Say $a, x$ and $y$ are three real number constants and $t$ is a real variable. Now define the complex number, $z = -y + i(a+x-t)$ and consider an integral of the form $\int_{t= f(x,y)}^{t = g(x,y)} z\text{ }tanh (\pi z) log (z^2 + a^2) dz$ for some real functions $f$ and $g$.
Obviously $dz = -i dt$ and hence its an integral over $t$. Assume that it is ensures that the range of $t$ over which the integration is being done is such that the integrand doesn't hit any of its poles or branch points/cuts of the integrand.
Now what I want is to calculate $\lim_{x,y \rightarrow 0} [\int_{t= f(x,y)}^{t = g(x,y)} z\text{ }\tanh (\pi z) \log (z^2 + a^2) dz] $
Clearly this integral isn't doable that I can just do the integral and then take the limit at the end.
Firstly if it is true that $\lim_{x,y \rightarrow 0} f(x,y) = \lim_{x,y \rightarrow 0} g(x,y) $ then can I immediately conclude that the limiting value of the integral is $0$ without any further analysis?
Can I say expand the integrand $z \text{ }\tanh(\pi z) \log(z^2 +a^2)$ in a Taylor series in $x$ and $y$ and then do the integration on the constant term thus obtained and then take the limit? (because any term of the series with a non-zero power of $x$ and/or $y$ will obviously die off in the eventual limit)
Can I also separately take the $x,y \rightarrow 0$ limit on the limits of the integrand and thus avoid calculating the full integral with the full $f$ and $g$ ?
Could I have just substituted $z = i(a-t)$ in the integrand from the beginning itself?