1. ## improper integrals

Suppose that

g(x)=1 for x<=0 and
g(x)=-1 for x>0

a) Find the limit as R-> infinity of integral (from -R to R) of g(x).
b) Repeat a), but from -R to R+n with n a positive number
c) Does the integral from -infinity to infinity of g(x) converge or diverge

2. I think the answers are:

a) 0 because of cancellation, but is there a good mathematical way to show this?

b) n, the -R and R region will essentially cancel leaving you with n, but I'm not sure if this is correct, and it is certainly not formal.

c) I think it diverges, but I'm not sure since the limit is 0...this seems contradictory.

A strange problem . . . I think I've solved it.
. . But my approach isn't very formal either.

Suppose that: .$\displaystyle g(x) \;=\;\begin{Bmatrix} 1 && x \leq 0 \\ -1 && x > 0 \end{Bmatrix}$

(a) Find: .$\displaystyle \lim_{R\to\infty} \int^R_{\text{-}R} g(x)\,dx$

The graph of $\displaystyle y \,=\,g(x)$ looks like this, for $\displaystyle x \in [-R,\,R]$
Code:
              |
|
* - - - *1
:       |       R
- + - - - + - - - + - -
-R       |       :
-1o - - - *
|
|

Since the integral represents area,
. . $\displaystyle \int^R_{\text{-}R} g(x)\,dx$ is the total area of the shaded regions below.
Code:
              |
|
* - - - *1
|:::A:::|       R
- + - - - + - - - + - -
-R       |:::B:::|
-1o - - - *
|
|

Since the area of $\displaystyle B$ is the negative of the area of $\displaystyle A$,
. . the total area is 0 (zero).

That is: .$\displaystyle \int^R_{\text{-}R}g(x)\,dx \;=\;0$

Therefore: .$\displaystyle \lim_{R\to\infty}\int^R_{\text{-}R} g(x)\,dx \;=\;0$

(b) Repeat (a), but: .$\displaystyle \lim_{R\to\infty} \int^{R+n}_{\text{-}R}\!\! g(x)\,dx$ . with $\displaystyle n > 0.$
The graph is a variation of the one in part (a).
Code:
              |
|
* - - - *1
|:::A:::|       R    R+n
- + - - - + - - - + - - * - -
-R       |:::B:::|::C::|
-1o - - - * - - *
|
|

Once again, the area of $\displaystyle B$ is the negative of the area of $\displaystyle A.$
. . And the area of $\displaystyle C$ is: .$\displaystyle -n$

Hence: .$\displaystyle \int^{R+n}_{\text{-}R}g(x)\,dx \;=\;-n$

Therefore: .$\displaystyle \lim_{R\to\infty}\int^{R+n}_{\text{-}R}\!\! g(x)\,dx \;=\;-n$

c) Does: .$\displaystyle \int^{\infty}_{\text{-}\infty} g(x)\,dx$ . converge or diverge?

As shown in part (a), it converges to 0.

4. I'm afraid Soroban is wrong about part (c). (Which is very, very unusual!).

The definition of $\displaystyle \int_{-\infty}^{\infty} f(x) dx$ is $\displaystyle \lim_{\alpha\to -\infty} \int_{-\infty}^a f(x) dx+ \lim_{\beta\to\infty}\int_a^\infty f(x)dx$
where $\displaystyle \alpha$ and $\displaystyle \beta$ go to infinity independently.

The integral in part (a), $\displaystyle lim_{R\to\infty}\int_{-R}^R f(x)dx$ is the "Cauchy principal value". If the integral itself exists, then so does the Cauchy principla value and they are equal. But the Cauchy principal value may exist when the integral itself does not.

The integral in (b) shows that the integral in (c) does NOT converge.

5. Thanks guys. I'm still a little confused on part c). So it diverges, essentially because the value of the limit that we found depended on the limit of integration, i.e., on n? I'm just not entirely sure how b) shows c) diverges...I think it is because of the dependence on n.