. Clearly diverges. I'm trying to prove it. – R.W Oct 26 '17 at 22:36
  • @Rafael: Check your inequality. And you mean $e^{-2}$? – Jonas Meyer Oct 26 '17 at 22:38
  • No, I just mislead the inequality. If $x^{-1}e^{-2} < x^{-1}e^{-x}$ then using that the integral preserves inequality we have that the integral in $x^{-1}e^{-2} $ goes to infinity and then we must have that the integral in $x^{-1}e^{-x}$ goes to infinity? – R.W Oct 26 '17 at 22:44
  • @JonasMeyer I think I get it. Post this hint as an answer and I'll accept it. It is a very usefull one and solves the problem completely. Thanks a lot for your hint! – R.W Oct 26 '17 at 22:47

4 Answers4

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Since $x^{-1}e^{-x} \ge x^{-1}e^{-1}$ on $(0,1),$ and $\int_0^1 x^{-1}\,dx = \infty,$ we have $\int_0^1 x^{-1}e^{-x}\,dx=\infty.$

zhw.
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Let $g(x)=\frac{e^{-x}-1}{x}$ for $x\in (0,2]$. Then $g$ can be made continuous at $x=0$ by defining $g(0)=-1$.

But then $$\int_{\xi}^2 x^{-1}e^{-x}\,dx = \int_{\xi}^2 g(x)\,dx + \int_{\xi}^2 \frac{dx}{x}$$

Since $g$ is continuous on $[0,2]$ you get:

$$\lim_{\xi\to 0+}\int_{\xi}^2 x^{-1}e^{-x}\,dx = \int_{0}^{2}g(x)\,dx + \lim_{\xi\to 0+} \int_{\xi}^{2}\frac{dx}{x}$$

Basically, $x^{-1}e^{-x}$ behaves "like" $\frac{1}{x}$ when $x$ is near zero.

Thomas Andrews
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Note that \begin{align} \lim_{\xi\to 0}\int_\xi^2x^{-1}e^{-x}\mathrm dx =& \lim_{\xi\to 0}\int_\xi^2x^{-1}D_x\big(-e^{-x}\big)\mathrm dx \\ =& \lim_{\xi\to 0}\left[ x^{-1}(-e^{-x})\Big|_{\xi}^{2}-\int_\xi^2D_x\big(x^{-1}\big)\big(-e^{-x}\big)\mathrm dx \right] \\ =& \lim_{\xi\to 0}\left[ \frac{1}{x}(-e^{-x})\Big|_{\xi}^{2} - \int_\xi^2\left(\frac{1}{x^2}\right)(-e^{-x})\mathrm dx \right] \\ =& \lim_{\xi\to 0}\left[ \underbrace{-\frac{1}{2}(e^{-2})+\frac{1}{\xi}(e^{-\xi})}_{\to \infty} + \underbrace{\int_\xi^2\left(\frac{1}{x^2}\right)(e^{-x})\mathrm dx}_{<\infty} \right] \end{align}

Elias Costa
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By convexity $e^{-x}\geq \max(1-x,0)$, hence $$ \int_{0}^{2}\frac{e^{-x}}{x}\,dx \geq -1+\int_{0}^{1}\frac{dx}{x} $$ trivially implies divergence.

Jack D'Aurizio
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