A question about improper integral form $-\infty$ to $\infty$

Question

In want to calculate the improper integral $$ \int_{-\infty}^{\infty}\frac{x}{e^{|x|}}dx $$ As one can show, we have that $$ \int_{0}^{\infty}\frac{x}{e^{|x|}}dx=1\qquad \int_{-\infty}^{0}\frac{x}{e^{|x|}}dx=-1 $$ Can I deduce that $$ \int_{-\infty}^{\infty}\frac{x}{e^{|x|}}dx=0 $$ or should I first prove that $$ \int_{c}^{\infty}\frac{x}{e^{|x|}}dx\qquad \int_{-\infty}^{c}\frac{x}{e^{|x|}}dx $$ converges for any arbitrary $c$ and that their sum does not depend upon $c$? Does in general it suffices to verify the converges of the integral for specific $c$? Thanks!

Answer 1

I think you have $$\int_{a}^{b}\frac{x}{e^{|x|}}dx = -\dfrac{(|b|+1)}{e^{|b|}} + \dfrac{(|a|+1)}{e^{|a|}}$$

and you can let $a \to -\infty$ and $b \to \infty$ independently to get a result of $0$ since each term tends towards $0$.

Your suggestion of adding together $$\int_{0}^{\infty}\frac{x}{e^{|x|}}dx=1 \text{ and } \int_{-\infty}^{0}\frac{x}{e^{|x|}}dx=-1$$ is sufficient, providing they are true (and they are)