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Consider the integral $\displaystyle \int^1_{-1} x^{-1/2} dx$

The function is discontinuous at $x=0$

So we use improper integral:

$\displaystyle \lim\limits_{x \to 0^-} \int^x_{-1} x^{-1/2} dx + \lim\limits_{x \to 0^+} \int^1_x x^{-1/2} dx \tag1$

Instead if we avoid the limits and do it in the following way, we will definitely get the exact same result:

$\displaystyle \int^0_{-1} x^{-1/2} dx + \int^1_0 x^{-1/2} dx \tag2$

QUESTION

$(1)$ So why are limits necessary for improper integrals?

$(2)$ Does equation $(2)$ work for higher dimensional improper integrals? (i.e. by dividing our domain and making the discontinuous point at corners of each sub domain)

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    $\begingroup$ No, $x^{-1/2}$ is continuous in the domain $x>0$, $x=0$ belongs not to the Domain. $\endgroup$ – Dr. Sonnhard Graubner Jul 16 '19 at 18:15
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    $\begingroup$ Riemann (and more generally Darboux) integration is built on the function being bounded on a finite interval. The base theory does not generalize well outside of these realms, so we choose to give certain integrals a meaning and that is what is called improper integration. $\endgroup$ – Cameron Williams Jul 16 '19 at 18:51
  • $\begingroup$ Wait you have to many $x$s . . . you have it for the variable of integration and the limit, you can’t do that $\endgroup$ – gen-z ready to perish Jul 16 '19 at 21:06
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Since $x^{-1/2}$ is undefined at $0$, the integral $\displaystyle\int_{-1}^0x^{-1/2}\,\mathrm dx$ doesn't exist in the sense of Riemann integration. And, by definition, we have$$\int_{-1}^0x^{-1/2}\,\mathrm dx=\lim_{t\to0^-}\int_{-1}^tx^{-1/2}\,\mathrm dx.$$That is, by definition $\displaystyle\int_{-1}^0x^{-1/2}\,\mathrm dx$ means $\displaystyle\lim_{t\to0^-}\int_{-1}^tx^{-1/2}\,\mathrm dx$.

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  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jul 16 '19 at 18:50
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If the problem were, instead, $\int_{-1}^1 \frac{1}{x}dx$ then, since [tex]\lim_{x\to 0}Ln(x)[/tex] does not exist, that integral does not exist. But ignoring the discontinuity, $\left[log|x|\right]_{-1}^1= 0$. They are NOT the same.

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In order to do what you want to do, you will likely try to invoke the part of the Fundamental Theorem of Calculus which tells us that if $f(x)$ is continuous over $[a,b]$ then $\int_{a}^{b}f(x)dx$ exists, and moreover, is equal to $F(b) - F(a),$ where $F(x)$ is any antiderivative of $f(x)$.

However, even though you want to, you will not be able to legitimately use this theorem as $x^{-1/2}$ is not continuous at the interval endpoint $0$.

Then what do you do?

You rewrite the problem in terms of limits so that $f(x) = x^{=1/2}$ is continuous over the closed intervals $[-1, x]$ and $[x,1]$.

And, when you take your left-hand and right-hand limits as $x \to 0$, you never get to $0$ (just arbitrarily close), and so, the Fundamental Theorem of Calculus is in force throughout the limit process until you arrive at the correct answer.

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