Determine whether the improper integral converges or diverges.
-$$\int _ { 1 } ^ { \infty } e ^ { - x } \sin x d x$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

We can use Integration by Parts to solve this problem:
Let u = sin(x) and dV = e^(-x) dx.
Then du = cos(x) dx and V = -e^(-x).
Using the formula for integration by parts, we have

∞
∫      e^(-x)sin(x) dx = [-e^(-x)sin(x)]_1^∞ - ∫  -e^(-x)cos(x) dx
       1

The limit as x approaches infinity of e^(-x)sin(x) is 0 because e^(-x) goes to 0 faster than sin(x) grows, and the same is true of e^(-x)cos(x), so our integral simplifies to 
∫      e^(-x)sin(x) dx = [-e^(-x)sin(x)]_1^∞ + ∫  e^(-x)cos(x) dx

Since the absolute value of e^(-x) times any trigonometric function is bounded above by e^(-x), we know that the integral on the right-hand side converges, which means the original integral converges.

Determine Whether the Improper Integral Converges or Diverges ∫1∞e−xsin⁡xdx\int _ { 1 } ^ { \infty } e ^ { - x } \sin x d x∫1∞​e−xsinxdx

Determine Whether the Improper Integral Converges or Diverges $\int _ { 1 } ^ { \infty } e ^ { - x } \sin x d x$