Currently, I'm working on a problem pertaining to certain integrals involving the modified Bessel function of the first kind. On p. 59 of this book by Rosenheinrich, it is stated that

$$\int e^{-x} I_{0}(x) dx = x e^{-x}[I_{0}(x) + I_{1}(x)] .$$

Moreover, Wolfram obtains: $$\int I_{0}(2 \sqrt{x}) = \sqrt{x} I_{1}(2 \sqrt{x}) + c. $$

Definite integrals are also of interest to me. For instance , we have: $$\int_{0}^{\infty} e^{-x} I_{0}(2 \sqrt{x}) dx = e. $$

Now, I'm looking for results on the following integral:

$$\int \Big{(} \frac{1}{e^{x}-1} - \frac{1}{e^{x}} \Big{)} \Big{(} I_{0}(2 \sqrt{x}) - 1 \Big{)} dx \qquad \qquad (*) $$

The definite version for $x$ ranging over the nonnegative reals amounts to about $0,761$.

Question: have any results been obtained on the integral stated in $(*)$ - with regards to both the indefinite and definite version?


1 Answer 1


Just a partial answer for the definite version, name it $\beta_0$.

Using Laplace Transform from this Maple 2021 output

 $$ I(p) = \int_{0}^{\infty} e^{-px} (I_{0}(2 \sqrt{x}) - 1) dx = \frac{e^{1/p}-1}{p}$$ Since $$\frac{1}{e^{x}-1}-\frac{1}{e^x} = \frac{e^{-2x}}{1-e^{-x}}= \sum_{p=2}^{\infty} e^{-px}$$ the integral reduces to the following series $$\beta_0 = \sum_{p=2}^{\infty} \frac{e^{1/p}-1}{p} = 0.76137869727328467225018681101291207836...$$ Logarithmically convergent monotone sum that can be obtained using either Levin's-u sequence transformation or Euler-MacLaurin. In this last case PARI/GP provides this value

enter image description here

Expanding $e^{1/p}-1$ as a Taylor series we get $$\beta_0 = \sum_{p=2}^{\infty} \sum_{n=1}^{\infty}\frac{1}{n!}p^{-n-1}=\sum_{n=1}^{\infty}\frac{1}{n!}\sum_{p=2}^{\infty}\frac{1}{p^{n+1}}$$

a much faster convergent series in terms of Riemann function $\zeta(z)$ $$\beta_0 = \sum_{n=1}^{\infty} \frac{\zeta(n+1)-1}{n!}$$

  • 1
    $\begingroup$ If you have Bessel functions $I_0$ in there, it seems confusing to call the integral also $I_0$. $\endgroup$ Nov 24, 2021 at 21:21
  • 1
    $\begingroup$ @Gerald_Edgar. Corrected. Thanks a lot for pointing this out. $\endgroup$ Nov 24, 2021 at 21:30
  • $\begingroup$ @JorgeZuniga thank you for you answer. Could you perhaps elaborate a little bit on why the definite integrals amounts to the sum of the laplace transform you've computed? $\endgroup$
    – Max Muller
    Nov 25, 2021 at 0:08
  • 1
    $\begingroup$ @Max_Muller. Ok, I will provide more details. $\endgroup$ Nov 25, 2021 at 1:38

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