Question

Question from real analysis (improper integrals).

Answer 1

Does the following integral exist?

Answer 2

\[\int\limits_{-\infty}^{\infty} dx\]

Answer 3

I'll answer yours if you answer mine

Answer 4

Whats your question

Answer 5

I'll send it in a message

Answer 6

@zepdrix please help

Answer 7

you have my answer?

Answer 8

It's gonna diverge. \[\int_{-\infty}^\infty dx = x|_{-\infty}^\infty = \infty -(-\infty) = \infty\]

Answer 9

@Kainui in the question it is given, Justify your answer/

Answer 10

@welshfella

Answer 11

well seems to me that Kainui's solution justifies the answer.

Answer 12

infinity + infinity = infinity

Answer 13

Its told, to justify.. But I dont think writing infinity+ infinity is allowed.. Though its the result.. I think we need to give a divergence test.

Answer 14

@Loser66

Answer 15

@HolsterEmission

Answer 16

@IrishBoy123

Answer 17

The integral indeed doesn't exist because it doesn't converge, but from a certain point of view ("certain" being the operative term) you could argue that its value is \(0\):
\[\int_{-\infty}^\infty x\,\mathrm dx=\lim_{\epsilon\to\infty}\int_{-\epsilon}^\epsilon x\,\mathrm dx=\lim_{\epsilon\to\infty}\left[\frac{x^2}{2}\right]^\epsilon_{-\epsilon}=\lim_{\epsilon\to\infty}\left(\frac{\epsilon^2}{2}-\frac{\epsilon^2}{2}\right)=0\]This limiting process is used to determine what's called the Cauchy principal value: https://en.wikipedia.org/wiki/Cauchy_principal_value

Answer 18

Its only dx

Answer 19

Oops, totally misread that.

In that case, yes, the integral still diverges. Even the CPV is indeterminate.

Answer 20

Great. I cannot recall but few years before I read something: If the Cauchy principal value diverges, then the improper integral will diverge. Thanks. :)