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.


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?

  • $\begingroup$ I changed the $n$ in the limit variable to a $t$ like it was in the integrals, please tell me if I changed your question to something it was not intended to say $\endgroup$ – Alice Ryhl Feb 17 '15 at 20:11
  • $\begingroup$ These aren't really functions as you call them, they're some possibly divergent expression attempping to represent real numbers. $\endgroup$ – Alice Ryhl Feb 17 '15 at 20:12
  • $\begingroup$ I just posted the question verbatim from the textbook, it really doesn't give more context than that. $\endgroup$ – user3776749 Feb 17 '15 at 20:16


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

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

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  • $\begingroup$ Perfect. Thanks for putting me on the right track. $\endgroup$ – user3776749 Feb 19 '15 at 20:37

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 $t$.

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