I am trying to evaluate the following integral numerically, however I find the value of integral drops to zero at large large $r$ values. Is there any technique to evaluate the integral at any value of $r$ without facing such a problem even using approximate techniques? The figure below shows the integral value versus r at different t values. The curve in black does not show the problem, however the one in red shows the problem; at $r = 4$, the integral drops to zero and the total output goes to one.

$$I =1 -\int_{0}^{\infty} du \left (u^{1 - \beta}\right) e^{-\left ( \frac{u^2 + r^2}{4t}\right)}I_\beta\left ( \frac{ur}{2t}\right)$$enter image description here

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    $\begingroup$ For what values of $\beta$ and $t$ do you get a problem? $\endgroup$ Aug 14, 2020 at 17:50
  • $\begingroup$ The problem appears for large $r$ values at any $t$ and $\beta$ values. In the figure, the integral works well before $r = 4$ and after that it drops to zero. $\beta$ can be any integer and $t$ is real number that should be higher than zero. In this plot, $t$ is around 1E-5 and $\beta$ is 14 @RobertIsrael $\endgroup$
    – Engineer
    Aug 14, 2020 at 18:01

1 Answer 1


It seems the integrand has a very sharp peak near $u=r$, and it's easy for numerical integrators to miss it entirely.

The exact value of the integral, according to Maple, is

$$ {\frac {2 t}{ r^\beta\; \Gamma \left( {\beta} \right) } \left( \Gamma \left( {\beta} \right) -\Gamma \left( {\beta},{\frac {{r}^{2}}{4t}} \right) \right) }$$

  • $\begingroup$ Do I need to exclude $r = u$ during the numerical integration? Is that what you mean? @Robert Israel $\endgroup$
    – Engineer
    Aug 14, 2020 at 23:14
  • $\begingroup$ No, I mean that most of the area under the curve seems to be in a small interval around $u=r$. $\endgroup$ Aug 14, 2020 at 23:43
  • $\begingroup$ Do you have any suggestions to solve the problem? I am thinking to divide the integral into two integrals; the first integral has boundaries from 0 to $r-$ and the second integral has boundaries from $r+$ to infinity.@Robert Israel $\endgroup$
    – Engineer
    Aug 15, 2020 at 19:11
  • $\begingroup$ That might help. $\endgroup$ Aug 16, 2020 at 1:56

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