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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?

Community
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მამუკა ჯიბლაძე
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    Can I suggest changing the title to something more descriptive? – Mark Grant Mar 25 '14 at 10:16
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    $\coth(ix)=-i\cot(x)$. The integral is thus undefined (except using principal value) for $z$ on the imaginary axis. The discontinuity is probably just a standard application of Cauchy's formula. –  Mar 25 '14 at 10:16
  • @MarkGrant I honestly thought about it but could not come up with anything better. Can you suggest something more descriptive? – მამუკა ჯიბლაძე Mar 25 '14 at 10:21
  • @Carl Yes you are most probably right. But I don't know how to use the Cauchy formula when one crosses infinitely many poles. – მამუკა ჯიბლაძე Mar 25 '14 at 10:24
  • Sum infinitely many contributions, taking care with interchange of sum and integral (though the $e^{-t^2}$ must help with this). –  Mar 25 '14 at 10:30
  • @Carl Thank you, I'll try. Still I do not see how to stay on the same branch when crossing the imaginary axis... – მამუკა ჯიბლაძე Mar 25 '14 at 10:51
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    @მამუკაჯიბლაძე: This is quite far from my area, so I'd rather leave that to someone else. I'd advise against the words "I don't understand..." and "help!" in the title though, as many people skimming the question list might mistake your question for an undergraduate plea for help with homework (which it clearly isn't). – Mark Grant Mar 25 '14 at 14:57
  • @MarkGrant Well in a sense it is :)

    I mean for a specialist in the field this is certainly a lamer type of question and a perfectly understandable reaction could be something like rtfm. So in this sense it is maybe even appropriate...

     – მამუკა ჯიბლაძე Mar 25 '14 at 15:03
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    Title: what about something like "Discontinuity in complex integral with infinitely many poles"? –  Mar 26 '14 at 10:02

2 Answers2

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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.

juan
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  • Thank you very much for your thorough work, but now I became even more confused. Are $G_1$ and $G_2$ on different branches of the same analytic function? Also, in any case either $G$ or $G_1$ or $G_2$ give something analytic separately in the upper and lower half-planes, can they be extended to the whole plane minus a discrete set of singular points (poles or (possibly infinite) branchings) or not? – მამუკა ჯიბლაძე Mar 28 '14 at 08:32
  • The integral define two analytic functions $G_1$ on the upper half plane and $G_2$ on the lower half plane. They are almost the same because $G_1(z)=-\overline{G_2(\overline z)}$. Each one is extended to a maximal region $G_1(z)$ to all the plane except for an angle of $\pi/2$ centered along the negative imaginary axis. The other in the symetrical region. These two functions differs as explicit in (2). The maximal region is that I said. This is shown because the difference can not be extended and one of the function yes on each border. – juan Mar 28 '14 at 11:44
  • So $G_1(z)$ and $G_2(z)$ are not branch of the same function. They can be extended, but each one only to an angle of $2\pi -\pi/2$. Each one a different angle. Their difference is the series in (2). – juan Mar 28 '14 at 11:50
  • The two functions seems to be "pretending" to be the same by the trick of being almost the same on the real axis. So they have the same asymptotic expansion in $z=\infty$. Also they used the trick of have the same integral defining them. – juan Mar 28 '14 at 11:54
  • Your data, refers to the point $z=1/7$ for my $G$. I compute $G_1(1/7)=0.4828593984740549016643202166100300361835498318469- i 6.5049410187201248215706080358441541206598301848206955269048210^{-21}$ $G_2(1/7)=-0.48285939847405490166432021661003003618354983184692000227 - i 6.50494101872012482157060803584415412065983018482069552690482921836826339857738435684108621277 10^{-21}$ and their difference divided by 2 is equal to $0.482859398474054901664320216610030036183549831846920002276456568$ Your number above. Near $0$ they are not so nearly equal. – juan Mar 28 '14 at 12:08
  • Sorry I still do not understand at least one thing. Take any of these $G$, $G_1$ or $G_2$. Suppose the integral gives some analytic function $G_U$ in some neighborhood $U$ of some nonzero $z$. Then if we look at the neighborhood $-U$ of $-z$, will not the same integral give $-G_U$ there? Because the integrand just gets multiplied by $-1$ and the path and all the poles remain the same, right? If this is OK, then as soon as you can extend any of the $G$, $G_1$, $G_2$ beyond some half-plane, seemingly it must give you an odd function on the whole plane because of that argument, no? – მამუკა ჯიბლაძე Mar 28 '14 at 14:28
  • What you say is true. Until the end, it is not an odd function. There is only two functions $G_1(z)$ and $G_2(z)$. $G_1$ has a maximal region of analyticity $\Omega$. It is formed of all $z$ with $-\pi/4< arg(z)<5\pi/4$. The other function is defined in the region conjugate, or if you want the region $-\Omega$. When your point $z$ moves along $\Omega$, $-z$ moves in $-\Omega$ when $z$ arrives to the border of $\Omega$, $-z$ arrives to the border of $-\Omega$. And as you said $G_1(z)=-G_2(-z)$. But you never find an open set where the two coincide. $0\not\in\Omega$. For more email me. – juan Mar 28 '14 at 18:20
  • Still struggling with it. I managed to check numerically several values. Everywhere I could do it $G_1(-z)=-G_1(z)$ holds for $-\pi/4<arg(z)<\pi/4$. – მამუკა ჯიბლაძე Apr 02 '14 at 20:37
  • I am sorry, I erased my last comment agreeing with you. In fact we have $G_1(-z)=\overline{G_1(\ \overline z\ )}$. This is true always. In particular for $x$ real $G_1(ix)$ is real. I tried to send you a mail, but I get a failure notice. Perhaps you may send me one. – juan Apr 03 '14 at 17:48
  • Yes you are right, I found some new values where it is evident that $G_1(-z)\ne-G_1(z)$ in general. I will try to send you mail. – მამუკა ჯიბლაძე Apr 03 '14 at 17:57
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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.

berlin
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    I feel that you say something important but cannot really use it - sorry, I lack expertise in the field. When I go ahead and either differentiate or integrate my stuff by $z$ under the integral, I get something that is equally impenetrable for me. In fact I even tried to expand it in the Maclaurin series but what I get is divergent. It seemingly gives an asymptotic expansion for large $z$ but it does not converge anywhere – მამუკა ჯიბლაძე Mar 25 '14 at 15:25
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    I'm confused. This integral doesn't converge for any $z$ on the imaginary axis: it oscillates between $-\infty$ and $0$ over and over. – David E Speyer Mar 25 '14 at 19:41