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Evaluate $$\int_0^{\infty } \frac{2^{\frac{r}{\delta }} \left(2^{\frac{r}{\delta }}-1\right) r\ e^{-\frac{\alpha ^2+\left(2^{\frac{r}{\delta }}-1\right)^2}{2 \beta ^2}}\log 2 }{\beta ^2 \delta } I_0\left(\frac{\alpha \left(2^{\frac{r}{\delta }}-1\right)}{\beta ^2}\right) \, dr$$

I arrived at the above equation while trying to obtain the Expectation of $r$.

I'm at my wits end with this problem. I don't know how to continue so either some hints, pushes in the right direction or the full evaluation would be highly appreciated. Thanks

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What is $I_0$? A constant? Without this, make a change of variables $$r\to r\ln 2/\delta$$to obtain up to a multiplicative constant

$$\int_0^{\infty } e^{r} \left(e^{ {r} }-1\right) r\ e^{-\frac{\alpha ^2+\left(e^{ {r} }-1\right)^2}{2 \beta ^2}} I_0\left(\frac{\alpha \left(e^{ {r} }-1\right)}{\beta ^2}\right) dr$$

Now make a change of variables $u=e^r-1$, $du = e^rdr$ to obtain (again, up to a multiplicative constant)

$$\int_0^{\infty } u \ln(u+1)\ e^{-\frac{\alpha ^2+u^2}{2 \beta ^2}} I_0\left(\frac{\alpha u}{\beta ^2}\right) du,$$ which seems to be more or less doable depending on the form of $I_0$.

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  • $\begingroup$ $I_0$ is the modified Bessel function of the first kind of order zero which is also a function of $r$: en.wikipedia.org/wiki/… $\endgroup$ – Afloz Jul 2 '14 at 17:36

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