Area under random walk The problem - we have random walk set of functions such that:
$$
f(0) = 0, f(x) = f(i) + \alpha_i(x-i), x \in (i,i+1], i=0,\dots,n-1
$$
where $\alpha_i \in \{-1, +1\}$ choosed uniformly with plobability $\dfrac{1}{2}$. 
Need to find probability that $P(\int\limits_0^n f(x) dx = 0)$.
Thank you
 A: As results above justify, consider sum $S = \sum\limits_{i=0}^{n-1}(n-i+\frac{1}{2})\alpha_i = (n+\frac{1}{2})\sum\limits_{i=0}^{n-1}\alpha_i - \sum\limits_{i=0}^{n-1}i \alpha_i$
Denote as $n^{+}$ and $n^{-}$ number of steps up and down respectively. Clear, $n=n^{+}+n^{-}$. Then:
$$
(n+\frac{1}{2})\sum\limits_{i=0}^{n-1}\alpha_i = (n+\frac{1}{2})(n^{+}-n^{-})
$$ 
Also, we can work around another term:
$$
\sum\limits_{i=0}^{n-1}i\alpha_i = \sum\limits_{i: \alpha_i = 1}i - \sum\limits_{j: \alpha_j = -1}j = \frac{n(n-1)}{2}-2\sum\limits_{j: \alpha_j = -1}j
$$
Therefore:
$$
S = (n+\frac{1}{2})[n^{+}-n^{-}] - \frac{n(n-1)}{2} + 2\sum\limits_{j: \alpha_j = -1}j 
$$
$$
P(S=0) = P\left(\frac{n(n-1)}{2} - (n+\frac{1}{2})[n^{+}-n^{-}]= 2\sum\limits_{j: \alpha_j = -1}j \right) = P\left(\frac{1}{4}(n(n-1) - (2n+1)(n^{+}-n^{-})) = \sum\limits_{j: \alpha_j = -1}j\right)
$$
Also, $P(S=0) = \sum\limits_{k=0}^{n-1}P(S=0|n^{-}=k)P(k)$
Fix $n^{-}=k$ first. Then we have:
$P(S=0|k) = P\left(\frac{1}{4}(n(n-1) - (2n+1)(2n-k)) = \sum\limits_{j: \alpha_j = -1}j\right)$
Denote $G(k) = \frac{1}{4}(n(n-1) - (2n+1)(2n-k)$, as $k$ fixed it's fixed number. And $\sum\limits_{j: \alpha_j = -1}j$ is sum of positive $k$ terms. The number of ways to present $G(k)$ in the form of sum positive k terms is $$\binom{G(k)+k-1}{G(k)-k}$$
Number of all possible ways to choose $k$ moments of decreasing from $n$ moments is $$\binom{n}{k}$$
Hence
$$
P(S=0|k) = \frac{\binom{G(k)+k-1}{G(k)-k}}{\binom{n}{k}}
$$
And 
$$
P(S=0) = \sum\limits_{k=0}^{n-1} \frac{\binom{G(k)+k-1}{G(k)-k}}{\binom{n}{k}} P(k) = \sum\limits_{k=0}^{n-1} \frac{\binom{G(k)+k-1}{G(k)-k}}{\binom{n}{k}} \binom{n}{k}\frac{1}{2^n} = \frac{1}{2^n}\sum\limits_{k=0}^{n-1}\binom{G(k)+k-1}{G(k)-k}
$$
A: Partial result:
Computation of the integral, first we compute the integral $$\int_i^{i+1} f(x)dx = f(i)+\frac12\cdot \alpha_i = \sum_{j=0}^i \alpha_j +\frac12\alpha_i$$ 
So the total integral is $$\sum_{i=0}^n \sum_{j=0}^i\alpha_j +\frac12\alpha_i=\sum_{i=0}^n(n-i+\frac12)\cdot\alpha_i$$
From this we see, that if $n$ is uneven, then the result is not an integer. Hence for those cases the probability is 0.
