# derivation of integral representations of Jacobi $\vartheta$ functions

By rearranging the Fourier series of $\cos(ux)$, one obtains the series

 ${\pi\cos(ux)\over 2u\sin(\pi u)}={1\over 2u^{2}}+\sum_{n=1}^{\infty}(-1)^{n}{% \cos(nx)\over u^{2}-n^{2}}$

This equation which is valid for all real values of $x$ such that $-\pi\leq x\leq\pi$ and all non-integral complex values of $u$. By comparison with the convergent series $\sum_{n=0}^{\infty}1/n^{2}$, it follows that this series is absolutely convergent. Note that this series may be viewed as a Mittag-Leffler partial fraction expansion.

Let $y$ be a positive real number. Multiply both by $2ue^{-yu^{2}}$ and integrate.

 $\int_{i-\infty}^{i+\infty}{\pi\cos(ux)e^{-yu^{2}}\over\sin(\pi u)}\,dv=2\int_{% i-\infty}^{i+\infty}e^{-yu^{2}}\left[{1\over 2u^{2}}+\sum_{n=0}^{\infty}(-1)^{% n}{\cos(nx)\over u^{2}-n^{2}}\right]\,u\,du$

Because of the exponential, the integrand decays rapidly as $u\to i\pm\infty$ provided that $\Re u>0$, and hence the integral converges absolutely. Make a change of variables $v=u^{2}$

 $=\int_{P}e^{-yv}\left[{1\over 2v}+\sum_{n=1}^{\infty}(-1)^{n}{\cos(nx)\over v-% n^{2}}\right]\,dv$

The contour of integration $P$ is a parabola  in the complex $v$-plane, symmetric about the real axis with vertex at $v=-1$, which encloses the real axis. Its equation is $\Re v+1=2(\Im v)^{2}$

Let $S_{m}$ ($m$ is an integer) be the straight line segment joining the points $v=(i+m+1/2)^{2}$ and $v=(i-m-1/2)^{2}$. Along this line segment  , we may bound the integrand in absolute value    as follows:

 $\left|\sum_{n=1}^{\infty}(-1)^{n}{\cos(nx)\over v-n^{2}}\right|\leq\sum_{n=1}^% {\infty}{(-1)^{n}\over|v-n^{2}|}\leq\sum_{n=1}^{\infty}{(-1)^{n}\over|v_{m}-n^% {2}|}$

where $v_{m}=m^{2}+m-3/4$ is the point of intersection  of $S_{m}$ with the real axis. To proceed further, we break up the last summation into two parts.

Since the squares closest in absolute value to $v_{m}$ are $m^{2}$ and $(m+1)^{2}=m^{2}+2m+1$, it follows that $|v_{m}-n^{2}|\geq|m-3/4|$ for all $m,n$. Hence, we have

 $\sum_{i=1}^{2m}{1\over|v_{m}-n^{2}|}\leq{2m\over m-3/4}\leq 8$

When $n>2m$, we have $n^{2}\geq(2m+1)^{2}=4m^{2}+4m+1>4m^{2}+4m-3=4v_{m}$. Hence, $|n^{2}-v_{m}|>3n^{2}/4$ and

 $\sum_{n=2m+1}^{\infty}{1\over|v_{m}-n^{2}|}<{4\over 3}\sum_{n=2m+1}^{\infty}{1% \over n^{2}}<{4\over 3}\sum_{n=1}^{\infty}{1\over n^{2}}={2\pi\over 9}$

Finally $1/(2v_{m})<1/2$ since $v_{m}>1$ when $m\geq 1$. Also, $|e^{-yv}|=e^{-y\Re v}-e^{-yv_{m}}. From these observations, we conclude that

 $\left|\int_{S_{m}}e^{-yv}\left[{1\over 2v}+\sum_{n=1}^{\infty}(-1)^{n}{\cos(nx% )\over v-n^{2}}\right]\,dv\right|

Note that this quantity approaches 0 in the limit $m\to\infty$.

Let $P_{m}$ be the arc of the parabola $P$ bounded    by the endpoints  of $S_{m}$. Together, $S_{m}$ and $P_{m}$ form a closed contour which encloses poles of the integrand. Hence, by the residue theorem  , we have

 $\int_{P_{m}}e^{-yv}\left[{1\over 2v}+\sum_{n=1}^{\infty}(-1)^{n}{\cos(nx)\over v% -n^{2}}\right]\,dv+\int_{S_{m}}e^{-yv}\left[{1\over 2v}+\sum_{n=1}^{\infty}(-1% )^{n}{\cos(nx)\over v-n^{2}}\right]\,dv=$
 $2\pi i\sum_{n=1}^{m}(-1)^{n}\cos(nx)e^{-n^{2}y}$

Taking the limit $m\to\infty$ we obtain

 $\int_{P}e^{-yv}\left[{1\over 2v}+\sum_{n=1}^{\infty}(-1)^{n}{\cos(nx)\over v-n% ^{2}}\right]\,dv=2\pi i\left({1\over 2}+\sum_{n=1}^{\infty}(-1)^{n}\cos(nx)e^{% -n^{2}y}\right)$

