A q triple product integral

Lemma: $\sum_{n=-\infty}^{\infty} {\dfrac{1}{\left(n-a\right)^2+b^2}}=\dfrac{\pi}{b}\dfrac{\tanh \pi b \sec^2 \pi a}{\tan^2 \pi a + \tanh^2 \pi b}$

Proof: Consider the function $f\left(z\right)=\dfrac{\pi}{\left(\left(z-a\right)^2+b^2\right)\tan{\pi z}}$ This function has simple poles at $z=a\pm bi$ and $n, \forall n\in\Z$ . The residues at these poles are $\text{Res}\left(f\left(z\right), a\pm bi\right)=\dfrac{\pm \pi}{2bi\tan{\left(\pi \left(a \pm bi\right)\right)}}$ $\text{Res}\left(f\left(z\right), n\right)=\dfrac{1}{\left(n-a\right)^2+b^2}$ Consider a square contour defined by the vertices $\left(N+\dfrac{1}{2}\right)\left(\pm1\pm i\right)$ , where $N\in\Z^+, N>\max\left(a,b\right)$ . We denote this contour as $\Gamma_N$ .

By Cauchy's Residue Theorem, we have $\oint_{\Gamma_N}f\left(z\right)=2\pi i\left(\text{Res}\left(f\left(z\right), a\pm bi\right)+\sum_{n=-N}^{N}\text{Res}\left(f\left(z\right), n\right)\right)$ $\oint_{\Gamma_N} f\left(z\right) =2\pi i\left(\dfrac{ \pi}{2bi\tan{\left(\pi \left(a +bi\right)\right)}}-\dfrac{\pi}{2bi\tan{\left(\pi \left(a -bi\right)\right)}}+\sum_{n=-N}^{N} {\dfrac{1}{\left(n-a\right)^2+b^2}}\right)$ $\oint_{\Gamma_N} f\left(z\right) =2\pi i\left(\dfrac{-\pi}{b}\dfrac{\tanh \pi b \sec^2 \pi a}{\tan^2 \pi a + \tanh^2 \pi b}+\sum_{n=-N}^{N} {\dfrac{1}{\left(n-a\right)^2+b^2}}\right)$ Now we prove that the limit of the LHS as $N\to \infty$ is 0. We can show that for all values of $z$ along $\Gamma_N$ , $\left|\dfrac{1}{\tan\left(z\right)}\right|$ is maximised at $z=\left(N+\dfrac{1}{2}\right)i$ , thus $\left|\dfrac{1}{\tan\left(z\right)}\right|\leq \left|\dfrac{1}{\tan\left(\left(N+\dfrac{1}{2}\right)i\right)}\right|=\coth{\left(N+\dfrac{1}{2}\right)}$ Hence by the estimation lemma, $\left|\oint_{\Gamma_N} f\left(z\right) \right|\leq \text{Length}\left(\Gamma_N\right)\text{sup}_{z \in \Gamma_N} \left|f\left(z\right)\right|\leq 4\left(2N+1\right)\text{sup}_{z \in \Gamma_N} \left|\dfrac{1}{\tan\left(z\right)}\right|\left|\dfrac{\pi}{\left(z-a\right)^2+b^2}\right|\leq4\left(2N+1\right)\coth{\left(N+\dfrac{1}{2}\right)}\mathcal{O}\left(\dfrac{1}{N^2}\right)$ Taking the limit as $N\to\infty$ , we get $\left|\oint_{\Gamma_\infty} f\left(z\right) \right|=0$ $0=2\pi i\left(\dfrac{-\pi}{b}\dfrac{\tanh \pi b \sec^2 \pi a}{\tan^2 \pi a + \tanh^2 \pi b}+\sum_{n=-\infty}^{\infty} {\dfrac{1}{\left(n-a\right)^2+b^2}}\right)$ $\sum_{n=-\infty}^{\infty} {\dfrac{1}{\left(n-a\right)^2+b^2}}=\dfrac{\pi}{b}\dfrac{\tanh \pi b \sec^2 \pi a}{\tan^2 \pi a + \tanh^2 \pi b}$ QED

