Say I have the function $f(x) = x \tanh(\pi x) \log (x^2 +a^2)$ where $a$ is some positive real number. Then it seems to be me that Mathematica when given such a Log[]
function implicitly puts a branch-cut along the positive imaginary axis starting at $x = ia$.
- Now if I ask Mathematica to integrate $f(x)$ along some semicircle in the upper half plane parametized as say, $x = A e^{i\phi}$ ($A>a$) for $0 \leq \phi \leq \pi$, then is Mathematica's answer trustworthy?
Like did a large $A$ expansion (Series about $A=\infty$) and found that $xf(x)$ (that would practically be the integrand when converted into a $\phi$ integral), this function asymptotes as, $e^{2i\phi} \log(e^{2i\phi})A^2 + 2A^2 \log(A) + a^2$ (+ terms which go to zero as $A$ goes to infinity). Now on this asymptotic form the $\phi$ is integrable from $0$ to $\pi$ giving a finite answer.
So relying on this Mathematica result can I say that this integral therefore diverges in the large $A$ limit as $ \text{(number)} A^2 + \text{(number)} A^2 \log(A)$ ?
Like for example I chose $a=2$ and it seems to me that on using "NIntegrate" Mathematica says that the integral of this function on a circle centered at $2i$ goes to zero as the radius goes to 0.
Is this correct?
Thinking analytically this seems to be right...