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?
 A: 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.
