The simplest and cleanest pre-ECMAScript-6 solution (which is also sufficiently robust to return false even if a non-numeric value such as a string or null is passed to the function) would be the following:
function isInteger(x) { return (x^0) === x; }
The following solution would also work, although not as elegant as the one above:
function isInteger(x) { return Math.round(x) === x; }
Note that Math.ceil() or Math.floor() could be used equally well (instead of Math.round()) in the above implementation.
Or alternatively:
function isInteger(x) { return (typeof x === 'number') && (x % 1 === 0); }
One fairly common incorrect solution is the following:
function isInteger(x) { return parseInt(x, 10) === x; }
While this parseInt-based approach will work well for many values of x, once x becomes quite large, it will fail to work properly. The problem is that parseInt() coerces its first parameter to a string before parsing digits. Therefore, once the number becomes sufficiently large, its string representation will be presented in exponential form (e.g., 1e+21). Accordingly, parseInt() will then try to parse 1e+21, but will stop parsing when it reaches the e character and will therefore return a value of 1. Observe:
> String(1000000000000000000000)
'1e+21'
> parseInt(1000000000000000000000, 10)
1
> parseInt(1000000000000000000000, 10) === 1000000000000000000000
false