Is this proof for divergency/convergency valid?

Question

$$\int_{-\infty}^{0}x^{2}e^{x}dx$$ -The answer is convergent.

Iv'e written it as : $$\int_{-\infty}^{0}e^{\ln x^{2}}e^{x}dx$$

which equals to : $$\int_{-\infty}^{0}e^{x+2\ln x}dx$$

It is "safe" to assume that $e^{x+2\ln x}$<$e^{x}$

in the Expression $\int_{-\infty}^{b}a^{x}dx$, whenever a>1 the integral is convergent. e>1 so the answer is convergent.

is that a legit proof?

Answer 1

You have transformed your integral into, $$ \int_{-\infty}^{0}e^{x+2\ln x}dx$$

As you know $\ln x $ is not defined for $x<0$ and the integration is over $(-\infty, 0).$

I recommend integration by parts to evaluate the original integral.

Answer 2

$$I=\int_{-\infty}^0x^2e^xdx=\int_0^\infty u^2e^{-u}du=\Gamma(3)=2!=2$$ so I would say it is convergent

I have used the substitution $u=-x$