By rearranging the Fourier series of $\cos (ux)$ , one obtains the series $${ \pi \cos (u x) \over 2 u \sin (\pi u)} = {1 \over 2 u^2} + \sum_{n=1}^\infty (-1)^n {\cos (nx) \over u^2 - n^2}$$ This equation which is valid for all real values of $x$ such that $-\pi \le x \le \pi$ 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 sides by $2 u e^{-y u^2}$ and integrate. $$\int_{i-\infty}^{i + \infty} {\pi \cos (u x) e^{-yu^2} \over \sin (\pi u)} \, dv = 2 \int_{i-\infty}^{i + \infty} e^{-y u^2} \left[ {1 \over 2 u^2} + \sum_{n=0}^\infty (-1)^n {\cos (nx) \over u^2 - n^2} \right] \, u \, du$$

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 $v = u^2$ $$= \int_P e^{-y v} \left[ {1 \over 2 v} + \sum_{n=1}^\infty (-1)^n {\cos(nx)\over v - n^2} \right] \, dv$$ 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: $$\left| \sum_{n=1}^\infty (-1)^n {\cos (nx) \over v - n^2} \right| \le \sum_{n=1}^\infty {(-1)^n \over |v - n^2|} \le \sum_{n=1}^\infty {(-1)^n \over |v_m - n^2|}$$ 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 $|v_m - n^2| \ge |m - 3/4|$ for all $m,n$ . Hence, we have $$\sum_{i=1}^{2m} {1 \over |v_m - n^2|} \le {2m \over m - 3/4} \le 8$$

When $n > 2m$ , we have $n^2 \ge (2m+1)^2 = 4m^2 + 4m + 1 > 4m^2 + 4m - 3 = 4 v_m$ . Hence, $|n^2 - v_m| > 3 n^2 /4$ 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 $1/(2 v_m) < 1/2$ since $v_m > 1$ when $m \ge 1$ . Also, $|e^{-y v}| = e^{-y \Re v} - e^{-y v_m} < e^{-y m^2}$ . From these observations, we conclude that $$\left| \int_{S_m} e^{-y v} \left[ {1 \over 2 v} + \sum_{n=1}^\infty (-1)^n {\cos(nx)\over v - n^2} \right] \, dv \right| < e^{-y m^2} \left( 1 + 8 + {2 \pi \over 9} \right) \int_{S_m} dv = (4m + 2) \left( 9 + {2 \pi \over 9} \right) e^{-y m^2}$$ 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 $$\int_{P_m} e^{-y v} \left[ {1 \over 2 v} + \sum_{n=1}^\infty (-1)^n {\cos(nx)\over v - n^2} \right] \, dv + \int_{S_m} e^{-y v} \left[ {1 \over 2 v} + \sum_{n=1}^\infty (-1)^n {\cos(nx)\over v - n^2} \right] \, dv =$$ $$2 \pi i \sum_{n = 1}^m (-1)^n \cos(nx) e^{-n^2 y}$$

Taking the limit $m \to \infty$ we obtain $$\int_{P} e^{-y v} \left[ {1 \over 2 v} + \sum_{n=1}^\infty (-1)^n {\cos(nx)\over v - n^2} \right] \, dv = 2 \pi i \left( {1 \over 2} + \sum_{n = 1}^\infty (-1)^n \cos(nx) e^{-n^2 y} \right)$$ 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 $$\int_{i-\infty}^{i + \infty} {\pi \cos (u x) e^{-yu^2} \over \sin (\pi u)} \, dv = 2 \pi i \left( {1 \over 2} + \sum_{n = 1}^\infty (-1)^n \cos(nx) e^{-n^2 y} \right)$$ Making a change of variables $x = 2z, y = -i \pi \tau$ and tidying up some, we obtain $$\int_{i-\infty}^{i + \infty} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \, dv = i \left( 1 + 2 \sum_{n = 1}^\infty (-1)^n e^{i \pi n^2 \tau} \cos(2 n z) \right) = i \vartheta_4 (z | \tau)$$

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 $\pi/2 < z < \pi/2$ . 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; $$\left| {1 \over \sin \pi u} \right| < c$$ for some constant $c$ whenever $\Im u = 1$ . The absolute value of the cosine in the numerator is easy to bound: $$|\cos (2 u z)| \le e^{2 |u| \, |z|}$$ 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 $\Im (\tau u^2) \ge \Im \tau \, (\Re u)^2 / 9$ , and so $$\left| e^{i \pi \tau u^2} \right| = e^{-\pi \Im (\tau u^2)} \le e^{-\pi \Im \tau \, (\Re u)^2 / 9}$$

Taken together, the estimates of the last paragraph imply that $$\left| \int_{i + R}^{i + \infty} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \right| < c \int_{i + R}^{i + \infty} e^{2 |u| |z| -\pi \Im \tau \, (\Re u)^2 / 9}$$ 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 $$\left| \int_{i + R}^{i + \infty} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \, du \right| < c \int_{i + R}^{i + \infty} e^{-\pi \Im \tau \, (\Re u)^2 / 10} \, du < {5 c \over \pi \, \Im \tau} R e^{-\pi \Im \tau \, R^2 / 10} \qquad.$$ Likewise, under the same restriction on $R$ , $$\left| \int_{i - \infty}^{i - R} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \, du \right| < c \int_{i + R}^{i + \infty} e^{-\pi \Im \tau \, (\Re u)^2 / 10} \, du < {5 c \over \pi \, \Im \tau} R e^{-\pi \Im \tau \, R^2 / 10} \qquad.$$

Since the contour of integration is compact and the integrand is analytic in a neighborhood of the contour, $$\int_{i - R}^{i + R} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \, du$$ 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 $$\int_{i - \infty}^{i + \infty} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \, du$$ 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 $$\int_{i-\infty}^{i + \infty} {\cos (2 u z) e^{i \pi \tau u^2} \over \sin (\pi u)} \, dv = i \left( 1 + 2 \sum_{n = 1}^\infty (-1)^n e^{i \pi n^2 \tau} \cos(2 n z) \right) = i \vartheta_4 (z | \tau)$$ 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 $z$ .