Question

Show that [infinity] 0 e−x9 is convergent. SOLUTION We can't evaluate the integral directly because the antiderivative of e−x9 is not an elementary function. We write [infinity] 0 e−x9 dx = 1 0 e−x9 dx + [infinity] 1 dx and observe that the first integral on the right-hand side is just an ordinar

Answer 1

Explanation:

Just remember that

`e^(-x^9) < e^(-x)`

That is the most important part, therefore,

`\int\limits_(0)^(\infty){e^(-x^9)} \, dx < \int\limits_(0)^(\infty){e^(-x)} \, dx\\`

Since

`\int\limits_(0)^(\infty){e^(-x)} \, dx = \lim\limits_(n \to \infty) -e^(-x) + e^(-1) = e^(-1)`

and the integral

`\int\limits_(0)^(1) e^(-x^9) \, dx`

is an integral over a bounded interval and

`\int\limits_(0)^(\infty) {e^(-x^9)} \, dx = \int\limits_(0)^(1) {e^(-x^9)} \, dx + \int\limits_(1)^(\infty) {e^(-x^9)} \, dx`

The integral

`\int\limits_(0)^(\infty) {e^(-x^9)} \, dx`

converges.