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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}]
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  • $\begingroup$ Please post the code you've already tried. $\endgroup$
    – Dr. belisarius
    Commented Mar 31, 2014 at 6:51
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    $\begingroup$ 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
    Commented Mar 31, 2014 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
    Commented Mar 31, 2014 at 10:30

1 Answer 1

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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}]

yields

-1.24122

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:

Manipulate[ 
 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.

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  • $\begingroup$ 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
    Commented Mar 31, 2014 at 9:17
  • $\begingroup$ @Methyl see update $\endgroup$
    – ubpdqn
    Commented Mar 31, 2014 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
    Commented Mar 31, 2014 at 9:56
  • $\begingroup$ Let me be the first to give well-deserved +1, a typically neat and clean answer by you. $\endgroup$
    – ciao
    Commented Mar 31, 2014 at 10:07

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