Integral: For What Values of p is Convergent?

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SUMMARY

The integral from 1 to infinity of the function \(\int \frac{1}{x^p}dx\) converges for values of \(p > 1\). The evaluation involves taking the limit as \(t\) approaches infinity, resulting in the expression \(\lim_{t \to \infty} \frac{1}{p-1} \left[\frac{1}{t^{p-1}} - 1\right]\). The term \(t^{p-1}\) appears in the denominator due to the integration process and the application of the Fundamental Theorem of Calculus. Correct identification of the integral is crucial, as the initial confusion stemmed from miswriting the integral as \(\int \frac{1}{x^{-p}}dx\).

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I got lost in an example in my book. Hoping someone could explain it to me.For what values of p is the intergral

from 1 to infinity [tex]\int \frac {1}{x^p}dx[/tex]

convergent?

from 1 to infinity [tex]\int \frac {1}{x^p}dx[/tex]

= lim (t -> infinity) [tex]\frac {x^-^p^+^1}{-p+1}[/tex] (from x = 1 to x = t)

= lim (t -> infinity) [tex]\frac {1}{p-1} [\frac {1}{t^p^-^1} - 1][/tex]

the only thing that confuses me about this is how the t^p-1 ended up in the denominator because after the 2nd sept I get the following:

= lim (t -> infinity) [tex]\frac {t^p^-^1}{p-1} - \frac {1}{p-1}[/tex]

Thanks!
 
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Isn't [tex]\int \frac{1}{x^{-p}}dx=\frac{x^{1+p}}{1+p}[/tex]?
 
Oh, I'm sorry the intital inegral is

[tex]\int \frac {1}{x^p}dx[/tex]

thanks for catching that mistake :)
 

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