An Impossible Integral?

Look at this cute little definite integral. You might say, why did you even bring up this question? The usual way to solve this is to first recall that the antiderivative of 1 over x squared is minus 1 over x, then directly apply the upper and lower limits, and substitute x equals 1, which gives minus 1, then substitute x equals minus 2, which gives minus 1 divided by minus 2, that is 1 over 2, and then subtract the second value from the first. Doing that gives minus 1 minus 1 over 2, which simplifies to minus 3 over 2, and that's it. But you will be surprised to know that this answer is incorrect. This definite integral is not as innocent as it looks. So, be careful. Now, this whole thing is wrong because it only works when the function, or this integrand, is well-defined and continuous on the entire interval from minus 2 to 1, which is clearly not the case here, since 1 over x squared blows up at x equals 0. The graph of 1 over x squared looks like this, from x equals minus infinity to 0. The curve stays above the x-axis and rises sharply upward as it approaches 0. Then the value shoots up as it gets close to 0, never touching it. And the same behavior repeats on the other side, because from 0 to x equals plus infinity, the graph again stays positive and shoots upward near 0, clearly showing a break at x equals 0. So the values of 1 over x squared are always positive, and the graph always stays above the x-axis. And therefore, even by just looking at this picture, it is clear that getting a negative answer like minus 3 over 2 is not possible. Now, because of this break, the problem must be treated as an improper integral, where the interval is split at 0 and limits are taken separately from the left and from the right. This means that this integral can be written as the limit of the integral of 1 over x squared from x equals minus 2 to some variable p, where p moves closer and closer to 0 from the left side, or it is 0 with minus as superscript. Then plus the limit of the integral of 1 over x squared, where x equals some other variable q to 1, where q moves closer and closer to 0 from the right side, or it is 0 with plus as superscript. Now, here is the deal. If both these limits converge, which means if we get finite values for both these limits, then and only then can we say that the original integral has a proper value, else we say it diverges, which means it doesn't approach a single finite number, and instead it either grows infinitely large or it oscillates wildly. Let us now focus only on the left part to understand what is really happening. First, we take the antiderivative in the usual way and get minus 1 over x. Next, we apply the upper and lower bounds by substituting x equals p, which gives minus 1 over p, and then substituting x equals minus 2, which gives minus 1 over minus 2, that is 1 over 2. After this, we subtract the lower value from the upper value, so we get minus 1 over p minus 1 over 2. Now comes the main part. As p moves closer and closer to 0 from the negative side, minus 1 over p becomes larger and larger in the positive direction and does not stop growing. Because of this, the whole expression keeps increasing without bound and never settles to a fixed number. This shows clearly that this left side limit does not converge. Now, as mentioned earlier, since one part already blows up, the full integral cannot exist. Hence, we conclude that this integral diverges. If you enjoyed this video, please don't forget to like, share, and subscribe to our channel. Goodbye.