<p>In class today we got into a debate as to whether the definite integral from -2 to 2 of [1/x (dx)] is divergent or zero, and why. You would think it would be zero but the math seems to lead to it being divergent as you are taking one-sized limits on each side of the y-axis which are each divergent. So can 2 divergent sides cancel out and be zero, or is it just doubly divergent and not zero? I am curious to see what my fellow AP calc scholars think of this issue!!</p>
<p>It would be 0 because you can seperate it into 2 seperate improper integrals, one from -2 to 0 and the other from 0 to -2 and limit the equation as x approaches 0 for each integral</p>
<p>I believe it would be divergent. You can break it up like in post #2, but it is -inf and +inf on each side, and infinite quantities don’t obey the laws of algebra. In particular, inf - inf ≠ 0.</p>
<p>Generally speaking, you cannot add or subtract infinities because they could be different values (to the extent that we can apply values to infinity), i.e. they could grow at different rates.</p>
<p>However, I believe that the definite integral of 1/x dx from -2 to 2 is indeed zero, and that you can indeed subtract these infinities. Why? Because the graph of the function is rotationally symmetrical, which means that the area under the function on the negative side from -2 to 0 must equal the area under the function on the positive side from 0 to 2.</p>
<p>It’s obviously zero. Graph it. The areas you’ll find are symmetrical about the y axis, thus one would be negative, there other positive, but (ignoring the sign) the same number. It would be like integrating y = x from -2 to 2. The areas are equal, so they would cancel out and yield you zero.</p>
<p>If memory serves, the integral diverges. The symmetry is largely irrelevant. Since neither of the limits technically exists (the limits each tend toward infinity), they cannot be subtracted. Keep in mind that the property that assumes that the definite integral from a to c of f(x) dx is only equal to the definite integral from a to b of f(x) dx plus the definite integral from b to c of f(x) dx if both a <= b <= c AND that the actual integral exists.</p>
<p>For this particular situation, if you can’t calculate the integral from -2 to 0 of 1/x dx, and you can’t calculate the integral from 0 to 2 of 1/x dx, you can’t assume that they cancel each other out.</p>
<p>Maybe I’m missing something here, but why can’t you just take the anti derivative of 1/x and evaluate it from -2 to 2? You just get ln2 - ln2 which equals 0…</p>
<p>^ because 0 is between -2 and 2 and you can’t take ln 0</p>
<p>Right. Forgot all this stuff I’ve been learning only applies to continuous functions.</p>
<p>^Exactly right, this integral doesn’t exist. It is correct to calculate the integral of 1/x by evaluating ln(abs(x)) at the endpoints of the interval only when 1/x is continuous everywhere within the interval (and at the endpoints as well), and 1/x is not continuous at x=0. It seems like you should just get a large negative number + the same large positive number, for a total of zero, but as someone pointed out, infinities do not add like regular numbers.</p>
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<p>Plain out untrue. Evaluate the integral of sin x dx from 0 to 2pi. You’ll get zero because two equal areas (one negative, and one positive) cancel out. What we have here is the same concept.</p>
<p>Because it is from -2 to 2 (as opposed to say, -5 to 2) the areas are equal. It IS the same infinity. We are not talking about lim x–> infinity of e^x vs lim x –> infinity of x^2. It’s the same thing heading at the same rate to infinity (it’s the same function, for pete’s sake). Because of the boundaries (-2 to 2), the areas are the same, which means you get zero. </p>
<p>The fact that they are equal areas makes all the different.</p>
<p>EDIT:
I plugged it into a TI-89 and I got an undef. so perhaps the whole “The integral doesn’t exist” might actually be true. If it diverged, it’d show the little infinity symbol so, according to the TI, it doesn’t do that either.</p>
<p>Hah, now I see why this topic came up. We just covered indefinite integrals today in class. Anyway, it would appear the the integral from -2 to 2 of 1/x is definitely divergent. While intuitively, the integral is obviously 0, it cannot be legitimately calculated because the limits do not exist.</p>
<p>So by that logic the integral should be undefined, not divergent.</p>
<p>I’ve changed my mind. It’s not zero, it’s not divergent, it’s undefined.</p>
<p>Well, the TI-89 agrees with you, Secret Azn Man.</p>
<p>Actually, Xav, the difference between your example and the question we’re looking at is the discontinuity.</p>
<p>What was interesting to me about this entire piece is that most of the textbooks that I’ve seen dodge the case of discussing the symmetry altogether. The examples they’ll give are more like Example 7 from [Pauls</a> Online Notes : Calculus II - Improper Integrals](<a href=“http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx]Pauls”>http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx) .</p>
<p>Maybe they feel that the non-symmetrical examples are not convincing (they certainly seem like they should equal 0). But even an example like that one could give a kid fits if he tried to break it up into integral(1/x^3 from -2 to 2 dx) + integral(1/x^3 from 2 to 3 dx)…</p>
<p>I think the reason for the “divergent” terimnology is that an integral is essentialy a summation of individual areas, and much like a series might diverge, a summation of those areas would diverge as well.</p>
<p>The integral of that is undefined for sure, but I wouldn’t label it divergent. Functions are divergent; integrals are numerical values. I think labeling it undefined or D.N.E. would be the best bet.</p>
<p>So this poses the question of whether the integral of 1/x dx from -1 to 2 is positive infinity (divergent) or undefined. I’m going to say that the integral of any discontinuous interval is undefined, even though the infinity from 0 to 2 is “bigger” than the infinity from -1 to 0.</p>