derivation of integral representations of Jacobi $\vartheta$ functions

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

$$\frac{\pi \cos (ux)}{2u\sin (\pi u)}=\frac{1}{2u^2}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{u^2-n^2}$$

This equation which is valid for all real values of $x$ such that $-\pi \le x\le \pi$ and all non-integral complex values of $u$. By comparison with the convergent series ${\sum }_{n=0}^{\mathrm{\infty }}1/n^2$, 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 $2u{e}^{-y{u}^2}$ and integrate.

$${\int }_{i-\mathrm{\infty }}^{i+\mathrm{\infty }}\frac{\pi \cos (ux)e^{-y{u}^2}}{\sin (\pi u)}𝑑v=2{\int }_{i-\mathrm{\infty }}^{i+\mathrm{\infty }}{e}^{-y{u}^2}\left[\frac{1}{2u^2}+\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{u^2-n^2}\right]u𝑑u$$

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

$$={\int }_P{e}^{-yv}\left[\frac{1}{2v}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{v-n^2}\right]𝑑v$$

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 $\mathrm{\Re }v+1=2{(\mathrm{\Im }v)}^2$

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:

$$\left|\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{v-n^2}\right|\le \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{|v-n^2|}\le \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{|v_m-n^2|}$$

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 $|v_m-n^2|\ge |m-3/4|$ for all $m,n$. Hence, we have

$$\sum _{i=1}^{2m}\frac{1}{|v_m-n^2|}\le \frac{2m}{m-3/4}\le 8$$

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

Finally since $v_m>1$ when $m\ge 1$. Also, . From these observations, we conclude that

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

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

$${\int }_{P_m}{e}^{-yv}\left[\frac{1}{2v}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{v-n^2}\right]𝑑v+{\int }_{S_m}{e}^{-yv}\left[\frac{1}{2v}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{v-n^2}\right]𝑑v=$$

$$2\pi i\sum _{n=1}^m{(-1)}^n\cos (nx)e^{-n^{2}y}$$

Taking the limit $m\to \mathrm{\infty }$ we obtain

$${\int }_P{e}^{-yv}\left[\frac{1}{2v}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{\cos (nx)}{v-n^2}\right]𝑑v=2\pi i\left(\frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\cos (nx)e^{-n^{2}y}\right)$$

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

$${\int }_{i-\mathrm{\infty }}^{i+\mathrm{\infty }}\frac{\pi \cos (ux)e^{-y{u}^2}}{\sin (\pi u)}𝑑v=2\pi i\left(\frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\cos (nx)e^{-n^{2}y}\right)$$

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

$${\int }_{i-\mathrm{\infty }}^{i+\mathrm{\infty }}\frac{\cos (2uz)e^{i\pi \tau u^2}}{\sin (\pi u)}dv=i(1+2\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n{e}^{i\pi n^2\tau }\cos (2nz))=i{\vartheta }_4(z|\tau )$$

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 . 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 $\mathrm{\Im }\tau >0$.

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

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

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

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

$$\mathrm{\Im }(\tau u^2)=2\mathrm{\Re }\tau \mathrm{\Re }u+\mathrm{\Im }\tau {(\mathrm{\Re }u)}^2-\mathrm{\Im }\tau =\mathrm{\Im }\tau \left({\left(\mathrm{\Re }u+\frac{\mathrm{\Re }\tau }{\mathrm{\Im }\tau }\right)}^2-1-{\left(\frac{\mathrm{\Re }\tau }{\mathrm{\Im }\tau }\right)}^2\right)$$

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

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

Taken together, the estimates of the last paragraph imply that

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

$$R>\frac{180|z|}{\pi \mathrm{\Im }\tau }\mathit{\hspace{1em}\hspace{1em}}{R}^2>\frac{180|z|}{\pi \mathrm{\Im }\tau }\mathit{\hspace{1em}\hspace{1em}},$$

it will be the case that

$$-\pi \mathrm{\Im }\tau {(\mathrm{\Re }u)}^2/10\mathit{\hspace{1em}\hspace{1em}},$$

and hence

Likewise, under the same restriction on $R$,

Since the contour of integration is compact and the integrand is analytic in a neighborhood of the contour,

$${\int }_{i-R}^{i+R}\frac{\cos (2uz)e^{i\pi \tau u^2}}{\sin (\pi u)}𝑑u$$

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 $R\to \mathrm{\infty }$, and hence

$${\int }_{i-\mathrm{\infty }}^{i+\mathrm{\infty }}\frac{\cos (2uz)e^{i\pi \tau u^2}}{\sin (\pi u)}𝑑u$$

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

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

$${\int }_{i-\mathrm{\infty }}^{i+\mathrm{\infty }}\frac{\cos (2uz)e^{i\pi \tau u^2}}{\sin (\pi u)}dv=i(1+2\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n{e}^{i\pi n^2\tau }\cos (2nz))=i{\vartheta }_4(z|\tau )$$

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
Canonical name	DerivationOfIntegralRepresentationsOfJacobivarthetaFunctions
Date of creation	2013-03-22 14:39:54
Last modified on	2013-03-22 14:39:54
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	26
Author	rspuzio (6075)
Entry type	Derivation
Classification	msc 33E05