∫ 1 ∞ 1 x p d x { p > 1 , i n t e g r a l c o n v e r g e s p ≤ 1 , i n t e g r a l d i v e r g e s {\displaystyle \int _{1}^{\infty }{\frac {1}{x^{p}}}dx{\begin{cases}p>1,integral\;converges\\p\leq 1,integral\;diverges\end{cases}}}
∫ 1 ∞ 1 x p d x = l i m a → ∞ ∫ 1 a 1 x p d x = { l i m a → ∞ [ x − p + 1 − p + 1 ] 1 a , p ≠ 1 l i m a → ∞ [ l n x p ] 1 a , p = 1 {\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 p > 1 , − p + 1 < 0 {\displaystyle p>1,-p+1<0} :
Let | − p + 1 | = k {\displaystyle |-p+1|=k} ,
[ x − k − k ] 1 a = a − k − k − 1 − k = 1 − k a k − 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, l i m a → ∞ [ 1 − k a k + 1 k ] = 1 k . {\displaystyle lim_{a\rightarrow \infty }\left[{\frac {1}{-ka^{k}}}+{\frac {1}{k}}\right]={\frac {1}{k}}.}
If p < 1 , − p + 1 > 0 {\displaystyle p<1,-p+1>0} :
Let − p + 1 = k {\displaystyle -p+1=k}
[ x k k ] 1 a = a k k − 1 k {\displaystyle \left[{\frac {x^{k}}{k}}\right]_{1}^{a}={\frac {a^{k}}{k}}-{\frac {1}{k}}}
Therefore, l i m a → ∞ [ a k k − 1 k ] → ∞ . {\displaystyle lim_{a\rightarrow \infty }\left[{\frac {a^{k}}{k}}-{\frac {1}{k}}\right]\rightarrow \infty .}
If p = 1 {\displaystyle p=1} :
[ l n x p ] 1 a = l n a p − l n 1 = l n a p {\displaystyle \left[lnx^{p}\right]_{1}^{a}=lna^{p}-ln1=lna^{p}}
Therefore, l i m a → ∞ [ l n a p ] → ∞ . ◻ {\displaystyle lim_{a\rightarrow \infty }\left[lna^{p}\right]\rightarrow \infty .\Box }
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b05304b472317d1688223ec69948ab80bdf2e6aa)
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![{\displaystyle \left[{\frac {x^{-k}}{-k}}\right]_{1}^{a}={\frac {a^{-k}}{-k}}-{\frac {1}{-k}}={\frac {1}{-ka^{k}}}-{\frac {1}{-k}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64b07352c71ce15f4bd6cafc08cbe30a3ecc1d69)
![{\displaystyle lim_{a\rightarrow \infty }\left[{\frac {1}{-ka^{k}}}+{\frac {1}{k}}\right]={\frac {1}{k}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9ce0ec8db6cf149cec954dce5b5c66ff3fa2194)
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![{\displaystyle \left[{\frac {x^{k}}{k}}\right]_{1}^{a}={\frac {a^{k}}{k}}-{\frac {1}{k}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9aa40f06360d180a6c2e80263711133951d5e8f)
![{\displaystyle lim_{a\rightarrow \infty }\left[{\frac {a^{k}}{k}}-{\frac {1}{k}}\right]\rightarrow \infty .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fcbfc8886a3f45e1b95f748fb7e1e38591fc89b7)
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![{\displaystyle \left[lnx^{p}\right]_{1}^{a}=lna^{p}-ln1=lna^{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b6aa325b75f2514f465ec86c143a22228b921d7)
![{\displaystyle lim_{a\rightarrow \infty }\left[lna^{p}\right]\rightarrow \infty .\Box }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f3856f9bc1be8c0ee3878cbfb44843a255157c3)
