# Difficulty finding Expectation of a special function

I have a special function given as:

f(r)=1βλ2r/βexp([2r/β−1]Kλ){\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 rr. Mathematica was not able to solve the associated Integral function. So it returns:

∫∞0{1βλ2r/βexp([2r/β−1]Kλ)} rdr \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}]

=================

– Dr. belisarius
Mar 31 ’14 at 6:51

1

It looks like unless K/lambda <0 your integral is not convergent. Also, it doesn't look like f[r] it's normalized. – b.gatessucks Mar 31 '14 at 10:04      Yes right. That's what @ubpdqn showed - Re[a3a2]<0Re[\frac{a3}{a2}]<0, that is Re[KÎ»]<0Re[\frac{K}{Î»}]<0. Unfortunately, I think I have a bigger problem, because both KK and Î»Î» are positive in my model. f(r)f(r) is actually correct. Although I removed a product term K Log[2]K Log[2] from f(r)f(r) which on testing does not seem to have any major effect. – Afloz Mar 31 '14 at 10:30 ================= 1 Answer 1 ================= The integral is conditionally convergent. You can progress using substitution: u=2^{\frac{r}{b}}\iff r= b\log_2 u u=2^{\frac{r}{b}}\iff r= b\log_2 u Hence,\frac{dr}{du}=\frac{b}{u\ln 2}\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}]

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}]

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

Great! Interesting work. It appears to me that \lambda\lambda implies a2a2 which appears to be < 0 < 0. This seems to contradict our original assumption that \lambda >= 0\lambda >= 0. Am I right?
– Afloz
Mar 31 ’14 at 9:17

@Methyl see update
– ubpdqn
Mar 31 ’14 at 9:25

You’re quite correct. The solution here is conditioned on Re[a3/a2] < 0Re[a3/a2] < 0, that is Re[\frac{K}{\lambda}] < 0Re[\frac{K}{\lambda}] < 0. Unfortunately, I think I have a bigger problem, both KK and \lambda\lambda are positive in my model. – Afloz Mar 31 '14 at 9:56      Let me be the first to give well-deserved +1, a typically neat and clean answer by you. – ciao Mar 31 '14 at 10:07