Assume we have non-negative variables $\alpha,\gamma,s$.

The following integral $$ \int_{\gamma}^{\infty}\left( 1 -\frac{1}{1+s\, v^{-1}}\right) \,v^{\frac{2}{\alpha}-1}\, dv = \alpha \, F(\gamma \, s^{-1})^{\frac{1}{\alpha}}$$ where we define the function $F$ to be $$F(x) = \int_x^\infty \frac{t}{1+t^\alpha}\, dr $$

This can be done by transformation $v=st^\alpha$ $\rightarrow dv=\alpha \, st^{\alpha-1}\,dt$ and making the necessary substitutions.

Now assume I have a slightly different integral $$ \int_{\gamma}^{\infty}\left( 1 -\frac{c}{1+s\, v^{-1}}\right) \,v^{\frac{2}{\alpha}-1}\, dv $$ with $c$ being positive constant. How can I simplify in a similar way to the example I gave above? Do i need a transformation too.? Note that the same transformation as the one above does not work...



The reason the first case simplifies so well is that $1-\frac1{1+x}=\frac{x}{1+x}$ and the constant term in the numerator 'vanishes'; in the case we have $1-\frac{a}{1+x}=\frac{(1-a)+x}{1+x}$ which is not as convenient.

  • $\begingroup$ Yes, i agree, so do you mean to say that there is no hope in simplifying the other example... $\endgroup$ – Henry Jan 24 '15 at 6:09

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