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

Step-by-step 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.