Most Efficient Code for Nth Prime

public static int nthprime(int n) {
if (n == 1) return 2;
if (n == 2) return 3;
int limit, root, count = 1;
limit = (int)(n * (Math.log(n) + Math.log(Math.log(n)))) + 3;
root = (int) Math.sqrt(limit) + 1;
limit = (limit - 1) / 2;
root = root / 2 - 1;
boolean[] sieve = new boolean[limit];
for (int i = 0; i < root; ++i) {
if (!sieve[i]) {
++count;
for (int j = 2 * i * (i + 3) + 3, p = 2 * i + 3; j < limit; j += p) {
sieve[j] = true;
}
}
}
int p;
for (p = root; count < n; ++p) {
if (!sieve[p]) {
++count;
}
}
return 2 * p + 1;
}