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?
 A: Your approach is correct, but if you want something more concise:

Since the integrand is odd, it suffices to only consider $t>0$. Over that interval, we have
$$\int_0^\infty te^{-t}~\mathrm dt$$
Now apply the ratio test, noting that $te^{-t}$ is bounded for $t\to0$.
A: 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 $t \, e^{-t} \in L^1(\mathbb R)$.
A: The integrand function is odd. We just need the nature near $+\infty$.
Observe that
$$\lim_{t\to +\infty}t^\color {red}{2}\cdot te^{-t}=\lim_{+\infty}t^3e^{-t}$$
$$=\lim_{+\infty}e^{-t (1-3\frac {\ln (t)}{t})}=e^{-\infty}=0$$
So for enough large $t $,
$$0 <t^\color {red }{2}\cdot te^{-t}<1 \implies  0 <te^{-t}<\frac {1}{t^\color {red}{2}} $$
the integral is then convergent by comparison test.
