I would like to understand if the following two infinite series can be further expressed in terms of known functions:

$$\sum_{n\geq 0} (-1)^nI_{2n+1}(A)\frac{\cos((2n+1)B)}{2n+1}$$


$$\sum_{n\geq 1} (-1)^nI_{2n}(A)\frac{\sin(2nB)}{2n}$$

where $A$ is a positive real constant and $B$ is real. $I_n(\cdot)$ is the nth order of the modified Bessel function of first kind.


Have a look at this question. We have that:

$$ e^{A\sin x}= I_0(A) + 2\sum_{n\geq 0}(-1)^n I_{2n+1}(A)\sin((2n+1) x)+2\sum_{n\geq 1}(-1)^n I_{2n}(A)\cos(2nx) $$ hence by considering the even part and replacing $x$ with $Bx$ we get: $$ \cosh(A\sin(Bx))= I_0(A) + 2\sum_{n\geq 1}(-1)^n I_{2n}(A)\cos(2nBx) $$ and: $$ I_0(A)+2\sum_{n\geq 1}(-1)^{n}I_{2n}(A)\frac{\sin(2nB)}{2nB}=\int_{0}^{1}\cosh(A\sin(Bx))\,dx $$ so, by rearranging, $$\sum_{n\geq 1}(-1)^n I_{2n}(A)\frac{\sin(2nB)}{2n}=\color{red}{-\frac{B}{2} I_0(A)+\frac{B}{2}\int_{0}^{1}\cosh(A\sin(Bx))\,dx} $$ and the other series can be managed in a similar way. From:

$$ \sinh(A\sin(Bx))= 2\sum_{n\geq 0}(-1)^n I_{2n+1}(A)\sin((2n+1)Bx) $$ it follows that: $$ \color{red}{\frac{B}{2}\int_{0}^{1}\sinh(A\sin(Bx))\,dx} = \sum_{n\geq 0}(-1)^n I_{2n+1}(A)\frac{1-\cos((2n+1)B)}{2n+1}.$$

  • $\begingroup$ Ok, there is then no closed formula for the series. Thanks anyway $\endgroup$ – JFNJr Jun 25 '16 at 19:08
  • $\begingroup$ Could you also explain how to manage the other series? By integrating from 0 to 1 I obtain the extra series $\sum_{n\geq0}(-1)^n\frac{I_{2n+1}(A)}{B(2n+1)}$. Is it equal to some known function? $\endgroup$ – JFNJr Jun 25 '16 at 19:50
  • $\begingroup$ @JFNJr: by integrating over $(0,1)$ which function? I got $\frac{\sin(2nB)}{2nB}$ exactly by integrating $\cos(2nB x)$ (with respect to $x$) on the $(0,1)$ interval. $\endgroup$ – Jack D'Aurizio Jun 25 '16 at 19:57
  • $\begingroup$ I refer to the other series I want to express. Could you please write down the steps also for that? $\endgroup$ – JFNJr Jun 25 '16 at 20:20
  • $\begingroup$ @JFNJr: they are pretty much the same, you just have to take the odd part. Anyway, updating. $\endgroup$ – Jack D'Aurizio Jun 25 '16 at 20:22

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