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I am badly stuck in some integration here and will appreciate any help out of it.

$$\int^\infty_0f(r) dr = \int^\infty_0 \frac{Ar}{1+Cr^\alpha} e^{-Br^2} dr$$

If I let $u = Br^2$, then I get

$$ = \frac{A}{2B} \int^\infty_0\frac{\exp(-u)}{1+(u/B)^{\alpha/2}} du$$

But I am stuck while proceeding further. Any idea?

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These are special values (at $1$) of Laplace transforms of $\frac1{1+(u/B)^{\alpha/2}},$ which for rational $\alpha$ gives the Meijer $G$ function, but the parameters (including their number) depend on the actual rational number, so it seems unlikely that there is a closed form.

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  • $\begingroup$ If I consider $\alpha = 4$, then? $\endgroup$
    – SJa
    Commented Feb 7, 2018 at 5:33
  • $\begingroup$ for $\alpha=4$ it's a combination of cosine and sine integrals, nothing simpler than that, I'm afraid. $\endgroup$
    – Carlo Beenakker
    Commented Feb 7, 2018 at 13:25

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