No, it isn't. The "jth integral of x" is , not .

Theexponentialfunction, , has the property you are talking about.

??? theAgain, this series can be expressed in the function

Now, my question is does there exist a function that expresses the series

where is the jth *derivative* of x?firstderivative of x is 1 and the "jth derivative of x" for j> 1 is 0. So such a series would be x+ 1+ 0+ 0+ 0...= x+ 1.

(...and I guess or something like that)

Thanks