Count valid unset bits in numbers from 1 to n in java

Java program for Count valid unset bits in numbers from 1 to n. Here mentioned other language solution.

/*
   Java program for
   Count valid unset bits in all numbers from 1 to n
*/
public class BitCount
{
	// Returns number of set bits of given number
	public int countSetBit(int num)
	{
		int x = num;
		x = x - ((x >>> 1) & 1431655765);
		x = (x & 858993459) + ((x >>> 2) & 858993459);
		x = (x + (x >>> 4)) & 252645135;
		x = x + (x >>> 8);
		x = x + (x >>> 16);
		x = x & 63;
		// Number of set bit
		return x;
	}
	// Handles the request of count the all 
	// unset bits from (1 to n)
	public void countUnsetBit(int n)
	{
		int result = 0;
		// Execute loop from 1 to n
		for (int i = 1; i <= n; ++i)
		{
			// Total bits - Active bits
			result += ((int)(Math.log(i) / Math.log(2.0) + 1.0)) - 
              countSetBit(i);
		}
		// Display result
		System.out.println(" Unset bits from (1 to " + n 
                           + ") : " + result);
	}
	public static void main(String[] args)
	{
		BitCount task = new BitCount();
		// Test A
		// number(1-7)  Un-setbit
		// ----------------
		// 1  = 1         0
		// 2  = 10        1
		// 3  = 11        0   
		// 4  = 100       2
		// 5  = 101       1
		// 6  = 110       1 
		// 7  = 111       0
		// --------------------
		//                5
		task.countUnsetBit(7);
		// Test B
		// number(1-10)  Setbit
		// ----------------
		// 1  = 1         0
		// 2  = 10        1
		// 3  = 11        0   
		// 4  = 100       2
		// 5  = 101       1
		// 6  = 110       1 
		// 7  = 111       0
		// 8  = 1000      3
		// 9  = 1001      2 
		// 10  = 1010     2 
		// --------------------
		//                12
		task.countUnsetBit(10);
	}
}

Output

 Unset bits from (1 to 7) : 5
 Unset bits from (1 to 10) : 12