I have a special function given as:

$${\rm f}\left(r\right) ={1 \over \beta\lambda}\,2^{r/\beta} \exp\left({\left[2^{r/\beta} - 1\right]K \over \lambda}\right)$$

I should find the Expectation of the random variable $r$. Mathematica was not able to solve the associated Integral function. So it returns:

$$ \int_0^{\infty}\left\{{1 \over \beta\lambda}\,2^{r/\beta} \exp\left({\left[2^{r/\beta} - 1\right]K \over \lambda}\right)\right\}\ r\,{\rm d}r $$

Does anyone recognize how I can reduce this function so I can solve it further?

==== Edit =====

This is the code I tried:

Integrate[ r*fr, {r, 0, \Infinity}, Assumptions->{K>=1, \lambda >=0, \beta >=0}]
  • $\begingroup$ Please post the code you've already tried. $\endgroup$ – Dr. belisarius Mar 31 '14 at 6:51
  • 1
    $\begingroup$ It looks like unless K/lambda <0 your integral is not convergent. Also, it doesn't look like f[r] it's normalized. $\endgroup$ – b.gates.you.know.what Mar 31 '14 at 10:04
  • $\begingroup$ Yes right. That's what @ubpdqn showed - $Re[\frac{a3}{a2}]<0$, that is $Re[\frac{K}{λ}]<0$. Unfortunately, I think I have a bigger problem, because both $K$ and $λ$ are positive in my model. $f(r)$ is actually correct. Although I removed a product term $K Log[2]$ from $f(r)$ which on testing does not seem to have any major effect. $\endgroup$ – Afloz Mar 31 '14 at 10:30

The integral is conditionally convergent. You can progress using substitution: $u=2^{\frac{r}{b}}\iff r= b\log_2 u $ Hence,$\frac{dr}{du}=\frac{b}{u\ln 2}$

You can do these substitutions in Mathematica:

f[r_, b_, la_, k_] := 2^(r/b) Exp[k (2^(r/b) - 1)/la]/(b la)
exp = f[x, a1, a2, a3] /. {2^(x/a1) -> u};
ex = D[a1 Log[2, u], u];
ans = Integrate[a1 Log[2, u] exp ex, {u, 1, Infinity}]

The symbolic integral is thence:

ConditionalExpression[-((a1 E^(-(a3/a2)) Gamma[0, -(a3/a2)])/(
  a3 Log[2]^2)), Re[a3/a2] < 0]

Now you can compare numerically:

N@Integrate[r f[r, 1, -1, 1], {r, 0, Infinity}]



and using the symbolic integral:

N[ans /. {a1 -> 1, a2 -> -1, a3 -> 1}]

yields: -1.24122

A small sample:

Grid[Table[{1, j, 1, N@ans /. {a1 -> 1, a2 -> j, a3 -> 1}, 
   Integrate[r f[r, 1, j, 1], {r, 0, Infinity}]}, {j, 
   Range[-1, -0.1, 0.1]}], 
 Dividers -> {{False, False, False, True, {False}}, None}]

enter image description here

I arbitrarily chose some parameters to illustrate. Further insights regarding convergence can be obtained:

 Plot[{r f[r, 1, j, 1], r f[r, 1, 1, j]}, {r, 0, 10}, 
  PlotRange -> {-1, 10}, 
  Epilog -> Text[Style[j, 20, Red], {6, 5}]], {j, -1, 1, 0.15}]

enter image description here

Obviously the parameter constraints (regions of interest) are up to users intention.

  • $\begingroup$ Great! Interesting work. It appears to me that $\lambda$ implies $a2$ which appears to be $ < 0$. This seems to contradict our original assumption that $\lambda >= 0$. Am I right? $\endgroup$ – Afloz Mar 31 '14 at 9:17
  • $\begingroup$ @Methyl see update $\endgroup$ – ubpdqn Mar 31 '14 at 9:25
  • $\begingroup$ You're quite correct. The solution here is conditioned on $Re[a3/a2] < 0$, that is $Re[\frac{K}{\lambda}] < 0$. Unfortunately, I think I have a bigger problem, both $K$ and $\lambda$ are positive in my model. $\endgroup$ – Afloz Mar 31 '14 at 9:56
  • $\begingroup$ Let me be the first to give well-deserved +1, a typically neat and clean answer by you. $\endgroup$ – ciao Mar 31 '14 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.