Is there a constant $c$ that $\frac{c}{1+x}$ be a probability density function for $x>0$?

Question

There are two conditions for probability density function: $f(x)\geq0$ and $\int_{-\infty}^{+\infty}f(x)dx=1$.

Here for the first condition we must have $c\geq 0$. for the second condition: $$\int_{0}^{+\infty} \frac{c}{1+x}=1 \Rightarrow \frac{1}{c} = lim_{t\to \infty} \int_{0}^{t} \frac{1}{1+x} \Rightarrow \frac{1}{c}=lim_{t \to \infty} (ln\vert1+t\vert - ln\vert1+0\vert) \Rightarrow \frac{1}{c}=\infty$$

so $c$ must be zero or there are no such $c$?

Answer 1

There is no such $c$. You can't even take $c=0$, because then the PDF is $0$ everywhere, so the integral is $0$, not $1$.