# Can $\int \Big{(} \frac{1}{e^{x}-1} - \frac{1}{e^{x}} \Big{)} \Big{(} I_{0}(2 \sqrt{x}) - 1 \Big{)} dx$ be evaluated?

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?

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

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 $$\zeta(z)$$ $$\beta_0 = \sum_{n=1}^{\infty} \frac{\zeta(n+1)-1}{n!}$$

• If you have Bessel functions $I_0$ in there, it seems confusing to call the integral also $I_0$. Nov 24, 2021 at 21:21
• @Gerald_Edgar. Corrected. Thanks a lot for pointing this out. Nov 24, 2021 at 21:30
• @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? Nov 25, 2021 at 0:08
• @Max_Muller. Ok, I will provide more details. Nov 25, 2021 at 1:38