09/14/2014, 05:07 PM

(09/13/2014, 11:49 PM)jaydfox Wrote:

Notice that I put "approximately equal". I haven't checked, but I assume it's exactly equal, but only in the sense that it should satisfy the functional equation exp'(x) = exp(x).

Now, set beta = 1/2 and truncate the negative powers of x:

Hey I wanted to post that !

Btw the equation is f ' (x) = f(x) + o(1) for x sufficiently large.

I had the exact same proof ...

I knew I was running out of time since it looked so familiar !

Did's answer on MSE is the classical way to prove it.

There are many proofs of this actually.

But Jay posted the shortest I think.

Gauss and Euler must have known this already.

...

However I still notice nobody has given a justification for sheldon's integral.

This is the second time it gives a correct solution.

It would also give the same solution for exp(x)ln(x+1) + exp(x)sqrt(x).

In general using sheldon's integral we can say

If +fake( f ) and +fake ( g ) exist then

+fake( f + g ) = +fake( f ) + +fake( g )

which makes sense.

One could generalize

If +fake( ln f ) and +fake ( ln g ) exist then

+fake( f * g ) = exp ( +fake( ln f ) + +fake( ln g ) )

Although the RHS is not optimal then and not equal to the integral.

I wonder what post 9 means to sums and products of f and g.

Still thinking.

regards

tommy1729