# Quick Improper Integrals Question

• Feb 21st 2010, 03:11 PM
Lord Voldemort
Quick Improper Integrals Question
"Determine whether each integral is convergent or divergent. Evaluate those that are convergent"
$\int_{-\infty}^{\infty}{\cos(\pi x)}dx=\int_{-\infty}^{0}{\cos(\pi x)}dx+\int_{0}^{\infty}{\cos(\pi x)}dx$
$=\displaystyle\lim_{r\to-\infty}{\int_{r}^{0}{\cos(\pi x)}dx}+\displaystyle\lim_{t\to\infty}{\int_{0}^{t} {\cos(\pi x)}dx}$
Sorry people, my LaTex skills break down here. Anyways,
$\int{\cos(\pi x)}dx=\frac{\sin(\pi x)}{\pi}$
But then from here I have to evaluate sine to infinity...
• Feb 21st 2010, 03:19 PM
TheEmptySet
Quote:

Originally Posted by Lord Voldemort
"Determine whether each integral is convergent or divergent. Evaluate those that are convergent"
$\int_{-\infty}^{\infty}{\cos(\pi x)}dx=\int_{-\infty}^{0}{\cos(\pi x)}dx+\int_{0}^{\infty}{\cos(\pi x)}dx$
$=\displaystyle\lim_{r\to-\infty}{\int_{r}^{0}{\cos(\pi x)}dx}+\displaystyle\lim_{t\to\infty}{\int_{0}^{t} {\cos(\pi x)}dx}$
Sorry people, my LaTex skills break down here. Anyways,
$\int{\cos(\pi x)}dx=\frac{\sin(\pi x)}{\pi}$
But then from here I have to evaluate sine to infinity...

What about writing it as a sum

$\int_{0}^{\infty}\cos(\pi\cdot x)dx=\sum_{n=0}^{\infty}\int_{2 n}^{2 (n+1)}\cos(\pi x)dx$
• Feb 21st 2010, 03:22 PM
Lord Voldemort
Well, you're getting a bit ahead of me; we haven't yet learned to write functions as sums, so I don't think it would be acceptable to do on the homework. But here, can I just say that because the limit does not exist, it is divergent? It's doesn't exist for a different reason than what is usual in improper integrals that I have seen thus the reservation.
• Feb 21st 2010, 03:26 PM
TheEmptySet
Nevermind I Think I have made an error.