Proof of divergence of $ \int_{0}^{2} x^3e^{{1} / {x}} dx $ using multiple partial integration

Question

I have to calculate this integral with a singularity at $x=0$ . $$ \int_{0}^{2} x^3e^{{1} / {x}} dx $$

I came to the conclusion that it diverges and I know it is right, but I am not sure about my way of doing it.

I first want to use partial integration multiple times:

$$ \int_{}^{} x^3e^{{1} / {x}} dx = x^3 \left(\int_{}^{} e^{{1} / {x}} dx \right) - \int_{}^{} 3x^2 \left(\int_{}^{} e^{1/x} dx \right) dx $$

Now I want to continue partial integration where $x^3$ gets derivived until it is a real number. Then we end up with $ \int_{}^{} e^{1/x} dx $ at at least one place.

So if I $\int_{0}^{2} e^{1/x} dx$ diverges or converges $\int_{0}^{2} x^3e^{{1} / {x}} dx$ also does.

Then I use the taylor series to check it.

$$\int e^{1/x} dx =\int1+\frac{1}{x}+\frac{1}{x^22!}+\frac{1}{x^33!}+\frac{1}{x^44!}+... \\ = C + x + \ln x-\frac{1}{2!}\frac{1}{x}-\frac{1}{3!}\frac{1}{2x^2}-\frac{1}{3!}\frac{1}{3x^3}-...$$

$$ lim_{a \to 0} \int_{a}^{2} e^{1/x} dx = " \left( x + \ln 2-\frac{1}{2!}\frac{1}{2}-\frac{1}{3!}\frac{1}{2 \cdot 2^2}-\frac{1}{3!}\frac{1}{3 \cdot 2^3}-... \right) - \left( 0 + \ln 0-\frac{1}{2!}\frac{1}{0}-\frac{1}{3!}\frac{1}{2 \cdot 0^2}-\frac{1}{3!}\frac{1}{3 \cdot 0^3}-... \right) " $$

This would end up in something like $Real Number - (-\infty) = \infty$

That means $\int_{0}^{2} e^{1/x} dx$ diverges so $ \int_{0}^{2} x^3e^{{1} / {x}} dx $ also diverges.

Is this a valid proof? I am especially not sure about the partial integration part.

Answer 1

Put

$$t=\frac1x\implies dx=-\frac{dt}{t^2}$$

and thus the integral is

$$\int_\infty^{1/2}\frac1{t^3}e^t\left(-\frac{dt}{t^2}\right)=\int_{1/2}^\infty\frac{e^t}{t^5}dt$$

and as everything ispositive we can use comparison, say: $\;\cfrac{e^t}{t^5}\ge 1\;$ and we're done. Or simpler:

$$\lim_{t\to\infty}\frac{e^t}{t^5}\neq0$$

Answer 2

Since $\displaystyle\;x^3e^{\frac1x}=\sum\limits_{n=0}^{\infty}\dfrac{x^3}{n!x^n}>\dfrac{x^3}{4!x^4}=\dfrac1{24x}\;$ for any $\;x>0\;,$

it results that

$\displaystyle\int_{\varepsilon}^2\!x^3e^{\frac1x}\mathrm dx>\!\int_{\varepsilon}^2\!\dfrac1{24x}\,\mathrm dx=\left[\dfrac1{24}\ln x\right]_{\varepsilon}^2=\dfrac1{24}\left(\ln2\!-\!\ln\varepsilon\right)\;,$

for any $\,\varepsilon\in(0,2)\,.$

Since $\;\lim\limits_{\varepsilon\to0^+} \dfrac1{24}\left(\ln2\!-\!\ln\varepsilon\right)=+\infty\;,\;$ it follows that

$\displaystyle\int_0^2\!\!x^3e^{\frac1x}\mathrm dx=\lim\limits_{\varepsilon\to0^+}\int_{\varepsilon}^2\!\!x^3e^{\frac1x}\mathrm dx=+\infty\,.$