Why are integrals well defined?

[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.}

 

[arildno]

Your sentence is, essentially meaningless:
"The integral behaves at zero"

Besides, your conclusion is wrong.

 

[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.

 

[arildno]

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

 

[VietDao29]

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.}

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)
-----------
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? :)

 

[jamesbob]

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