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Jun 13, 2017 at 10:20 history edited Alexandre CC BY-SA 3.0
Answer to one of the questions
Jun 12, 2017 at 12:19 history edited Alexandre CC BY-SA 3.0
Corrections and more details
Jun 12, 2017 at 12:14 history edited Alexandre CC BY-SA 3.0
Corrections and more details
Jun 12, 2017 at 8:20 history edited Alexandre CC BY-SA 3.0
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Jun 11, 2017 at 16:44 answer added Carlo Beenakker timeline score: 4
Jun 11, 2017 at 15:53 comment added Alexandre Great! I found the incomplete gamma for small $\alpha$ from the first infinite series proposed, and I can also get $\alpha=1$ from the second infinite series but it is expressed with the hypergeometric function ${}_2F_2$. Do you have more detail about this closed expression in terms of $I_0$ and $I_1$?
Jun 11, 2017 at 14:56 comment added Carlo Beenakker and for small $\alpha$ it's an incomplete gamma function, are these asymptotics of interest?
Jun 11, 2017 at 14:29 comment added Carlo Beenakker for $\alpha=1$ there is a closed form expression in terms of $I_0(t)$ and $I_1(t)$
Jun 11, 2017 at 13:32 comment added Alexandre You are right, it may have no closed form. I asked because this last integral seems to be an unexploited approach. Maybe I should look at the asymptotic behaviour and be satisfied with it.
Jun 11, 2017 at 13:23 comment added Igor Rivin I don't understand the downvotes. On the other hand, I don't understand why the OP expects a closed form, either...
S Jun 11, 2017 at 11:02 history suggested Glorfindel CC BY-SA 3.0
grammar corrections, thanks removed as per https://meta.stackexchange.com/q/2950/295232
Jun 11, 2017 at 11:00 review Suggested edits
S Jun 11, 2017 at 11:02
Jun 11, 2017 at 10:57 review First posts
Jun 11, 2017 at 11:11
Jun 11, 2017 at 10:54 history asked Alexandre CC BY-SA 3.0