In new mathematica 12 there is a new function AsymptoticIntegrate. However, it seems that it gives me incorrect results in some cases. To be completely honest, it is hard to tell if those results are indeed incorrect because the mathematical expressions are quite complex. So, I'm interested if someone else from this website has had similar experience? I want to know with what degree of certainty can I believe that the mathematica output in this case is indeed correct, having in mind that this function is quite new.

I do know that there are some examples where mathematica gives the wrong result, that's why I'm posting this question in the first place.

For example:

    Integrate[Cos[x]*BesselJ[0, 2 x], {x, 0, Infinity}]

gives $\frac{1}{\sqrt 3}$. However

AsymptoticIntegrate[Cos[x]*BesselJ[0, 2 x], {x, 0, t}, {t, Infinity, 1}]

gives BesselJ[0, 2 t] Sin[t] which is zero in the limit $t\to \infty$.

But these two examples should give the same result in the limit $t \to \infty$, right?

  • 6
    $\begingroup$ You should post one or two examples. Otherwise there is nothing concrete to analyze and respond to. $\endgroup$ Commented Jan 2, 2021 at 15:24
  • $\begingroup$ @DanielLichtblau I added an example now. Thanks for the suggestion. $\endgroup$
    – RedGiant
    Commented Jan 2, 2021 at 16:46
  • $\begingroup$ The improper integral Integrate[Cos[x]*BesselJ[0, 2 x], {x, 0, Infinity}] is not found as the limit of Integrate[Cos[x]*BesselJ[0, 2 x], {x, 0, t}, Assumptions -> t > 0] as t tends to infinity in Mathematica since the latter command returns the input. $\endgroup$
    – user64494
    Commented Jan 2, 2021 at 19:03
  • 1
    $\begingroup$ @RedGlant: A quit good asymptotics for the integral Integrate[Cos[x]*BesselJ[0, 2 x], {x, 0, t}] is 1/Sqrt[3] - (-(Cos[t]/(2 Sqrt[2 \[Pi]] Sqrt[t])) - Sin[t]/( 2 Sqrt[2 \[Pi]] Sqrt[t])) as t approaches infinity along the real axis, This can be found with help of Mathematica. $\endgroup$
    – user64494
    Commented Jan 2, 2021 at 19:47
  • 1
    $\begingroup$ @RedGlant: You are right. Thank you for your notice. The leading terms are Cos[t]/(2 Sqrt[2 \[Pi]] Sqrt[t]) - Cos[3 t]/( 6 Sqrt[2 \[Pi]] Sqrt[t]) + Sin[t]/(2 Sqrt[2 \[Pi]] Sqrt[t]) + Sin[3 t]/(6 Sqrt[2 \[Pi]] Sqrt[t]). It does not change the matter. $\endgroup$
    – user64494
    Commented Jan 2, 2021 at 20:35

1 Answer 1



Cos[x]*MellinTransform[BesselJ[0, a*x], a, s], {x, 0, t}, {t, 
Infinity, 1}, Assumptions -> s > 0] // FullSimplify, s, a] /. a -> 2

(*1/Sqrt[3] - 2 BesselJ[1, 2 t] Cos[t] + BesselJ[0, 2 t] Sin[t]*)

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