In v. Mathematica for Linux

Integrate[Sin[Cosh[t]], {t, 0, Infinity}]


Integrate::idiv: Integral of Sin[Cosh[t]] does not converge on {0,[Infinity]}.

while the integral is clearly bounded.

NIntegrate[Sin[Cosh[t]], {t, 0, Infinity}]

seems to return a finite value.

Any idea of what is going on?

  • 2
    $\begingroup$ this is very highly oscillatory function. Are you sure it is NIntegrate which got it right, and not Integrate? Intergate can not integrate the indefinite integral also. So it must have done some checking before. $\endgroup$
    – Nasser
    Commented May 18, 2020 at 22:23
  • 2
    $\begingroup$ @Nasser I think it converges, but I don't know whether it can be computed by Mathematica in terms of known functions. $\endgroup$
    – Michael E2
    Commented May 18, 2020 at 23:07
  • 1
    $\begingroup$ In summary: please report this to Support. $\endgroup$ Commented May 19, 2020 at 11:17

2 Answers 2


I believe the integral in question converges conditionally and can be calculated after substituting $x = \cosh(t)$, giving (on the integration domain) $$t = \text{arccosh}(x), \;\; dt = \frac{dx}{\sqrt{x^2-1}}.$$

Integrate[Sin[x]/Sqrt[x^2 - 1], {x, 1, ∞}]
1/2 π BesselJ[0, 1]

We can numerically integrate over a large finite range, forcing each zero to be integrated over. This will ensure we're not skipping over any peaks or valleys.

NIntegrate[Sin[Cosh[t]], {t, ##}] & @@ Prepend[ArcCosh[π Range[10000]], 0]

This extended syntax of NIntegrate is discussed here.


Using the undocumented function Integrate`InverseIntegrate[]:

Integrate`InverseIntegrate[Sin[Cosh[t]], {t, 0, ∞}]
   1/2 π BesselJ[0, 1]

N[%, 20]

which is consistent with Chip's reformulation:

NIntegrate[Sin[x]/Sqrt[x^2 - 1], {x, 1, ∞}, 
           Method -> "DoubleExponentialOscillatory", WorkingPrecision -> 20]

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