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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 $8/3$.

It had been suggested on MathOverflow, that the formula has a mistake, and the right-hand part should contain $ba^{-p/s}$ instead of $ba^{1/s}$. In this case with the above-mentioned data the equality holds, but still it does not work for $a=-b$; $p=s$.

Is there some error in the formula? Is this formula even well-known so to look somewhere for the correct form?

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  • $\begingroup$ It's not a standard formula, but the derivation should be straightforward. You expand the LHS as a Taylor series in $b$, which forces you do to integrals of the form $\int_0^\infty dx \, x^{n p} \exp(i a x^s)$. This one is standard, see Gradsteyn-Ryzhik 3.326. $\endgroup$ – user159249 Aug 4 '17 at 16:43
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It is correct that there is a typo in the right hand side series expression. The term in the numerator of the series inside of parenthesis raised to the n power should be $$ \left(iba^{-\frac{p}{s}}\right)^n. $$

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  • $\begingroup$ The expression, with the typo taken into account, gives the correct pointwise equivalence for both a=b=i/2 with s=p=1 and for a=-b with s=p=1. In the first case, the result is 1. In the second case, the result is infinity on both sides. The original context is to show that pointwise equality leads to misleading statements for arbitrary a, b, s, and p. The point was to understand that although $x\delta(x) "=" sin(x)\delta(x)$ in the sense of distributions, without test functions the equal sign is not justified. Otherwise, one concludes $\pi\delta(x) = 0\cdot\delta(x)$, which is of course false. $\endgroup$ – Michael Maroun Aug 8 '17 at 16:35
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Taylor series Expansion of $e^{ibx^p}$ yields

$\sum_{n=0}^\infty \frac{(ib)^n}{n!}x^{np}$.

Then you have to evaluate every term of the series by a Substitution. You can use the Substitution $u = -iax^s$ which yields $x = (i\frac{u}{a})^{1/s}$ and

$dx = e^{\frac{i \pi}{2s}}a^{-1/s}u^{1/s-1}/s$.

The integral $I_n = \int_0^\infty dx x^{np}e^{iax^s}$ turns to $I_n = \int_0^\infty \frac{du}{su} (i\frac{u}{a})^{np/s+1/s} e^{-u} = \int_0^\infty \frac{du}{su}(ia^{-1})^{np/s+1/s}u^{np/s+1/s} e^{-u} = e^{i \pi \frac{np+1}{2s}}a^{-np/s}/(a^{1/s}s) \int_0^\infty du u^{np/s+1/s-1} e^{-u} = \frac{e^{i \pi \frac{np+1}{2s}}{ \Gamma(\frac{np+1}{s})}a^{-np/s}}{a^{1/s}s}$.

Conclusion: You see that this series Expansion will take constants $(iba^{-p/s})^n$ that belong to the term $I_n$; the integral in your question has a small mistake.

The correct formula should be:

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

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  • $\begingroup$ Changing the sign still does not change the second test case... It still won't pass. $\endgroup$ – Anixx Aug 4 '17 at 17:19
  • $\begingroup$ Sorry, I have not completed it. I will write the correct one now. $\endgroup$ – kryomaxim Aug 4 '17 at 17:20
  • $\begingroup$ Taking $a=-b$ and $p=s=1$ gives $\sum 2$ on the right, regularized value is 1 while it should be 0. $\endgroup$ – Anixx Aug 4 '17 at 17:21
  • $\begingroup$ The last variant is what they suggested me on Mathoverflow. Yet it does not pass the second test. $\endgroup$ – Anixx Aug 4 '17 at 17:26

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