User:Mmmooonnnsssttteeerrr/Sandbox

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

My Proof
$$\int_{1}^{\infty}\frac{1}{x^p}dx=lim_{a\rightarrow\infty}\int_1^a\frac{1}{x^p}dx=\begin{cases}lim_{a\rightarrow\infty}\left[\frac{x^{-p+1}}{-p+1}\right]^a_1,p\ne1\\lim_{a\rightarrow\infty}\left[ln x^p\right]^a_1,p=1\end{cases}$$

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