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I would like to know Whether the integration

$\int_0^\infty\frac{s^{N_1+N_2}(2s^{N_1+1}-1)}{(1+s^{N_1+1})^4(1+s^{N_2+1})^2}ds$

is positive or negative? where $N_1,N_2$ are positive integers.

I am very grateful if anyone can give some ideas or solutions.

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    $\begingroup$ You probably want to add some context as to why you are interested in this. $\endgroup$ Jan 12, 2016 at 17:10
  • $\begingroup$ Thanks a lot. in fact, this integral arises from the calculation of the vortex nunbers. $\endgroup$ Jan 12, 2016 at 20:22
  • $\begingroup$ Do you need an answer for all $N_1,N_2$? $\endgroup$ Jan 13, 2016 at 10:30
  • $\begingroup$ Yes. Thanks. Maybe the sign is fixed for all positive integers $N_1, N_2$. $\endgroup$ Jan 13, 2016 at 11:06
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    $\begingroup$ According to WolframAlpha N[ integral_0^infinity (s^(3+2) (2 s^(3+1)-1))/((1+s^(3+1))^4 (1+s^(2+1))^2) ds] = -0.000858013 N[ integral_0^infinity (s^(3+4) (2 s^(3+1)-1))/((1+s^(3+1))^4 (1+s^(4+1))^2) ds] = 0.00247042 $\endgroup$ Jan 13, 2016 at 11:49

1 Answer 1

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No, the integral $$I(N_1,N_2)=\int_0^\infty\frac{s^{N_1+N_2}(2s^{N_1+1}-1)}{(1+s^{N_1+1})^4(1+s^{N_2+1})^2}ds$$ does not have a fixed sign for positive integers $N_1,N_2$; for example $$I(1,1)=-\frac{\pi}{512},\;\;I(1,2)=\frac{11}{8}+\frac{87\pi}{64}-\frac{28\pi}{9\sqrt{3}}\approx 0.00267$$ notice also that $I(2,2)=0$. More generally, $I(N_1,N_2)$ is negative for $N_2\ll N_1$ and positive for $N_2\gg N_1$.

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  • $\begingroup$ Hi Carlo Beenakker, Thanks a lot for your answer. Is there a formula to calculate the integral? How about the case when $N_1$ and $N_2$ are comparable? Does it has a fixed sign? Thanks again. $\endgroup$ Jan 13, 2016 at 12:19
  • $\begingroup$ It is nice that $I(2,2)=0$. Maybe there are more pairs of parameters when it equals 0? Then we have chance to describe a sign in general situation. $\endgroup$ Jan 13, 2016 at 12:34
  • $\begingroup$ the sign change happens approximately when $N_1\approx 3N_2$, but only $I(2,2)$ is identically zero. $\endgroup$ Jan 13, 2016 at 12:38
  • $\begingroup$ This looks interesting. Only $I(2,2)=0$. $\endgroup$ Jan 13, 2016 at 12:48
  • $\begingroup$ How about the sign for $I(2,1)$, $I(1,3)$? $\endgroup$ Jan 13, 2016 at 13:16

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