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Apr
2
comment About asymptotic behaviour of a divergent integral.
Thanks! I will I guess email you sometime! May be you can help in this other question I asked related to my complex analytic viewpoint of this integral, math.stackexchange.com/questions/736038/…
Apr
2
revised About the branch-cut in the complex logarithm
added 67 characters in body
Apr
2
asked About the branch-cut in the complex logarithm
Apr
1
comment About asymptotic behaviour of a divergent integral.
If the function depends on mod(a) then the function is non-analytically depended on the value of a and that is quite different than a simple dependency on a^2. There are other examples also - like $\sum _{n \in \mathbb{Z}} log (n^2 + a^2)$ depends on mod(a) and NOT on $a^2$ and this leads to non-trivial physics effects. (....i wonder if you not being aware of the physics origins of the situation is causing the confusion...) (BTW this infinite sum example uses Riemann zeta-function regularization)
Apr
1
comment About asymptotic behaviour of a divergent integral.
It isn't that simple - like I said consider the integral $\int _0 ^A x tanh (\pi x) \frac{1}{x^2 + a^2}$ - the value of this integral depends on $mod(a)$ and not just on $a^2$ - this wasn't obvious to me till a lot of effort - I would have hoped for a similar thing here.
Apr
1
comment About asymptotic behaviour of a divergent integral.
I guess that in the limit each hump will just produce just the residue at each point - right? Or is there any more subtlety here? I guess the usual "Residue" function on Mathematica does give the right answer!
Apr
1
comment About asymptotic behaviour of a divergent integral.
Any thoughts on the complex analytic version? I am thinking - does the answer depend on whether the semi-circle is closed in the UHP or in the LHP - of course its not just a semi-circle - it has to have a key-hole contour about the branch-points ( $\pm ia$ depending on the U/LHP) and a semi-circular hump around each pole (at i (n + 1/2) and n is positive or negative integer depending on whether its the U/LHP) on each side and then take the limit of each hump and the hole going to zero - right?
Apr
1
comment About asymptotic behaviour of a divergent integral.
Thanks! As I said - this answer is kind of bit different from what I am looking for - for various reasons I would have loved to see the finite part depend on "a" and not "a^2" - but that is not coming here - like if you replaced the log term by 1/(x^2 +a^2) then thats what would happen.
Mar
31
comment About asymptotic behaviour of a divergent integral.
By your argument this graph should have been asymptoting to a constant (C(1)) - but it doesn't.
Mar
31
comment About asymptotic behaviour of a divergent integral.
See this problem - I put in this command into Mathematica, "Plot [ ( NIntegrate [ x Tanh [ [Pi] x] Log [ x^2 + 1] , {x, 0, A}] ) - ( A^2 Log [A] - (A^2/2 ) + Log [A] ) , {A, 10000, 100000}]"
Mar
31
comment About asymptotic behaviour of a divergent integral.
Is my argument correct? Infact now I see that by my argument the coefficient of $A^2 Log [A]$ is infact $2i e^{2 i \phi }$ (and the integral of this coefficient from $\phi = 0 $ to $\pi$ is 0) SO the divergence is only $A^2$...
Mar
31
comment About asymptotic behaviour of a divergent integral.
I wonder if there is any issue starting in your asymptotics proclaimed for $log (x^2 + a^2)$ - for which Mathematica claims that the asymptotics about x = infinity is $ 2Log[x] + (a/x)^2 + O(1/x^3) $ - which is not the same as yours...
Mar
31
comment About asymptotic behaviour of a divergent integral.
Then one sees that this series expansion has two terms which diverge in the $A$ going to $\infty$ limit and those terms are $A^2Log(A)$ and $A^2$ (and there is a constant). Now one can integrate the $\phi$ dependence of these coefficients from $0 \leq \phi \leq \pi$ - for these functions of $\phi$ there is no branchcut in the UHP - is this argument right? [...this doesn't get the extra $log(A)$ term that you are getting...]
Mar
31
comment About asymptotic behaviour of a divergent integral.
Let me think about this $a^2$ dependence of $C(a)$ a bit more - there are various other reasons (given the physical source of this question!) as to why $C$ should have probably depended on $a$ and not $a^2$. BUT, my one try at an asymptotic argument was like if you think of this as a complex function $z Tanh (\pi z) Log (z^2 + a^2)$ and try integrating it on a $A$ radius semicircle (centered at the origin) over the UHP - now substitute $z = A e^{i\phi}$ and find the large $A$ asymptotics of it (using "Series" on Mathematica about $A = \infty$)
Mar
30
comment About asymptotic behaviour of a divergent integral.
[...also any thoughts why the coefficients of "A^2 log(A)" and "A^2" are not matching with my argument for getting those terms? .. wonder my argument misses this extra "Log A" term that you see...]
Mar
30
comment About asymptotic behaviour of a divergent integral.
Thanks for your insights - but I really wanted to understand it in the complex integral form because in this picture its not clear how the value of the integral depends on "a" - which I believe it does - this way of writing the integral makes it a priori feel as if $C(a)$ depends on "a^2" but probably it depends on "a". So to understand how $C(a)$ would change if the sign of a is changed is hard unless one can explicitly do the integrals in $C(a)$.
Mar
29
comment About asymptotic behaviour of a divergent integral.
@AntonioVargas Yes!
Mar
29
revised About asymptotic behaviour of a divergent integral.
added 25 characters in body
Mar
29
asked About asymptotic behaviour of a divergent integral.
Mar
29
awarded  Disciplined