Re: integral zeta

quasi wrote:

To tommy1729:

If you insist that primitives are uniquely determined
simply by the
concept of "just set C=0", you are going to run into
some problems.

For example, let f(x )= e^x - x.

Let's integrate f(x) two different ways ....

method (1) [standard Calc 1 approach]:

int f(x) dx = e^x - x^2/2 + C

Setting C=0 yields e^x - x^2/2

method (2) [using power series]:

e^x = 1 + x + x^2/2 + x^3/6 + x^4/24 + ...

f(x) = e^x - x = 1 + x^2/2 + x^3/6 + x^4/24 + ...

int f(x) dx = (x + x^3/6 + x^4/24 + x^5/120 + ...
... ) + C

Setting C=0 yields x + x^3/6 + x^4/24 + x^5/120 +
0 + ...

which equals e^x - x^2/2 - 1.

So which one is the unique primitive?

quasi

intresting argument. :-)

first note : i have already defined "integral zeta" by a series.

although perhaps there might be discussion about the analytic continuation or riemann surface ....

but anyways "integral zeta" is already defined...

in your argument, perhaps a good way to clarify is this

f(0) = INTEGRAL f(0)dx if possible by changing C.

so f(x) = exp(x) - x

f(0) = 1 = integral f(0)

therefore the integral is method(1).

another reason for avoiding series is the taylor expansion of exp(x^2) at 0 ...

you know what happens there dont you ?

a bit off topic perhaps but can you give nice examples of an integral which can take 3 or more values ...

just for fun :-)

regards
tommy1729
.