Exit time of a Gaussian random walk on $[-a,a]$

Let $S_n=\sum_{i=1}^n X_i$ be a standard Gaussian random walk with i.i.d. increments $X_i \sim \mathcal{N}(0,1)$. Define the first exit time $\tau$ as $\tau=\inf_n\{|S_n|\geq a \}$. I'm trying to work out the moments of $\tau$ and also the distribution of $\tau$.

I find this question extremely difficult for me. I know that for simple random walks or Brownian motions, $E[\tau]=a^2$ by either solving the differential equation or using the reflection principle. However I simply do not know how to start with this question. Really appreciate any help you can provide.

Let's pretend for convenience that $\tau=\inf_n\{|S_n|> a \}$ (it won't change the answer). Let $X_0\in [-a,a]$ be the starting point of the random walk, and let $f(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}$ be the pdf of $\mathcal N(0,1)$.

If you let $t(x)=E[\tau|X_0=x]$ for $x\in[-a,a]$, then $$t(x)=\int_{-a}^af(u-x)(1+t(u))du=\int_{-a-x}^{a-x}f(z)(1+t(z+x))dz$$ so $$t'(x)=(t(a)+1)(f(x+a)-f(x-a))$$ And since $t(-a)=t(a)$, $$t(x)=(t(a)+1)\int_{-a}^x(f(u+a)-f(u-a))du+t(a)$$ $$t(x)=\frac12(t(a)+1)(erf(\frac{a-x}{\sqrt2})+erf(\frac{a+x}{\sqrt2})-erf(\sqrt2 a))+t(a)$$ $$t(0)=\frac12(t(a)+1)(2erf(\frac{a}{\sqrt2})-erf(\sqrt2 a))+t(a)$$

Unfortunately I don't know how to figure out what $t(a)$ is. Maybe I'm missing some initial condition or clever symmetry.