Question

Let f(x)=k/(x^3+x) if x>=1,
and f(x)=0 otherwise
where k is const.

(a) For what value of k is f a probability density function?
(b) Find the mean of this density function.

I'm having a hard time finding my limits of integration, I think it's 1 to infinity. But if that's the case I end up with: k[ln(x) - (1/2)ln(x^2+1)] from 1 to infinity
which looks like k(infinity - infinity) - k[0-(1/2)ln(2)] and... i'm lost.

Answer 1

are you still there?

Answer 2

@Vadatajs

Answer 3

Yes?

Answer 4

so do you know what is the criterion to normalize the probability function?

Answer 5

\[\int\limits_{-\infty}^{\infty} f(x) dx = 1\]

Answer 6

absolutely. but in your case. what should be the limit?

Answer 7

1 to infinity is my guess since x >= 1

Answer 8

correct. now solve the equation.

Answer 9

\[ k \int\limits_{1}^{\infty} 1/x - x/(x^2+1) dx = k[\ln (x) - (1/2)\ln(x^2+1)] from 1 \to \infty\]i use partial fractions to come out with

Answer 10

ok..

Answer 11

This is all in my description...

Answer 12

now can you solve it?

Answer 13

No, k(infinity - infinity) - k[0-(1/2)ln(2)]

Answer 14

ln(x) from 1 to infinity -0.5ln(x^2 +1) from 1 to infinity

Answer 15

Once you integrate you have
\[ \frac{k}{2} \ln\left( \frac{x^2}{x^2+1} \right) \]
Evaluating that at x = 1, we clearly have \( -\frac{k}{2}\ln(2) \). In the limit as x --> infinity, it is
\[ \frac{k}{2} \ln(1) = 0 \]
Therefore the entire integral which must be equal to 1 gives us the equation:
\[ \frac{k}{2} \ln 2 = 1 \]

Now it's easy to find \( k \).

Answer 16

\[\int\limits_{1}^{\infty}(k/(x^3+x))dx=\int\limits_{1}^{\infty}(k/x)dx-\int\limits_{1}^{\infty}(kx/(x^2+1))dx \]
that's by using partial fractions

Answer 17

factor k out and solve the integrals separately as...

Answer 18

\[\lim_{R \rightarrow \infty}(\int\limits_{1}^{R}(1/x)dx-\int\limits_{1}^{R}(x/(x^2+1)dx)=\lim_{R \rightarrow \infty}(\ln R-0.5\ln(R^2+1))\]
this gives
\[\lim_{R \rightarrow \infty}(lnR-\ln(R^2+1)^(1/2))\]
this gives
\[\lim_{R \rightarrow \infty}((lnR)/0.5\ln(R^2+1))\]
this gives \[\infty/\infty\]
hence you think of L'Hopital's Rule

Answer 19

so you differentiate the numerator and the denomenator and simplify...this gives...
\[\lim_{R \rightarrow \infty}(R^2/(R^2+1))\]

Answer 20

this still gives
\[\infty/\infty\]
so you do the differential again... and we have...
\[\lim_{R \rightarrow \infty}(2R/2R)\]
this is simple to evaluate as the 2R's cancel out to give...
\[\lim_{R \rightarrow \infty}(1)=1\]

Answer 21

Note that it's...
\[\int\limits_{1}^{\infty}(1/(x^3+x))dx =1\]
therefore for an equation of the form...
\[\int\limits_{1}^{\infty}(k/(x^3+x))dx=1,\]
\[k(1)=1,\]
hence k=1...hope this answers your question...

Answer 22

What brightantwiboasiako has written is not correct
- his solution has dropped the value of the integral end point x = 1
- the way the limit of ln(x^2/(x^2+1) is evaluated is incorrect and gives the wrong answer.