3.7 Improper integrals  (Page 12/6)

Definition

1. Let $f(x)$ be continuous over an interval of the form $[a,\text{+}\infty ).$ Then
   
   ${\displaystyle {\int }_a^{+\infty }f\left(x\right)dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_a^{t}f\left(x\right)dx,}}$
   
   provided this limit exists.
2. Let $f(x)$ be continuous over an interval of the form $(\text{−}\infty ,b].$ Then
   
   ${\displaystyle {\int }_{\text{−}\infty }^{b}f\left(x\right)dx=\underset{t\to \text{−}\infty }{\text{lim}}{\displaystyle {\int }_t^{b}f\left(x\right)dx,}}$
   
   provided this limit exists.
   In each case, if the limit exists, then the improper integral    is said to converge. If the limit does not exist, then the improper integral is said to diverge.
3. Let $f(x)$ be continuous over $(\text{−}\infty ,\text{+}\infty ).$ Then
   
   ${\displaystyle {\int }_{\text{−}\infty }^{+\infty }f\left(x\right)dx={\displaystyle {\int }_{\text{−}\infty }^{0}f\left(x\right)dx+{\displaystyle {\int }_0^{+\infty }f\left(x\right)dx,}}}$
   
   provided that ${\displaystyle {\int }_{\text{−}\infty }^{0}f\left(x\right)dx}$ and ${\displaystyle {\int }_0^{+\infty }f\left(x\right)dx}$ both converge. If either of these two integrals diverge, then ${\displaystyle {\int }_{\text{−}\infty }^{+\infty }f\left(x\right)dx}$ diverges. (It can be shown that, in fact, ${\displaystyle {\int }_{\text{−}\infty }^{+\infty }f\left(x\right)dx={\displaystyle {\int }_{\text{−}\infty }^{a}f\left(x\right)dx+{\displaystyle {\int }_a^{+\infty }f\left(x\right)dx}}}$ for any value of $a.)$