User:Mmmooonnnsssttteeerrr/Sandbox

TheoremEdit

\int_1^\infty\frac{1}{x^p}dx\begin{cases}p>1,integral\;converges\\p\le1,integral\;diverges\end{cases}

My ProofEdit

\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]^a_1,p\ne1\\lim_{a\rightarrow\infty}\left[ln x^p\right]^a_1,p=1\end{cases}

If p>1, -p+1<0:

Let |-p+1|=k,

\left[\frac{x^{-k}}{-k}\right]^a_1=\frac{a^{-k}}{-k}-\frac{1}{-k}=\frac{1}{-ka^k}-\frac{1}{-k}

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

If p<1, -p+1>0:

Let -p+1=k

\left[\frac{x^k}{k}\right]^a_1=\frac{a^k}{k}-\frac{1}{k}

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

If p=1:

\left[ln x^p\right]^a_1=ln a^p-ln 1=ln a^p

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

Last modified on 27 October 2006, at 01:49