Does $\int_{-\infty}^{+\infty}te^{-|t|} \, dt$ converge? Good evening

Does $\displaystyle \int_{-\infty}^{+\infty}te^{-|t|} \, dt$ converge?

I have got a series of exercices without corrections so I carry on with a new exercice.
My solution :
$te^{-|t|}=\dfrac{t}{e^{|t|}}= \dfrac{t}{e^{\frac{|t|}{2}}}\times\dfrac{1}{e^{\frac{|t|}{2}}}$
So there exists $a>0$ such that $\forall |t|>a,\quad \dfrac{|t|}{e^{\frac{|t|}{2}}}<1\iff\dfrac{|t|}{e^{|t|}}<\dfrac{1}{e^{\frac{|t|}{2}}}$
Thus $\displaystyle \int_{-\infty}^{+\infty}e^{-\frac{|t|}{2}} \, dt=2\int_{0}^{+\infty}e^{-\frac{t}{2}} \, dt$
Let $F(x):=\displaystyle 2\int_{0}^{x}e^{-\frac{t}{2}} \, dt=-4\left[e^{-\frac{t}{2}}\right]_0^x=-4\left(e^{-\frac{x}{2}}-1\right)\underset{x\to+\infty}{\longrightarrow}4$
As $\displaystyle \int_{-\infty}^{+\infty}e^{-\frac{|t|}{2}} \, dt$ converges, then $\displaystyle \int_{-\infty}^{+\infty}te^{-|t|} \, dt$ converges.
Is it correct and is there something more concise?

You can also integrate by parts to an upper bound $R$ and take limits as $R \to \infty$:
$$
\int_0^R t \, e^{-t} \, dt 
= \left[ t \, \left( -e^{-t} \right) \right]_0^R - \int_0^R \left( -e^{-t} \right) \, dt
= -R \, e^{-R} + \int_0^R e^{-t} \, dt \\
= -R \, e^{-R} + \left[ -e^{-t} \right]_0^R
= -R \, e^{-R} - e^{-R} + 1 \\
\to 0 - 0 + 1 = 1
$$
Since $t \, e^{-t} \geq 0$ for $t \geq 0$ this shows that $t \, e^{-t} \in L^+([0, \infty))$ and symmetry implies $t \, e^{-t} \in L^1(\mathbb R)$.
