strubenuff

Is it possible? Remember, it's not defined at 0.

[ October 20, 2002: Message edited by: strubenuff ]</p>

Goliath

Delete me.

[ October 20, 2002: Message edited by: Goliath ]</p>

livius drusus

Five'll get you ten this belongs in <a href="http://iidb.org/cgi-bin/ultimatebb.cgi?ubb=forum&f=57" target="_blank">Science & Skepticism</a>. I'll take them odds.

Goliath

Oooohkay, I can't edit my post for some reason, so lemme try this:

Strubenuff,

First of all, I'll assume that by "integral," you mean Riemann integral (ie the integral that is taught in second semester calculus). In that case:

It most certainly is! There's a handy little theorem (can't remember who it's attributed to...Lebesque, I think?) that says that a real-valued function is Riemann integrable if and only if it has a countable number of discontinuities.

So, using \int_a^b f(x) dx to mean the (Riemann) integral from a to b of f(x) with respect to x, and remembering that |x|/x is 1 for x greater than 0, and -1 for x less than 0, we see that:

\int_-1^1 |x|/x dx=\int_-1^0 |x|/x dx + \int_0^1 |x|/x dx = \int_-1^0 -1 dx + \int_0^1 1 dx = -1+1 =0.

Sincerely,

Goliath

Edited to fix a small boo-boo, and to say: You can also see that it's Riemann integrable over (-1,0) and (0,1) via the definition of Riemann integral (ie taking limits of Riemann sums).

Or, you could be kinda slick and look at the following function: g(x)=|x|/x for x nonzero, and g(0)=0. Then it's easy to see that the Lebesque integral of g(x) on (-1,1) is zero, whence the Lebesque integral of |x|/x on (-1,1) is zero (since g(x)=|x|/x almost everywhere). So, since g(x) is continuous almost everywhere, it is Riemann integrable, and since it is bounded, the Riemann and Lebesque integrals must agree, whence the Riemann integral of |x|/x on(-1,1) must be the Lebesque integral of g(x) on (-1,1) (namely zero).

However, that's a bit pedantic.

(edited again to fix another small oops...now, back to grading!!!)

[ October 20, 2002: Message edited by: Goliath ]</p>

Friar Bellows

Draw the graph of this function. On the left side of the zero axis the "area under the curve" is -1, while on the right side it's 1, and so it cancels out to zero. Yes, I know this ignores the problem at x=0, but you sort of get the gut feeling that this can be ignored. See Goliath's post for the details, but I like looking at things graphically first.

ohwilleke

I'm in accord with Goliath. Intuitively, the function is defined over all put an infintessimally small point, and since integration is a measure of area, and since area is length times width, and width is infinitessimally small, the discontinuity shouldn't matter.

Friar Bellows

That's not always reliable. What if the "length" is infinitely large?

Shadowy Man

I agree with Friar Bellows that if you just look at the plot you can see that it is trivially zero, but to do it technically correct, I think you follow Goliath and to the contour integration excluding the pole at zero.

I remember learning contour integration in my math methods class. Ugh.

Feather

It's not an integration over the field of complex numbers, I think.

The function |x| is piecewise. You can only integrate over its proper domain.

Thus you must integrate over the interval (-1, 0) and add it to the integral over the interval (0, 1).

|x| is defined as -x, x &lt; 0; x, x&gt; 0. So this integral turns out to be zero.

Or you could just be a physicist and notice you're integrating symmetric even and odd functions over a symmetric interval and call the whole thing zero (as Friar pointed out).

Jesus Tap-Dancin' Christ

Irrlevan .Even if there was, given an absolute width of 0, the area must therefore be zero. Integrals work because you are taking an infinitely small, but still non-zero slice. On either side of zero on this graph, any and all values are +/-1, leaving no infinitely small yet non-zero width to multiply by.