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Does anyone have advice on how to go about finding (if it exists) a closed form for

$\sum_{n=0}^{\infty}\frac{J_n(n)}{n!}$?

Where $J_n$ represents the Bessel function of the first kind; numerically it appears to converge to $\approx 1.68226...$

I tried playing around with the recurrence relations for the Bessel functions in an attempt to find an exponential generating function for $J_n(n)$, but this lead to nonsensical results. Any other ideas?

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1 Answer 1

up vote 4 down vote accepted

We have that $$J_n(x) = \dfrac1{2 \pi} \int_{-\pi}^{\pi} e^{-i(n t-x \sin(t))} dt$$ Hence, $$J_n(n) = \dfrac1{2 \pi} \int_{-\pi}^{\pi} e^{-in(t-\sin(t))} dt = \dfrac1{2 \pi} \int_{-\pi}^{\pi} (a(t))^{n} dt$$ where $e^{-i(t-\sin(t))} = a(t) = \cos(t-\sin(t))-i\sin(t-\sin(t))$. We then have $$\sum_{n=0}^{\infty} \dfrac{J_n(n)}{n!} = \dfrac1{2\pi} \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \dfrac{(a(t))^{n}}{n!} dt = \dfrac1{2 \pi} \int_{-\pi}^{\pi} \exp(a(t)) dt\\ = \dfrac1{2 \pi} \int_{-\pi}^{\pi} \exp(\cos(t-\sin(t))) \cdot \cos(\sin(t-\sin(t))) dt$$which approximately evaluates to $1.68226$.

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Thanks! It's nice how that works out. –  Bitrex Jan 13 '13 at 5:27
    
@Bitrex There might be some neat "closed" form way to evaluate the integral as well. Nothing comes to my mind now though. –  user17762 Jan 13 '13 at 5:28

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