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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?Read more