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Apr 19, 20252 min read

An integral that is not evaluatable by normal means.

Not bounded on $[a, b]$ (Type 1)
If it is bounded, but it has a VA (Type 2) Refer to the solving process for how to solve. Improper integrals can either:
Converge to a finite value
Diverge to have infinite area

$\int_{1}^{\infty} tan^{- 1} (x) d x$

$\int_{2}^{8} \frac{1}{x - 2} d x$

Solving Improper Integrals

$\int_{1}^{\infty} (3 x + 1)^{- 2} d x$
$= lim_{A \to \infty} \int_{1}^{A} (3 x + 1)^{- 2} d x$
By inspection, $\int (3 x + 1)^{- 2} d x = - \frac{1}{3} (3 x + 1)^{- 1}$
$= lim_{A \to \infty} F (A) - F (1)$
$= - \frac{1}{3} (0 - \frac{1}{4}) = \frac{1}{12}$ Therefore, the integral converges to $\frac{1}{12}$

$\int_{0}^{5} \frac{l n ( x )}{x} d x$
Examine where the problematic endpoint is:
1. If the lower endpoint (0) is problematic, then evaluate $lim_{A \to 0^{+}}$
2. If the upper endpoint $(5)$ is problematic, then evaluate $lim_{A \to 5^{-}}$
$= lim_{A \to 0^{+}} \int_{A}^{5} f (x) d x$
By u-sub, choose $u = ln (x)$
Then, $= lim_{A \to 0^{+}} \int_{A}^{5} f (x) d x \frac{1}{u} d u$
$= (\frac{u ^{2}}{2}) + C = \frac{( l n ( x ) ) ^{2}}{2} + C$
Thus, $F (5) - F (A) = \frac{1}{2} (ln (5)^{2} - ln (A)^{2})$
$= - \infty$ , thus the limit does not exist and diverges

$\int_{- 1}^{1} x^{- 2} d x$
$= \int_{1}^{0} x^{- 2} d x + \int_{0}^{1} x^{- 2} d x$
$= lim_{A \to 0^{-}} \int_{- 1}^{A} x^{- 2} d x + lim_{B \to 0^{+}} \int_{B}^{1} x^{- 2} d x$
Note that the second part with $B$ diverges, and that the first part converges, then the entire integral diverges

$\int_{1}^{\infty} \frac{c o s ^{5} ( 3 x ^{2} + 1 ) + 1}{( x ^{2} + x ^{7} + 1 ) ^{2}} d x$
Solve with Comparison Theorem for Integrals