Solve $$\int_0^1 \int_0^{2\pi}\frac{ax-x^2\sin(\theta)}{\sqrt{a^2-2ax\sin(\theta)+x^2}}d\theta dx$$ This integral is from the following paper : Frictional coupling between sliding and spinning motion and can be understood as an integration of unit vector field parallel to the velocity of a spinning disk which is sliding with the speed of $a$ over a unit disk. The evaluation of the integral is claimed to be

$$\frac{4}{3}\frac{(a^2+1)E(a)+(a^2-1)K(a)}{a}$$ for $a<1$, and

$$\frac{4}{3}((a^2+1)E(1/a)+(a^2-1)K(1/a))$$ for $a\geq1$ in the paper, where $K$ and $E$ are the first and second kind of complete elliptic integrals respectively. But for me this integration is not clear since if I do the integration for $x$ first then it results in a term with natural log which is irrelevant to elliptic integral, and doing it for $\theta$ first results in the terms including $E(\frac{4ax}{(a-x)^2})$ and $K(\frac{4ax}{(a-x)^2})$ which seems hard to be reduced to $E(a)$ and $K(a)$. What approach would give the most clarified evaluation? Any advice or help will be appreciated.

  • $\begingroup$ strange...Have u checked there result numerically? $\endgroup$ – tired Feb 1 '16 at 11:05
  • $\begingroup$ a small addition: i doubt that u can exchange integration limits freely here because the integral is singular for some choices of parameters (for example $\Theta=0, a<0$) and one has to be a little bit carefule $\endgroup$ – tired Feb 1 '16 at 11:39
  • $\begingroup$ I would try the change of variables $x\sin\theta=aw$ and $x\cos\theta=at$. $\endgroup$ – Pierpaolo Vivo Feb 1 '16 at 13:16
  • $\begingroup$ a quick check in mathematica suggests that the results they obtained are incorrect. For example setting $a=1.5$ i obtain $2.95567$ using NIntegrate and a value of $11.1862$ using the result they give. values of $a$ lead to similar descripancies $\endgroup$ – tired Feb 1 '16 at 14:42
  • $\begingroup$ @tired Thank you for you check. But I think the claimed evaluation value should not exceed $\pi$ for any value of $a$. (There is a graph for it in the paper above) $\endgroup$ – generic properties Feb 1 '16 at 15:27

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.