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Alexandre
  • Member for 7 years, 6 months
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Closed form for $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$
Thanks! I would appreciate any reference or indication on how you got this. The coefficients $b_{n,0}$ seems to be $(-2)^nn!$.
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Closed form for $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$
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 $I_1$?
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Closed form for $\int_0^T e^{-x}\frac{I_n(\alpha x)}{x}dx$
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.
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