# Evaluate $\int_1^{\infty} \frac{3}{x^2}dx$

I need to do the integral:

$$\int_1^{\infty} \frac{3}{x^2}dx.$$

So I integrates and got that I need to work out

$$\left[-\frac{1}{x^3}\right]_1^{\infty}$$

Which I then evaluate to say

$$=-\frac{1}{\infty^3} + 1$$

From here we then get that

$$\frac{1}{\infty^3} \rightarrow 0 \implies -\frac{1}{\infty^3} + 1 \rightarrow 1.$$

Is this correct?

EDIT: I see my mistake. Can't believe I just did that...

• Careful: take a look at your integral: you have $\int 3{x^{-2}}\,dx$. Did you take the derivative instead of integrating? – amWhy Nov 26 '13 at 21:14
• @amWhy Embarrassingly, no, I integrated. I thought that -2 + 1 = -3. Awkward... – Kaish Nov 26 '13 at 21:36
• Don't worry. I overlooked (missed) the error, initially. We all make silly mistakes. Better a silly mistake than being totally clueless, which you're clearly not! – amWhy Nov 26 '13 at 21:38

Better: $$\int_1^\infty 3x^{-2}\,dx=\lim_{b\to\infty}\int_1^b 3x^{-2}\,dx=\lim_{b\to\infty}\left[-3x^{-1}\right]_1^b=\lim_{b\to\infty}\left[-3b^{-1}+3\right]=0+3=3$$
Note that for $p\lt-1$, \begin{align} \int_1^\infty ax^p\,\mathrm{d}x &=\left[\frac{a}{p+1}x^{p+1}\right]_1^\infty\\ &=-\frac{a}{p+1} \end{align} Here, $a=3$ and $p=-2$.
\begin {align*} \displaystyle\int_{1}^{\infty} \dfrac {3}{x^2} \, \mathrm{d}x &= \left[ - 3x^{-1} \right]_{1}^{\infty} = \lim_{u \to \infty} \left( -3b^{-1} + 3 \right) = 0 + 3 = \boxed {3}. \end {align*}