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In this paper (Maroun's PhD dissertation, 2013) at page 46 the following formula is given (apparently without a reference):

$$\int_0^{\infty } e^{i a x^s+i b x^p} \, dx=\sum _{n=0}^{\infty } \frac{\left(i b a^{\frac{1}{s}}\right)^n \exp \left(\frac{(i \pi ) (n p+1)}{2 s}\right) \Gamma \left(\frac{n p+1}{s}\right)}{n! a^{\frac{1}{s}} \left| s\right| }$$

Now I am trying to verify the formula. If I take $a=b=i/2$, $s=p=1$ the left hand side becomes $1$ while right hand side becomes $8/3$. Is there some error in the formula? Is this formula even well-known?

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  • $\begingroup$ @Stefan Kohl I fixed the link, thanks for pointing. $\endgroup$
    – Anixx
    Commented Aug 4, 2017 at 13:49
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    $\begingroup$ A formula with parameters requires quantifiers on the parameters. It is possibly understood that the parameters are reals, or positive reals. I don't know about your specific formula. $\endgroup$
    – YCor
    Commented Aug 4, 2017 at 14:05
  • $\begingroup$ related question: mathoverflow.net/questions/277580/… $\endgroup$
    – Abdelmalek Abdesselam
    Commented Aug 4, 2017 at 14:38

2 Answers 2

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We may change the variables as $xa^{1/s}=t$, this proves that $a^{1/s}\cdot \int$ is a function of $ba^{-p/s}$, not of $ba^{1/s}$.

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  • $\begingroup$ Should I understand it that there is a typo, and the formula should have $b a ^{-p/s}$ instead? Then the right hand side apparently diverges... $\endgroup$
    – Anixx
    Commented Aug 4, 2017 at 14:28
  • $\begingroup$ Well, following regularization it gives now correct result, but I expected the right hand side to converge where the left had side does. $\endgroup$
    – Anixx
    Commented Aug 4, 2017 at 14:36
  • $\begingroup$ But, if a and b have opposite signs $a=1 b=-1$, while $p=s$ the result should be $0$, which is still not the case. $\endgroup$
    – Anixx
    Commented Aug 4, 2017 at 15:34
  • $\begingroup$ @Anixx, it is very common that different expressions for the same thing have different regions of convergence... and of convergence in various senses, perhaps not pointwise... $\endgroup$
    – paul garrett
    Commented Aug 4, 2017 at 20:56
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I am verifying that it is true that there is a typo. The corrected expression should be a change to the right hand side series term in parenthesis.

It should read, $$ \left(iba^{-\frac{p}{s}}\right)^n. $$ The original context was that the convergence is only in the sense of distributions.

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