This issue's title has been changed, as the implementation of delayWhen is not correct - see the discussion below.

The original issue's description follows.

RxJS version: 6.1.0

Code to reproduce:

import { EMPTY, of } from "rxjs";
import { delay, delayWhen } from "rxjs/operators";

of(0).pipe(
    delayWhen(() => EMPTY.pipe(delay(0)))
).subscribe(console.log);

Expected behavior:

0 should be emitted and logged to the console.

Actual behavior:

Nothing is emitted and logged to the console.

If the of function is passed 1 instead of 0, 1 is logged to the console.

Additional information:

Looking at delayWhen, I can see another problem (other than #3663) with the implementation. If the value is falsy, and the notifier completes - rather than nexts - the value won't be emitted.