Why can I not integrate $\frac{1}{x^2}$ over $0$?

Question

The question is: $$\int_{-2}^{1} \frac{1}{x^2} dx$$

My solution is by fundamental rule is : $$\tfrac{1}{-2+1} \cdot x^{-2+1} \Big|_{-2}^1 = \frac{-3}{2}$$): But the solution is said that is a diverges.Why it is ?

Answer 1

Since you are integrating over an infinite discontinuity at $x=0$, you must split the limit into separate integrals over each interval over which the function is continuous. Thus, $\displaystyle I=\int_{-2}^{-h}\frac{1}{x^{2}}\,\mathrm{d}x+\int_{h}^{1}\frac{1}{x^{2}}\,\mathrm{d}x$, where $h$ is a number close to $0$. This allows you to treat each of these two integrals as you normally would.

So you can see that the integral is actually $\displaystyle -\frac{3}{2}+\frac{2}{h}$. This shows you that even though the "finite part" of the solution is what you have shown, there is still a divergent part which overwhelms it.

In effect, the curve $\cfrac1{x^2}$ does not approach infinity fast enough around $0$ so it has infinite area around the discontinuity. In general, you cannot ignore discontinuities when integrating over them.

Answer 2

First, do you understand what an integral is? You don't seem to give any indication that you do! Take a "Riemann sum" to approximate the integral using intervals of length $\Delta x$. One of those intervals must include x= 0. We can take the endpoints of that interval to be $x_a> 0$ and $x_a- \Delta x< 0$. We can, arbitrarily, choose $x_0= \Delta x/2$ as the point in that interval at which to evaluate the interval. The contribution to the integral of that interval is $f(x_0)\Delta x$$= \frac{1}{x_0^2}\Delta_x$$= \frac{4}{(\Delta x)^2}\Delta x$$= \frac{4}{\Delta x}$. Now, as $\Delta x$ goes to 0, to give the actual integral, that goes to infinity.

That is why the integral does not exist.

Answer 3

The answer to your question,

Why can I not integrate $\frac1{x^2}$ over $0$?

is quite simple: Because the function is not integrable on an interval that includes $0$.

Answer 4

$f(x)=1/x^2$ is even, thus we can consider the improper integrals $\displaystyle{\int_{0}^{2}}(1/x^2)dx$ and $\displaystyle{\int_{0}^{1}}(1/x^2)dx$.

$G(a):=\displaystyle{\int_{a}^{1}}(1/x^2)dx$ for $a >0$.

$G(a)=-1/x \big ]^{1}_{a}= -1+1/a$.

$\lim_{a \rightarrow 0}G(a) =$

$\lim_{a \rightarrow 0} (-1+/a)$, does not exist.

Hence the improper integral

$\displaystyle{\int_{0}^{1}}(1/x^2)dx= \lim_{a \rightarrow 0} \int_{a}^{1}(1/x^2)dx$ does not exist.

What about $\displaystyle{\int_{0}^{2}}(1/x^2)dx$?

https://en.m.wikipedia.org/wiki/Improper_integral

Answer 5

Well, actually you can. Moreover, you can do it in two different ways. There is a theory of hyperfunctions that allows this and other theories as well...

But I will try to explain the thing in simple language.

You can integrate it either including the very point $x=0$ and not including it. And you will get different results.

You may ask, why? Why inclusion or not of just one point can change the result? The answer is that the function's singularity at zero behaves somewhat similar to how Dirac Delta "function" behaves: integral over this single point is not zero (and this is connected to the behavior of the function on the complex plane as well).

If you include the singularity the integral is called the Hadamard finite part: $$\mathcal{H}\int_{a}^b 1/x^2 dx= \lim_{\varepsilon \to 0^+} \left\{ \int_a^{x-\varepsilon}\frac{1}{x^2}\,dt + \int_{x+\varepsilon}^b\frac{1}{x^2}\,dt -\frac{1}{\varepsilon}\right\}$$ If you exclude the singularity, it will not be denoted by any special symbol: $$\int_{a}^b 1/x^2 dx= \lim_{\varepsilon \to 0^+} \left\{ \int_a^{x-\varepsilon}\frac{1}{x^2}\,dt + \int_{x+\varepsilon}^b\frac{1}{x^2}\,dt\right\}$$

So, the function in the first case behaves as if exactly at zero it had infinitely negative area. You can look at its antiderivative $-1/x$ and see that at zero it jumps from being infinitely positive to infinitely negative.

In other words, if you include the strange behavior of this function at zero, your conjectured solution would be correct.

Now, if you exclude the singularity, then... The area will be infinite. Moreover, it will be not just "infinite", but the following two divergent integrals will be equal: $\int_{-\infty}^\infty 1/x^2 dx=\int_{-\infty}^\infty 1 dx$. And, it will also behave as $2\pi\delta(0)$ in integral transforms... So, the integral $\int_{-2}^1 1/x^2 dx=2\pi\delta(0)-3/2$.

Answer 6

Analyticwise you can take any path connecting $-2$ and $1$ while omitting $0$, that is you could integrate like this $$\int_{-2}^{-\epsilon} \frac{1}{x^2} \, {\rm d}x + \int_{\gamma_\epsilon} \frac{1}{x^2} \, {\rm d}x + \int_\epsilon^1 \frac{1}{x^2} \, {\rm d}x$$ where $\gamma_\epsilon$ is the half-circle of radius $\epsilon>0$ starting at $-\epsilon$ and going clockwise to $\epsilon$. Parametrizing this circle by $x=\epsilon \, e^{it}$ for $t\in(\pi,0)$, the above line evaluates to $$ \left(\frac{1}{\epsilon} - \frac{1}{2}\right) - \frac{2}{\epsilon} + \left(-1+\frac{1}{\epsilon}\right) = -\frac{3}{2} \, .$$

So your result does make sense!