# Show that integral $\int_{-t}^tx\,dx=0$ limit as t approaches infinity

I've got it down to the the end but it ends up becoming infinity-infinity which is indeterminate?

• That integral is identically zero for all $t$. Hence, the limit is trivially zero. – Crostul Feb 20 '17 at 8:09
• You are right, you get an indeterminate form. But you can very well compute its limit. – Yves Daoust Feb 20 '17 at 8:23

## 1 Answer

We have $$\int_{-t}^t x \; \mathrm d x = \frac{1}{2} \left((-t)^2 -t^2 \right)=0.$$ Hence $$\lim_{t\to \infty} \int_{-t}^t x \; \mathrm d x =0.$$

Edit: Let $f:\mathbf R \to \mathbf R$ be a function. We set

$$\int_{-\infty}^\infty f(x) \; \mathrm{d} x := \lim_{\alpha \searrow-\infty} \int_\alpha^c f(x) \; \mathrm{d} x + \lim_{\beta \nearrow \infty} \int_c^\beta f(x) \; \mathrm{d} x$$ where $-\infty < c <\infty$ if and only if the both integrals on the right side are finite. That means that we have $$\lim_{t\to \infty} \int_{-t}^t x \; \mathrm d x \neq \int_{-\infty}^\infty x \; \mathrm d x$$ in your case because both integrals (on the right side) would be divergent.

• You mean limit as $t \to \infty$ So $\lim_{t\to \infty} \int_{-t}^t x \; \mathrm d x =\lim_{t\to \infty}0=0$. Also, for the questioner, let me add that $\lim_{t\to \infty} \int_{-t}^t x \; \mathrm d x \neq \int_{-\infty}^{\infty} x \; \mathrm d x.$ – Aaron Meyerowitz Feb 20 '17 at 7:35
• @AaronMeyerowtiz Thank you! – Niklas Feb 20 '17 at 7:36