Generating 10 digits unique random number in java

So you want a fixed length random number of 10 digits? This can be done easier:

long number = (long) Math.floor(Math.random() * 9_000_000_000L) + 1_000_000_000L;

Note that 10-digit numbers over Integer.MAX_VALUE doesn't fit in an int, hence the long.

I think the reason you're getting 8/9 digit values and negative numbers is that you're adding fraction, a long (signed 64-bit value) which may be larger than the positive int range (32-bit value) to aStart.

The value is overflowing such that randomNumber is in the negative 32-bit range or has almost wrapped around to aStart (since int is a signed 32-bit value, fraction would only need to be slightly less than (2^32 - aStart) for you to see 8 or 9 digit values).

You need to use long for all the values.

   private static void createRandomInteger(int aStart, long aEnd, Random aRandom){
    if ( aStart > aEnd ) {
      throw new IllegalArgumentException("Start cannot exceed End.");
    }
    //get the range, casting to long to avoid overflow problems
    long range = aEnd - (long)aStart + 1;
    logger.info("range>>>>>>>>>>>"+range);
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    logger.info("fraction>>>>>>>>>>>>>>>>>>>>"+fraction);
    long randomNumber =  fraction + (long)aStart;    
    logger.info("Generated : " + randomNumber);

  }