# User:Mmmooonnnsssttteeerrr/Sandbox

### Theorem

${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{p}}}dx{\begin{cases}p>1,integral\;converges\\p\leq 1,integral\;diverges\end{cases}}}$

#### My Proof

${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{p}}}dx=lim_{a\rightarrow \infty }\int _{1}^{a}{\frac {1}{x^{p}}}dx={\begin{cases}lim_{a\rightarrow \infty }\left[{\frac {x^{-p+1}}{-p+1}}\right]_{1}^{a},p\neq 1\\lim_{a\rightarrow \infty }\left[lnx^{p}\right]_{1}^{a},p=1\end{cases}}}$

If ${\displaystyle p>1,-p+1<0}$:

Let ${\displaystyle |-p+1|=k}$,

${\displaystyle \left[{\frac {x^{-k}}{-k}}\right]_{1}^{a}={\frac {a^{-k}}{-k}}-{\frac {1}{-k}}={\frac {1}{-ka^{k}}}-{\frac {1}{-k}}}$

Therefore, ${\displaystyle lim_{a\rightarrow \infty }\left[{\frac {1}{-ka^{k}}}+{\frac {1}{k}}\right]={\frac {1}{k}}.}$

If ${\displaystyle p<1,-p+1>0}$:

Let ${\displaystyle -p+1=k}$

${\displaystyle \left[{\frac {x^{k}}{k}}\right]_{1}^{a}={\frac {a^{k}}{k}}-{\frac {1}{k}}}$

Therefore, ${\displaystyle lim_{a\rightarrow \infty }\left[{\frac {a^{k}}{k}}-{\frac {1}{k}}\right]\rightarrow \infty .}$

If ${\displaystyle p=1}$:

${\displaystyle \left[lnx^{p}\right]_{1}^{a}=lna^{p}-ln1=lna^{p}}$

Therefore, ${\displaystyle lim_{a\rightarrow \infty }\left[lna^{p}\right]\rightarrow \infty .\Box }$