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

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

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    $\begingroup$ 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 $\lim_{t\to \infty} \int_{-t}^t x \; \mathrm d x \neq \int_{-\infty}^{\infty} x \; \mathrm d x.$ $\endgroup$ – Aaron Meyerowitz Feb 20 '17 at 7:35
  • $\begingroup$ @AaronMeyerowtiz Thank you! $\endgroup$ – Niklas Feb 20 '17 at 7:36

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