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anonymous

  • 4 years ago

can someone explain to me comparison of improper integrals / convergence and divergence?

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  1. anonymous
    • 4 years ago
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    This concept is confusing me

  2. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    ya the comaprison test :D

  4. anonymous
    • 4 years ago
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    LOL where did you just disappear to?

  5. anonymous
    • 4 years ago
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    :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
    • 4 years ago
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    the converse doesn't hold though

  7. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    If f diverges, you can't say anything about g. Similarly if g converges, you can't say anything about f.

  9. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    Thanks :D I really appreciate that :D

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