Is the following function real analytic in $t>0$:

$$F(t)=\int_0^1\frac{\exp(ctx)}{\sqrt{(\exp(bt)-1)(1-\exp(atx))-(1-\exp(at))(\exp(btx)-1)}} dx,$$

where $-a$ and $b$ are positive, and $c\not=a$?

I have consulted a large table of integrals looking for a closed form (for $t=1$), but without success.


This question arised during my efforts to show non-degeneracy of certain integrable systems. Real analyticity would make showing the non-degeneracy quite easy.

  • 2
    $\begingroup$ You don't need a closed form to show analyticity. You could try and prove a power series expansion. $\endgroup$ Aug 28, 2015 at 17:28
  • $\begingroup$ @AnthonyQuas Thank you very much for this comment. I have edited the question, which is now more explicit. $\endgroup$ Aug 28, 2015 at 17:41

1 Answer 1


Yes. As Anthony points out, you just expand in a power series at the singularities (everything is fine away from $0$ and $1$) and see what happens. What you see is that the integrand has a $1/\sqrt{x}$ singularity at both ends, so all is well. (as for actual expanding in a power series, I was lazy and used Mathematica)

  • 2
    $\begingroup$ Could you please explain, how to link it to the real-analyticity in $t$, $t>0$? It is possible to subtract $f_0(t)/\sqrt{x}+f_1(t)/\sqrt{1-x}$ ($f_0$ and $f_1$ real analytic) from the integrand, so to make it bounded. But then it appears to me (I have plotted the graph for some choice of parameters) that the remainder looks like $\sqrt{x}$, and $\int_0^1\sqrt{x+t^2}dx$ is not real analytic in $t$, as @Robert Israel pointed out, in the answer to my other question. $\endgroup$ Aug 29, 2015 at 6:59
  • $\begingroup$ @H.Berbeleque The singularity (in both cases) is at $0,$ and you only ask for real analyticity for $t>0.$ $\endgroup$
    – Igor Rivin
    Aug 29, 2015 at 12:10

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