Divergence Test: Evaluating I$_{17}$ at $\tiny{206.8.8.17}$

[karush]

$\tiny{206.8.8.17}$
$\textsf{Evaluate the following integral,
or stat that it diverges.}\\$
\begin{align*}
\displaystyle
&& I_{17}&
=\int_{0}^{\infty}36x^8 e^{-x^9}\, dx& &(1)&\\
&& &=36\int_{0}^{\infty}\frac{x^8}{e^{x^9}}\, dx & &(2)&\\
\end{align*}
$\textit{first what be the recommended divergence test to use on this?}$

 

[topsquark]

I don't know that you have to use a test in order to do the problem as stated but you can integrate it by inspection.

-Dan

 

[karush]

well it goes to 4 but we are supposed to test it for divergence first?
I was going to try the ratio test?

 

[MarkFL]

As it is am improper integral, I would write it like this

\(\displaystyle I=4\lim_{t\to\infty}\left(\int_0^t e^{-x^9}9x^8\,dx\right)\)

Now, let:

\(\displaystyle u=x^9\implies du=9x^8\,dx\)

And we have:

\(\displaystyle I=4\lim_{t\to\infty}\left(\int_0^t e^{-u}\,du\right)\)

Apply the FTOC:

\(\displaystyle I=4\lim_{t\to\infty}\left(-\left[e^{-u}\right]_0^t\right)\)

\(\displaystyle I=4\lim_{t\to\infty}\left(1-e^{-t}\right)\)

The limit exists, and so the integral converges:

\(\displaystyle I=4(1)=4\)

 

[karush]

so don't factor out 36 but use the factor 9 of it get a derivative

cool tool

 

[MarkFL]

I suppose you could look at the integrand in an integral like this and if it does not go to zero as the dummy variable of integration grows without bound, then you would know it diverges, but if it does go to zero, you still don't know if it converges or not.