## anonymous 4 years ago can someone explain to me comparison of improper integrals / convergence and divergence?

1. anonymous

This concept is confusing me

2. anonymous

Related to the comparison test for integrals I think, which is sort of similar to sums, I'm working on this sort of thing at the moment actually, we'll let someone else answer :p

3. anonymous

ya the comaprison test :D

4. anonymous

LOL where did you just disappear to?

5. anonymous

:D I can tell you a theorem about this... If we let $f(x) \geq g(x) \geq 0$ where $f\hspace{0.3cm} \&\hspace{0.3cm} g$ are defined on the interval $[a, \infty )$ then the following theorem holds: $\displaystyle\int_{a}^{\infty}f(x)dx \hspace{0.2cm} converges \Longrightarrow \displaystyle\int_{a}^{\infty}g(x)dx \hspace{0.2cm} converges$ and $\displaystyle\int_{a}^{\infty}g(x)dx \hspace{0.2cm} diverges \Longrightarrow \displaystyle\int_{a}^{\infty}f(x)dx \hspace{0.2cm} diverges$ This is a theorem and can be proved...

6. anonymous

the converse doesn't hold though

7. anonymous

You know how integrals can be associated with areas under curves? Well if the 'larger' function $f$ converges then the 'smaller' one $g$ must also converge (if it didn't, it would diverge and there would be some point after which where g>f, violating the conditions). Also, if the 'smaller' function $g$ diverges, then the 'larger' one $f$ must diverge (if it didn't, it would be finite and where would be some point after which f<g, again violating the conditions).

8. anonymous

If f diverges, you can't say anything about g. Similarly if g converges, you can't say anything about f.

9. anonymous

Shall we do an example? Does the following integral converge or diverge? $\displaystyle\int_{a}^{\infty}\frac{1}{x+e^x}dx.$ Solution. One might be tempted to do this: $\frac{1}{x+e^x} < \frac{1}{x} \forall x \in \mathbb{R}/\{0\}$ $\displaystyle\int_{3}^{\infty}\frac{1}{x}dx$ however this does not converge because we can write it as a sum: $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n}$ and we know this is the p-series with p=1 which can be shown to converge for $p > 1$ and diverge for $p \leq 1$ So we are in a position where the 'bigger' function diverges, but so what, because this tells us nothing about our 'smaller' one which is what we want to find out about. We need a function which is bigger, but converges and this will show convergence. Note that $\frac{1}{x+e^x} < \frac{1}{e^x} = e^{-x} \forall x \in \mathbb{R}.$ $\lim_{t\rightarrow \infty} \displaystyle\int_{3}^{t}e^{-x}dx = \lim_{t\rightarrow \infty} -e^{-x}\vert_{3}^{t} = \lim_{t\rightarrow \infty}\left(-\frac{1}{e^{t}} + \frac{1}{e^3}\right) = \frac{1}{e^3} < +\infty$ hence the 'larger' integral converges. Therefore $\displaystyle\int_{a}^{\infty}\frac{1}{x+e^x}dx$ must converge.

10. anonymous

Thanks :D I really appreciate that :D