Derivative and integral of the exponential e^t

[DiracPool]

TL;DR Summary

Is the integral of e to the t the same as the derivative?

If I take the exponential function e^t and take the derivative, I think I get the same e^t. Even if I keep doing it over and over, second, third derivative, etc. My admittedly naive question, though, is this symmetric? Meaning...if I take the the integral of e^t, do I just get the reverse or do I have t deal with an infinity of constants because it is an indefinite integral?

 

[PeroK]

Summary: Is the integral of e to the t the same as the derivative?

If I take the exponential function e^t and take the derivative, I think I get the same e^t. Even if I keep doing it over and over, second, third derivative, etc. My admittedly naive question, though, is this symmetric? Meaning...if I take the the integral of e^t, do I just get the reverse or do I have t deal with an infinity of constants because it is an indefinite integral?

Yes, an indefinite integral produces an equivalence class of functions, not a unique function.

This is important as you can see from the following example.

We want to find functions, $f(t)$ that satisfy $f''(t) = e^t$. Clearly $f(t) = e^t$ is one such function. But, $f(t) = e^t + At + B$ is also a solution to this differential equation, where $A, B$ are any two constants.

Another way to look at this is to say that if you integrate $f''(t) = e^t$ you do not get a unique result.

Differentiation, on the other hand, always gives a unique function as a result.

 

[HallsofIvy]

Yes, the derivative of $e^x$ is again $e^x$ and so the nth derivative, for all n, of $e^x$ is $e^x$ and all integrals of $e^x$ are again $e^x$. In fact, that is why "e" is defined as it is. One can show that the derivative or $a^x$, for a any positive number, is $C_aa^x$ where "$C_a$" is a constant (independent of x) that depends upon a. "e" is the unique number such that $C_e= 1$.