The integral is .
.
If is continuous on the interval
, then
is convergent.
Solve the integral by using integration by parts of integration.
\Apply integration by parts formula .
Consider .
Apply derivative on each side with respect to .
.
Consider .
Apply integral on each side.
\.
Substitute corresponding values in .
.
Therefore, the integral is converges to
.
The improper integral is converges to
.