# Help finding k. Issue with integration

Let the continuous random variable $X$ have a probability density function $f(x)$ such that

$$f(x) = k(1+x)^{-3}, x>0$$

$=0$ elsewhere

Find k

This is what I tried:

$\int_0^\infty k(1+x)^{-3}dx = 1$

since it's a probability density function

$k\int_0^\infty(1+x)^{-3}dx = 1$

$\int_0^\infty(1+x)^{-3}dx = \frac{1}{k}$

Let $u = 1+x$

$du = dx$

$\int_0^{\infty}u^{-3}du = {\frac{1}{k}}$

$[-\frac{u^{-2}}{2}]_0^{\infty} = \frac{1}{k}$

$[-\frac{(1+x)^{-2}}{2}]_0^{\infty} = \frac{1}{k}$

Now I'm stuck. Was I supposed to do something related to L'Hopitals Rule?

• Why L'Hospitals Rule? That rule governs finding the limit of a function. Here if you evaluate by putting $\infty$, it becomes zero and the only thing that is left is evaluation of 0. Commented Sep 9, 2014 at 9:18
• Yeah. That was a stupid suggestion... Moving on ^^) Commented Sep 9, 2014 at 12:18

You're doing it very well. Continuing your work, we will obtain \begin{align} -\frac12\lim_{a\to\infty}\left[\frac1{(1+x)^2}\right]_{x=0}^a&=\frac1k\\ \frac12&=\frac1k\\ k&=2 \end{align} since $$\lim_{a\to\infty}\frac1{(1+a)^2}=0.$$
Note that $1\le u$ and that $$\left[-\frac{u^{-2}}{2}\right]_1^{\infty}=\left[-\frac{1}{2u^2}\right]_{1}^{\infty}=\left(\lim_{u\to \infty}\left(-\frac{1}{2u^2}\right)\right)-\left(-\frac{1}{2\cdot 1^2}\right)=0+\frac 12.$$ Hence, we have $\frac 12=\frac 1k\Rightarrow k=2$.