# Why limits are necessary for improper integrals?

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)

• No, $x^{-1/2}$ is continuous in the domain $x>0$, $x=0$ belongs not to the Domain. – Dr. Sonnhard Graubner Jul 16 '19 at 18:15
• 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. – Cameron Williams Jul 16 '19 at 18:51
• Wait you have to many $x$s . . . you have it for the variable of integration and the limit, you can’t do that – gen-z ready to perish Jul 16 '19 at 21:06

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 $$\displaystyle\lim_{t\to0^-}\int_{-1}^tx^{-1/2}\,\mathrm dx$$.

• I've edited my answer. Thank you. – José Carlos Santos Jul 16 '19 at 18:50

If the problem were, instead, $$\int_{-1}^1 \frac{1}{x}dx$$ then, since $$\lim_{x\to 0}Ln(x)$$ does not exist, that integral does not exist. But ignoring the discontinuity, $$\left[log|x|\right]_{-1}^1= 0$$. They are NOT the same.

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.