Integral to Infinity of Reciprocal of Power of x

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Theorem

The improper integral

$\ds \int_1^\infty \dfrac {\d t} {t^x}$

exists if and only if $x > 1$.


Proof

First let $x \ne 1$.

Then:

\(\ds \int_1^\infty \dfrac {\d t} {t^x}\) \(=\) \(\ds \lim_{P \mathop \to \infty} \int_1^P t^{-x} \rd t\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \lim_{P \mathop \to \infty} \intlimits {\dfrac {t^{-x + 1} } {-x + 1} } 1 P\) Primitive of Power
\(\ds \) \(=\) \(\ds \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} - \dfrac {1^{1 - x} } {1 - x} }\)
\(\ds \) \(=\) \(\ds \lim_{P \mathop \to \infty} \dfrac 1 {x - 1} \paren {1 - \dfrac 1 {P^{x - 1} } }\) simplifying


If $x > 1$, then $x - 1 > 0$.

Hence from Sequence of Powers of Reciprocals is Null Sequence, $\dfrac 1 {P^{x - 1} } \to 0$ as $P \to +\infty$.


If $x < 1$, then $x - 1 < 0$.

Hence $P^{x - 1} \to 0$ as $P \to +\infty$.

Then from Reciprocal of Null Sequence it follows that $\dfrac 1 {P^{x - 1} } \to \infty$ as $P \to +\infty$.


Finally we have that from Integral of Reciprocal is Divergent:

$\ds \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} t \to \infty$

All cases are covered, and the result follows.

$\blacksquare$


Sources