# For what $d$ does $\sum\limits_{m=-\infty}^{\infty}\int_0^\infty e^{-t}\left[I_{|m|}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt$ converge?

For what $d\in\mathbb{N}$ does the following expression have a finite value?

$$u(d)=\sum_{m=-\infty}^{\infty}\int_0^\infty e^{-t}\left[I_{|m|}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt$$

$I_n(t)$ is a modified Bessel function of the first kind. For the context of the question, in case this is useful, I was attempting to solve the problem 'If $d$ people stand next to each other in a line and play a game where at each step they uniformly randomly choose one person who tosses a fair coin and if it comes up heads then he moves forwards $1$ metre and if tails he moves backwards $1$ metre, then choose another person to do the same; what is the probability $p(d)$ that they will ever be in a line next to each other again?'.

My solution used a similar technique to the method presented here for finding an expression for Polya's random walk constants. I got $p(d)=1-\frac{1}{u(d)}$ (although I'm not 100% sure it's correct), which would mean that for $d$ such that $u(d)$ diverges (since the functions are positive) I would expect $p(d)$ to be $1$, and when $u(d)$ converges I would expect it to be in $[1,\infty)$ giving a legitimate value for $p(d)$. Is this the case, and for what $d$ will $u(d)$ diverge?

I get that $u(d) =\sum\limits_{m=-\infty}^{\infty}\int_0^\infty e^{-t}\left[I_{|m|}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt$ diverges for all $d$.

I'll use the simplest asymptotics for large $z$ of $I_a(z) \approx \dfrac{e^z}{\sqrt{2\pi z}}$ (note that this is independent of $a$) and split each integral into two parts of $(0, d)$ and $(d, \infty)$.

Since $I_a(z) = \sum\limits_{j=0}^{\infty} \dfrac{(z/2)^{2j+a}}{j!(j+a)!}$, for small $z$, $I_{|m|}(z) \approx \dfrac{1}{|m|!}+\dfrac{(z/2)^{|m|+2}}{(|m|+1)!}$.

If $|z| \le 1$, $|I_a(z)| \le \sum\limits_{j=0}^{\infty} \dfrac{(1/2)^{2j+a}}{j!(j+a)!} = \sum\limits_{j=0}^{\infty} \dfrac{1}{j!(j+a)!2^{2j+a}} < \dfrac1{a!2^a}$.

Let $v(d) =\sum\limits_{m=-\infty}^{\infty}\int_0^d e^{-t}\left[I_{|m|}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt =\sum\limits_{m=-\infty}^{\infty} v_{|m|}(d)$ where $v_{m}(d) =\int_0^d e^{-t}\left[I_{m}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt$

Similarly, let $w(d) =\sum\limits_{m=-\infty}^{\infty}\int_d^{\infty} e^{-t}\left[I_{|m|}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt =\sum\limits_{m=-\infty}^{\infty} w_{|m|}(d)$ where $w_{m}(d) =\int_d^{\infty} e^{-t}\left[I_{m}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt$.

Then:

$$|v_{m}(d)|\le\int_0^d e^{-t}\left|I_{m}\left(\frac{t}{d}\right)\right|^d \;\mathrm dt$$ $$\le\frac1{(m!2^m)^d}\int_0^d e^{-t} \;\mathrm dt$$ $$\le\frac1{(m!2^m)^d}$$ so: $$v(d)\le2\sum_{m=0}^{\infty} \frac1{(m!2^m)^d}$$

which converges rather violently.

Now comes the interesting part.

$$w_{m}(d)=\int_d^{\infty} e^{-t}\left[I_{m}\left(\frac{t}{d}\right)\right]^d \;\mathrm dt$$ $$\approx\int_d^{\infty} e^{-t}\left[ \dfrac{e^{t/d}}{\sqrt{2\pi t/d}}\right]^d \;\mathrm dt$$ $$= \int_d^{\infty} \dfrac{1}{(2\pi t/d)^{d/2}} \;\mathrm dt$$ $$= \left(\dfrac{d}{2\pi}\right)^{d/2}\int_d^{\infty} \dfrac{\mathrm dt}{t^{d/2}}$$

If $d \le 2$, the integral diverges.

If $d > 2$, $w_m(d) \approx \left(\dfrac{d}{2\pi}\right)^{d/2}\dfrac{d^{1-d/2}}{d/2-1} =\dfrac{d}{(2\pi)^{d/2}(d/2-1)} =\dfrac{1}{(2\pi)^{d/2}(1/2-1/d)}$.

Since all the $w_m(d)$ are asymptotically independent of $m$ (which surprises me), the sum of them diverges.

Therefore, the overall sum diverges.

• Nice answer. +1, including for describing convergence as violent. Could you add some links e.g. for the Bessel function asymptotics? – Anon Jan 9 '17 at 0:50
• I just got it from good old Wikipedia: en.wikipedia.org/wiki/Bessel_function – marty cohen Jan 9 '17 at 1:10