An interesting integral

[trigkid213]

What is [attachment=1123:attachment] (The area between y=1/x and the x-axis from -1 to 2)? Please support your answer; there's a reason I posted this.

[DarkerLine]

Undefined.

You'd think it were simple, integrate 1/x and evaluate at the endpoints. However, there is an obvious discontinuity at 0, so you split the integral up into two and use limits to avoid the 0. However, neither of the limits converges, so you can't really do anything.

Basically, you end up with an "infinity - infinity" undefined form, and there's really not a lot you can do with that. If the infinities could just be kind to us and cancel, we'd be left with ln 2 as the answer. Life isn't that kind to us, however.

If you ever actually have to evaluate this integral, I'd take a step back and reconsider the approach you used to get that expression - it may also help you to figure out how to proceed.

[trigkid213]

My question is, in this case, why can't the infinities be nice to us and cancel. Since it is an odd function, [attachment=1124:attachment] should exactly equal -[attachment=1125:attachment], so they would cancel out.

[DarkerLine]

You cant say "A is exactly equal to B" when neither A nor B are even numbers.

The problem is that adding a number to either A or B doesn't change A or B, so assigning a value to their difference is rife with contradictions.

Which this guy didn't appear to understand.

[trigkid213]

Whoa...what?

[alexrudd]

[trigkid213]

[Weregoose]

[trigkid213]

[chipmaster]

photobucket.com is a good choice. imageshack.us is also nice.

[trigkid213]

Can this problem be done with contour integration? (I barely know anything about it)

[DarkerLine]

Okay. I've figured out what bothers me about letting the infinities "cancel" in this integral.

Say you split it up into two integrals, from -1 to 0 and from 0 to 2. Say, furthermore, that you let b->0 and integrate from -1 to -b and from b to 2. You now get ln b - ln 1 + ln 2 - ln b = ln 2. So far, so good.

But say you set the intervals up as -1 to -b and 2b to 2, letting b go to zero in the same way. Since if b were 0, these would be the same intervals, this should equal the same number, right?

But instead you get ln b - ln 1 + ln 2 - ln (2b) as your answer, which as b approaches 0 (or indeed even if it doesn't) becomes 0. Something of a problem?

I haven't done calculus in a couple years, I'm currently in a backwards math class which consists of flipping a coin ten times to see how it lands. So excuse me if I've used a taboo calculus method.

[thornahawk]

@trigkid, about that thought of yours as to why shouldn't the infinities cancel, try researching the concept of the "Cauchy principal value". ;)

thornahawk

[luby]

1/0 is infinity right? so if you divide 1/0 by 1/0 you get 1/0 * 0/1 = 0/0
it is a paradox

[Weregoose]

1/0 is undefined. The limit of 1/x as x approaches 0 is infinity. Don't confuse them with one another, and—for the sake of all that is peachy with the world—don't try to disprove them, either. There are already way too many absurd arguments on that subject.

[trigkid213]

OK, another way of looking at it. From -1 to 0 and from 0 to 1 the areas are going to cancel, up to the point where x actually equals 0. So they would cancel if we could explain that point. The problem is that there is a 0 in the denominator at that point, so it is undefined. Now, why is that any different from an integral like (x+1)(x-1)/(x+1) from, say, -3 to 3. This is going to be -6, even though the function at x=-1 is undefined. Why can't we likewise ignore the undefined point at 0 for 1/x?

[todlangweilig]

[trigkid213]

But why does it matter that it's removable?

[todlangweilig]

It matters because to find a definite integral, one needs a definite area. If I was to shoot say a pool ball straight up at x = 0 on the graph of f = 1/x, it would never hit a boundry. But with the second graph even though there is a hole, there is a definite place where the area under the curve is and a definite place where the area under the curve isn't.

Lets say we where to take a container, and for a minute pretend that the volume of this container was analogous to the area under our curve. How can we figure out how much our container holds if it has no bottom? In theory, it would have an infinite volume since it has no bottom bound. Let's imagine this container in space for a minute, just the container, all around this contain is just the void of space. What does the size of the container matter if it has no bottom. Lets say there are now two containers but both have infinite space in which to reside. Does it matter that one container is ln 2 bigger than the other? No, because as soon as I try to rationalize infinity as a number or a bound, I can make the other container bigger than the first. ie infinity + ln2 = infinity. Since the area of the right part of the graph is equal to infinity, and the left part is also equal to infinity, as far as I can reason, the area of the two graphs is equal.

Im not sure this is all valid but there are definitally some problems with trying to come up with a definite value for the integral. This kind of makes sense with the answer of infinity over infinity. But that being said, there maybe be some way to do the indefinite integral of f(x) = 1/x.

[leofox]

blablabla if this has been said already dont shoot me. I'm too lazy to read the thread ok.
The answer is 0.693.

The integral from -1 to 1 is 0, because the simmetry of the formula. Leaving us with the integral between 1 and 2, which can easily be calculated.