Q. 4 - Calculus

Question

What it means for an improper integral to converge or to diverge?

Step-by-Step Solution

Answer

If the involved limits exist, then we say that the improper integral converges to the value determined by those limits, and if the limits are infinite or fail to exist, then we say that the improper integral diverges. 

Step 1. Definition for Converges and diverges in the improper integral.

A  definite integral ${\int }_a^{b}f\left(x\right)dx$ is improper if

(i) At least one of the limit $a \mathrm{or} b$are infinity or

(ii)$f\left(x\right)$ is discontinuous at least at one point$c\in \left[a,b\right]$.

The point of discontinuity may or may not include the boundary point. Hence the graph of$f\left(x\right)$ has a vertical asymptote at that point.

If the limits exist, this means that the improper integral converges. If the limits are infinite or fail to exist, then the improper integral diverges. 

Step 2. Example

Examples for convergent and divergent.

The example for Divergent improper integral is,

${\int }_1^{\infty }\frac{1}{\sqrt{x}}dx$

The value of this integral is $\infty$ , which is infinite thus not define.

The example for Convergent improper integral is,

${\int }_1^{\infty }\frac{1}{x^2}dx$

The value of this integral is$1$, which is not finite.

The example for divergent improper integral with definite limit  is,

${\int }_{\frac{\pi }{3}}^{\frac{\mathrm{\pi }}{6}}se{c}^{2}xdx$

The integrand $f\left(x\right)$is discontinuous at $\frac{\mathrm{\pi }}{2}$ and $f\left(x\right)$ has asymptote at  point $\frac{\mathrm{\pi }}{2}$.