How to check if a variable is an integer in JavaScript?

That depends, do you also want to cast strings as potential integers as well?

This will do:

function isInt(value) {
  return !isNaN(value) && 
         parseInt(Number(value)) == value && 
         !isNaN(parseInt(value, 10));
}

With Bitwise operations

Simple parse and check

function isInt(value) {
  var x = parseFloat(value);
  return !isNaN(value) && (x | 0) === x;
}

Short-circuiting, and saving a parse operation:

function isInt(value) {
  if (isNaN(value)) {
    return false;
  }
  var x = parseFloat(value);
  return (x | 0) === x;
}

Or perhaps both in one shot:

function isInt(value) {
  return !isNaN(value) && (function(x) { return (x | 0) === x; })(parseFloat(value))
}

Tests:

isInt(42)        // true
isInt("42")      // true
isInt(4e2)       // true
isInt("4e2")     // true
isInt(" 1 ")     // true
isInt("")        // false
isInt("  ")      // false
isInt(42.1)      // false
isInt("1a")      // false
isInt("4e2a")    // false
isInt(null)      // false
isInt(undefined) // false
isInt(NaN)       // false

Here’s the fiddle: http://jsfiddle.net/opfyrqwp/28/

Performance

Testing reveals that the short-circuiting solution has the best performance (ops/sec).

// Short-circuiting, and saving a parse operation
function isInt(value) {
  var x;
  if (isNaN(value)) {
    return false;
  }
  x = parseFloat(value);
  return (x | 0) === x;
}

Here is a benchmark:
http://jsben.ch/#/htLVw

If you fancy a shorter, obtuse form of short circuiting:

function isInt(value) {
  var x;
  return isNaN(value) ? !1 : (x = parseFloat(value), (0 | x) === x);
}

Of course, I’d suggest letting the minifier take care of that.