使用Rx.NET从另一个可观察对象通知异步对象

我想通知订阅的听众 Subject _sub 从另一个可观察到的,然后记录一些信息 Do _sub 不是异步的。这里的问题是,在第一个可观察到的情况下,没有我会等待的OnNextAsync函数。最好的办法是什么?

 class Program
        {
            private static Subject<int> _sub = new Subject<int>();

            static void Main(string[] args)
            {
                _sub.SelectMany(async _ =>
                {
                    Console.WriteLine("SUB START: " + _);
                    await Task.Delay(3000);
                    Console.WriteLine("SUB END: " + _);
                    return 1;
                }).Subscribe();


                Start();
            }

            public static void Start()
            {
                int count = 0;
                Observable.Interval(TimeSpan.FromSeconds(5)).Select(x =>
                {
                    Console.WriteLine("START INTERVAL");
                    _sub.OnNext(count++); //onNext is not awaitable
                    Console.WriteLine("END INTERVAL");
                    return 1;
                })
                .Do(_ => Console.WriteLine("ALL FINISHED"))
                .Subscribe();

                Console.WriteLine("READLINE");
                Console.ReadLine();
            }

        }

结果:

READLINE
START INTERVAL
SUB START: 0
END INTERVAL
ALL FINISHED
SUB END: 0

READLINE
START INTERVAL
SUB START: 0
SUB END: 0
END INTERVAL
ALL FINISHED