I don't understand behavior of this integral, help! In an answer to a question I needed the following integral:
$$
f(z):=\int\limits_0^\infty t\coth(zt)e^{-t^2}dt;
$$
it represented deviation from modularity of some other function. However I noticed that this only works in one area and fails to work in another. After some confusion I discovered its source: it seems that $f(z)$ as defined suffers a discontinuity. For example, here is the result of a numeric calculation:
$$
\begin{array}{r|rl}
x&f(x+7i)&\approx\\
\hline
.005&.482859&-.000025 i\\
.004&.482859&-.000020i\\
.003&.482859&-.000015 i\\
.002&.482859&-.000010i\\
.001&.482859&-.000005 i\\
0&&(?)\\
-.001&-.482859&-.000005 i\\
-.002&-.482859&-.000010i\\
-.003&-.482859&-.000015 i\\
-.004&-.482859&-.000020i\\
-.005&-.482859&-.000025 i
\end{array}
$$
Discontinuity is evident in the real part; the imaginary part seems to be extendable to something continuous but non-differentiable along the imaginary axis. The integral seemingly should converge also for (purely) imaginary $z$ but I could not obtain any reliable numerical approximations with methods known to me.
It seems like $f(z)$ represents a many-valued analytic function, and the integral jumps from one branch of it to another.

Is there a name for this phenomenon? 
Is it possible to modify the representation in such a way that it stays on the same branch? 
How can I obtain an alternative representation of the same analytic function which would clarify many-valuedness and show the branching points?

 A: First observe that $t\coth(tz) e^{-t^2}$ is an even function of $t$. So the function
can be written as 
$$f(t)={1\over 2}\int_{-\infty}^{+\infty} t\coth(tz) e^{-t^2} \,dt$$
The properties of the function 
$$G(z):= \int_{-\infty}^{+\infty} t\coth(it/z) e^{-t^2} \,dt\qquad 
\text{($G$ of georgian)}$$
are easier to state. We easily get  $f(z)=G(i/z)/2$, whenever $f(z)$ is defined.
$G(z)$ is analytic at the upper half plane (and at the lower one).
Observe that the poles of the integrand are only at $t=k\pi z$ for $k\in\bf Z^*$.
So $t=0$ is not a pole.  Given a non real value of $z$, the line of integration, i.e. 
the real axis can be tilted  a little, depending on $z$ without changing the value of 
the integral. In this way we may extend the function $G(z)$ on the upper semi plane,
until the the region  with $-\pi/4 <\arg(z)<5\pi/4$.  For this you have to use
different positions of the line of integration. Each position give you the function 
on a half plane.  But you can not pass the region given by the angles above, because
you would have to pass the poles.  (Recall the poles are changing with $z$). This is 
a little difficult to explain here.  
In the same way you can extend the function $G(z)$ defined on the half plane $\Im(z) <0$,
to a region that is an angle of $2\pi-\pi/2$. The one from $-5\pi/4<\arg(z)<\pi/4$.
We shall call these two functions $G_1(z)$ and $G_2(z)$.  They are easily computed
by the definition I have given.  For example when $x>0$ we will have
$$G_1(x)=\int_{-\infty e^{-\pi i/6}}^{+\infty e^{-\pi i/6}} t\coth(it/z) e^{-t^2} \,dt,\qquad  G_2(x)=\int_{-\infty e^{\pi i/6}}^{+\infty e^{\pi i/6}} t\coth(it/z) e^{-t^2} \,dt$$
We may compute the difference 
$$G_1(x)-G_2(x)=2\pi i \Bigl(\sum_{k=1}^\infty \text{Res}_{t=k\pi x}h(t,z)-
\text{Res}_{t=-k\pi x}h(t,z)\Bigr)$$
This is a relatively easy application of the Residue Theorem. This in fact gives us
$$G_1(x)-G_2(x)=4\pi^2x^2\sum_{k=1}^\infty k e^{-k^2\pi^2x^2}.\qquad{(2)}$$
Since 
$$4\pi^2x\sum_{k=1}^\infty k e^{-k^2\pi^2x}$$
is only analytic on a half plane, the region of analyticity of the difference 
is only the angle $-\pi/4<\arg(z)<\pi/4$.   So the region where we have $G_1(z)$
and $G_2(z)$ are maximal.
I have checked with Mathematica the equality (2).
I get 
$$G_1(2)-G_2(2)=1.1302143268204567473\times 10^{-15}$$
on the two sides.
A: There is a standard way to deal with such integrals.  Consider the expression
$$
\int_0^\infty 
\ln(\sinh z t) dt $$
which converges in the classical sense  for all $z$ apart from the zeroes of hyperbolic sine function  (Lebesgue or improper Riemann) and so defines a meromorphic function there.  Then differentiate with respect to $z$ to see  that the above integral represents a meromorphic function of $z$.  One can regard this as a purely formal computation but on a forum for research level mathematics one would probably expect a more rigorous approach. This is provided by the theory of parametric integrals for distributions of the portuguese mathematician J. Sebastião e Silva where it is ALWAYS allowed to differentiate under the integral sign (something which is not permitted in the classical sense in the example under consideration).  One also requires the fact that EVERY meromorphic function on the plane can be regarded as a distribution there---a far reaching generalisation of the Hadamard pricipal value.
