Improper Integral conceptual question: show that one diverges and the other converges.

Question

I'm asked to show that the first function is divergent, and the second is convergent.

$$\int_{-\infty}^{\infty} x dx $$

$$\lim\limits_{t \to \infty} \int_{-t}^t x dx$$

I'm sure I have all the pieces in front of me, but I'm just not putting them together.

EDIT:

Following the suggestion to try and compute the limits/integrals I've accomplished the following for the first expression:

$$\int_{-\infty}^{\infty} x dx = \int_{-\infty}^0 x dx+\int_0^{\infty} x dx$$ $$=\lim\limits_{t \to \infty} \int_{-t}^0 x dx + \lim\limits_{t \to \infty} \int_0^t x dx$$ $$=\lim\limits_{t \to \infty}[\frac{x^2}{2}|_{-t}^0] + \lim\limits_{t \to \infty}[\frac{x^2}{2}|_0^t]$$

Since $$\lim\limits_{t \to \infty}[\frac{(0)^2}{2}-\frac{(-t)^2}{2}]=-\infty$$ and $$\lim\limits_{t \to \infty}[\frac{(t)^2}{2}-\frac{0^2}{2}] =\infty$$

The entire integral is divergent.

Am I on the right track?

Answer 1

hint

try evaluating the integrals and compute limits, also note that the upper and lower infinity must have seperate limits: $$ \int_{-\infty}^{\infty} x\,dx=\lim_{a\to\infty}\lim_{b\to\infty}\int_{-a}^{b} x\,dx $$

edit:
Your solutions to the integrals you proposed in the questions show exactly why the first integral are divergent.

Answer 2

In the first integral, you have no conception of how the infinities were achieved; for example, the integral could be equal to

$$\lim_{M,N \to \infty} \int_{-M}^N dx \, x$$

In the second integral, however, $M = N=t$ and the integral is zero for all $t$.