Going back to the beginning of the proof, where the integral on the left hand side was expressed as an integral with respect to $u$, we obtain

 $\int_{i-\infty}^{i+\infty}{\pi\cos(ux)e^{-yu^{2}}\over\sin(\pi u)}\,dv=2\pi i% \left({1\over 2}+\sum_{n=1}^{\infty}(-1)^{n}\cos(nx)e^{-n^{2}y}\right)$

Making a change of variables $x=2z,y=-i\pi\tau$ and tidying up some, we obtain

 $\int_{i-\infty}^{i+\infty}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\,dv=i% \left(1+2\sum_{n=1}^{\infty}(-1)^{n}e^{i\pi n^{2}\tau}\cos(2nz)\right)=i% \vartheta_{4}(z|\tau)$

Because of the initial assumption  about the Fourier series, we only know that this formula   is valid when $\tau$ is purely imaginary with strictly positive imaginary part and $z$ is real and $\pi/2. However, we can use analytic continuation to extend the domain of its validity. On the one hand, the theta function on the right-hand side is analytic  for all $z$ and all $\tau$ such that $\Im\tau>0$.

On the other hand, I claim that the integral on the left hand side is also an analytic function of $z$ and $\tau$ whenever $\Im\tau>0$. To validate this claim, we need to examine the behaviour of the integrand as $u\to i\pm\infty$. The contribution of the denominator is bounded;

 $\left|{1\over\sin\pi u}\right|

for some constant $c$ whenever $\Im u=1$. The absolute value of the cosine in the numerator is easy to bound:

 $|\cos(2uz)|\leq e^{2|u|\,|z|}$

To bound the remaining term, let us examine the argument of the exponential carefully:

 $\Im(\tau u^{2})=2\Re\tau\,\Re u+\Im\tau(\Re u)^{2}-\Im\tau=\Im\tau\left(\left(% \Re u+{\Re\tau\over\Im\tau}\right)^{2}-1-\left({\Re\tau\over\Im\tau}\right)^{2% }\right)$

Therefore, if $|\Re u|>1+3|\Re\tau|/(\Im\tau)$, it will be the case that $\Im(\tau u^{2})\geq\Im\tau\,(\Re u)^{2}/9$, and so

 $\left|e^{i\pi\tau u^{2}}\right|=e^{-\pi\Im(\tau u^{2})}\leq e^{-\pi\Im\tau\,(% \Re u)^{2}/9}$

Taken together, the estimates of the last paragraph imply that

 $\left|\int_{i+R}^{i+\infty}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\right% |

when $R>1+3|\Re\tau|/(\Im\tau)$. If we impose the further conditions

 $R>{180|z|\over\pi\,\Im\tau}\qquad R^{2}>{180|z|\over\pi\,\Im\tau}\qquad,$

it will be the case that

 $2|u||z|-\pi\Im\tau\,(\Re u)^{2}/9<2\Re u\,|z|+2|z|-\pi\Im\tau\,(\Re u)^{2}/9<$
 $\left(2\Re u\,|z|-\pi\Im\tau\,(\Re u)^{2}/180\right)+\left(2|z|-\pi\Im\tau\,(% \Re u)^{2}/180\right)-\pi\Im\tau\,(\Re u)^{2}/10<$
 $-\pi\Im\tau\,(\Re u)^{2}/10\qquad,$

and hence

 $\left|\int_{i+R}^{i+\infty}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\,du% \right|
 $\left|\int_{i-\infty}^{i-R}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\,du% \right|
 $\int_{i-R}^{i+R}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\,du$

will be an analytic function of $z$ and $\tau$. Suppose that $z$ and $\tau$ are restricted to bounded regions of the complex plane and that, furthermore, $Im\tau$ is positive and bounded away from zero. Then the inequalities of the last paragraph imply that the integral converges uniformly as $R\to\infty$, and hence

 $\int_{i-\infty}^{i+\infty}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\,du$

is an analytic function of $u$ and $z$ in the domain $\Im\tau>0$.

Thus, by the fundamental theorem of analytic continuation, we may conclude that

 $\int_{i-\infty}^{i+\infty}{\cos(2uz)e^{i\pi\tau u^{2}}\over\sin(\pi u)}\,dv=i% \left(1+2\sum_{n=1}^{\infty}(-1)^{n}e^{i\pi n^{2}\tau}\cos(2nz)\right)=i% \vartheta_{4}(z|\tau)$

throughout this domain.

Finally, integral representations of the remaining three theta functions may be easily obtained from this one by adding the appropriate half-quasiperiods to $z$.

Title derivation of integral representations of Jacobi $\vartheta$ functions DerivationOfIntegralRepresentationsOfJacobivarthetaFunctions 2013-03-22 14:39:54 2013-03-22 14:39:54 rspuzio (6075) rspuzio (6075) 26 rspuzio (6075) Derivation msc 33E05