Find a real entire function $f(z)$ asymptotic to $\ln(x^2+1)$ for real $x$. Find a real entire function $f(z)$ asymptotic to $\ln(x^2 +1)$ for real $x$.
More specific I want $f(0)=0$ and $\frac{1}{2} \ln(x^2+1) < f(x) < 2 \ln(x^2+1)$.
Or prove it does not exist.
 A: As said in the comments I will write a partial "answer".
I do this because its to much for a comment and I feel its not an edit to the question.
I might edit the "answer" a few times.
I believe there is a real entire $f(x) = O(ln(x^2+1))$ where we use big-O notation.
Here is the assumed "proof sketch" :
$f(x) = \sum_{n=1}^\infty a_n x^{2n} = O(ln(x^2+1))$
Where $a_n$ alternates sign starting with $a_1 < 0$.
Also $|a_{2n}| > |a_{2n-1}|$.
$x > 0 \implies a_{2n-1}x^{4n-2} + a_{2n} x^{4n} < O(ln(x^2+1))$
$x > 1 \implies ln(a_{2n-1}x^{4n-2} + a_{2n} x^{4n}) < O(ln(ln(x^2+1)))$
$x > 1 \implies (4n-2) ln(x) + ln(a_{2n-1} + a_{2n} x^2) < O(ln(ln(x^2+1)))$
Now relaxing the equation :
$x > 1 \implies (4n-2) ln(x) + ln(a_{2n-1}) + ln(1 + A_n x^2) = C(ln(ln(x^2+1)))$
$x > 1 \implies  ln(a_{2n-1})  = C(ln(ln(x^2+1))) - (4n-2) ln(x) - ln(1 + A_n x^2)$
Now $a_{2n-1} = exp ( min ( C(ln(ln(x^2+1))) - (4n-2) ln(x) - ln(1 + A_n x^2) )$
The minimum must occur where the derivative is $0$.
Let the minimum be attained at $x= y$ then 
$  a_{2n-1}  = exp( C(ln(ln(y^2+1))) - (4n-2) ln(y) - ln(1 + A_n y^2)$
SO we must compute $y$ by setting the derivative to 0 and solving for x.
I use an approximation of the derivative ;
$\frac{d}{dx} C(ln(ln(x^2+1))) - (4n-2) ln(x) - ln(1 + A_n x^2) = $
( approx )
$-(4n-2)/x + C/(x ln(x))$
So we need to solve : $(-(4n-2) + C/ln(x))/x = 0$.
Or equivalently : $C = (4n-2) ln(x)$ 
$y = x = exp(C/(4n-2))$
(end proof sketch)
Numerically it seems pretty good.
I must thank tommy1729 and Sheldon (https://math.stackexchange.com/users/43626/sheldon-l)
I got the inspiration from them at the tetration forum : http://math.eretrandre.org/tetrationforum/showthread.php?tid=863
(they call this " fake function theory " )
I also wonder about least squares methods.
And integral representations.
This does not constitute a proof , I know.
But its too big for a comment.
EDIT :
Actually this might be a disproof ??
It seems the sequence $a_n$ might not be shrinking fast enough to give an entire function ?
Or did I make a big mistake ? 
But the sine function seems to suggest alternating sequences are very powerfull in approximating.
Is this the wrong way to compute it ? Did I make wrong assumptions ?
A bit confused.
EDIT 2 :
Tommy suggested finding an asymptotic to $f(x) = ln(x^2+1) exp(x^2)$.
The solution would then be $f(x) exp(-x^2)$.
That might work.
