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

Question

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?

Answer 1

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 $t \, e^{-t} \in L^1(\mathbb R)$.

Answer 2

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.

Answer 3

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

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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 $t\to0$.