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
.

Apply derivative on each side with respect to
.

.
Consider
.
Apply integral on each side.
\
.
Substitute corresponding values in
.


Again apply integration by parts.
\



Apply the limits.
\



Therefore, the integral
is converges to
.
The improper integral
is converges to
.