Suppose that, for $v$ near to $1$, we have that $$ g(v)=\mu^2(1-v)^2+O((v-1)^2),\qquad (*) $$ i.e. $g(v)\approx \mu^2(1-v)^2$ for $v$ near to $1$.

Now, let $$ S(v)=\int_0^v\left(g^{1/2}(s)\int_0^s g^{-3/2}(t)\, dt\right)\, ds. $$

It is claimed that, for $v$ near $1$, $$ S(v)=-\frac{1}{2\mu^2}(\ln(1-v)+O(1)).~\qquad (**) $$ I cannot see this!

I think they are using the approximation $(*)$ for the integrand, i.e. something like (if neglecting the domain of integration): \begin{align*} g^{1/2}(v)&=\mu(1-v),\\ g^{-3/2}(v)&=\mu^{-3}(1-v)^{-3}. \end{align*}

Then, the inner integral gives $$ \int g^{-3/2}(t)\, dt=\frac{1}{\mu^3}\int\frac{1}{(1-t)^3}\, dt=\frac{1}{2\mu^3(1-t)^2}+C. $$ If I now use this in the integrand of the outer integral, what I get is $$ \frac{1}{2\mu^2}\int \frac{(1-s)}{(1-s)^2}\, ds=\frac{1}{2\mu^2}\int\frac{1}{1-s}\, ds=-\frac{1}{2\mu^2}\ln(\lvert 1-s\rvert). $$


(1) I guess the $O(1)$ term in $(**)$ is a correction for terms that behave slower than logarithmic?

(2) So far so good. But I dont see why I can apply $(*)$ for the integrand because for $v$ near $1$ the domain we integrate over is not near to $1$ by definition of $S$...

  • $\begingroup$ I don't think that you may use the expansion $g(x) \approx \mu(1-x)^2$ in the integral, since it's range is mostly NOT near $1$. $\endgroup$ – denklo Nov 9 '18 at 10:26
  • $\begingroup$ Your first line should read $o(v-1)^2$ as otherwise you cannot make the deduction about the behaviour of $g$ near 1. $\endgroup$ – Richard Martin Nov 9 '18 at 10:27
  • $\begingroup$ @denklo You are right. But this was the only idea I had in order to have a chance to get the claimed result. $\endgroup$ – Salamo Nov 9 '18 at 10:29
  • $\begingroup$ @Salamo Ok then, i just wanted to make sure you are aware of this. $\endgroup$ – denklo Nov 9 '18 at 10:30
  • $\begingroup$ Lets assume that we are integrating from $v$ to$1$ with $v$ near to $1$ and that we can use the approximation. Does this then give the claimed result? $\endgroup$ – Salamo Nov 9 '18 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.