Well defined intergrals

1. May 2, 2006

jamesbob

Im having trouble explaining why intergrals are well defined. For instance:

$$\int_{0}^{\infty} \frac{1}{(x + 16)^{\frac{5}{4}}}dx.$$

Here do i say something like:

$$\mbox{The integral behaves at zero, and at } \infty, (x + 16)^{\frac{5}{4}} > x^{\frac{5}{4}} \mbox{ therefore the integral diverges.}$$

Last edited: May 2, 2006
2. May 2, 2006

arildno

"The integral behaves at zero"

3. May 2, 2006

AKG

It depends on what you mean by "well-defined". If you mean "is a real number" then look at what happens when you apply the definition of the improper integral to what you have, and show that you really do get a real number.

4. May 2, 2006

arildno

Actually, the value of your integral equals a most special prime.

5. May 2, 2006

VietDao29

No, it does not make sense. This is an Improper integral. As you know:
$$\mathop {\int} \limits ^ {b}_{a} f(x) dx = F(b) - F(a)$$, so what if b is not finite, i.e b tends to infinity? We have:
$$\mathop {\int} \limits ^ {\infty}_{a} f(x) dx = \lim_{b \rightarrow \infty} \mathop {\int} \limits ^ {b}_{a} f(x) dx = \lim_{b \rightarrow \infty}(F(b)) - F(a)$$
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Say you want to evaluate:
$$\mathop {\int} \limits ^ {\infty}_{1} \frac{dx}{x ^ 2}$$
We have:
$$\mathop {\int} \limits ^ {\infty}_{1} \frac{dx}{x ^ 2} = \left( \lim_{x \rightarrow \infty} - \frac{1}{x} \right) + \frac{1}{1} = 1$$.
Can you get this? :)

6. May 3, 2006

jamesbob

Yeah, thanks. Its just the way my silly university asks questions and explains things that confused me. Thanks for the help