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Using Jacobi's triple product, we can simplify the integral as such $\int_0^1 \prod_{k=1}^{\infty} \left(1-q^{6k}\right)\left(1+q^{6k-1}\right)\left(1+q^{6k-5}\right)dq =\int_0^1 \sum_{n=-\infty}^{\infty}q^{3n^2+2n}dq$ $=\sum_{n=-\infty}^{\infty} \dfrac{1}{3n^2+2n+1}$ $=\dfrac{1}{3}\sum_{n=-\infty}^{\infty} \dfrac{1}{\left(n+\dfrac{1}{3}\right)^2+\left(\dfrac{\sqrt{2}}{3}\right)^2}$ By our lemma, this equals $\dfrac{\pi}{\sqrt{2}}\dfrac{\tanh \left(\dfrac{\sqrt{2}\pi}{3}\right) \sec^2 \left(-\dfrac{\pi}{3}\right)}{\tan^2 \left(\dfrac{-\pi}{3}\right) + \tanh^2 \left( \dfrac{\sqrt{2}\pi}{3}\right)}$ After simplifying, we get $\dfrac{\pi}{\sqrt2}\dfrac{e^{\frac{4\sqrt2}{3}\pi}-1}{e^{\frac{4\sqrt2}{3}\pi}+e^{\frac{2\sqrt2}{3}\pi}+1} =\dfrac{\pi}{\sqrt2}\dfrac{e^{\frac{2^\frac{5}{2}}{3}\pi}-1}{e^{\frac{2^\frac{5}{2}}{3}\pi}+e^{\frac{2^\frac{3}{2}}{3}\pi}+1}$ Hence $a+b+c=2+3+1=\boxed{6}$ .

$\begin{aligned} \int_{0}^{1} {\prod _{ k=0 }^{ \infty }{ \left( 1-{ q }^{ 6k+6 } \right) \left( 1+{ q }^{ 6k+5 } \right) \left( 1+{ q }^{ 6k+1 } \right) } dq}&=\int_{0}^{1}{{ \left( { q }^{ 6 };{ q }^{ 6 } \right) }_{ \infty }{ \left( { -q }^{ 5 };{ q }^{ 6 } \right) }_{ \infty }{ \left( { -q };{ q }^{ 6 } \right) }_{ \infty }dq}&&&&&&& \textrm{Where} \quad (a;q)_{\infty}&=\prod_{k=0}^{\infty}{(1-aq^{k})}\\&=\int_{0}^{1}{\sum _{ n=-\infty }^{ \infty }{ { q }^{ 3{ n }^{ 2 }+2n } }dq }\\&=\sum _{ n=-\infty }^{ \infty }{\frac{1}{3n^{2}+2n+1}} \\&=\frac { \pi }{ \sqrt { 2 } } \frac { \sinh { \left( \frac { 2\pi \sqrt { 2 } }{ 3 } \right) } }{ \left( \cosh { \left( \frac { 2\pi \sqrt { 2 } }{ 3 } \right) -\cos \left( \frac { 2\pi }{ 3 } \right) } \right) }\\&= { \frac { \pi }{ \sqrt { 2 } } \frac { { e }^{ \frac { \pi { 2 }^{ \frac { 5 }{ 2 } } }{ 3 } }-1 }{ { e }^{ \frac { \pi { 2 }^{ \frac { 5 }{ 2 } } }{ 3 } }+{ e }^{ \frac { \pi { 2 }^{ \frac { 3 }{ 2 } } }{ 3 } }+1 } }\\\textrm{therefore}\quad\boxed{a+b+c=6} \end{aligned}$

For the series of the form $\displaystyle\sum _{ n=-\infty }^{ \infty }{\frac{1}{an^{2}+bn+1}}$ refer this paper .

proof for $\sum _{ n=-\infty }^{ \infty }{{ { q }^{ 3{ n }^{ 2 }+2n } }dq }={ { \left( { q }^{ 6 };{ q }^{ 6 } \right) }_{ \infty }{ \left( { -q }^{ 5 };{ q }^{ 6 } \right) }_{ \infty }{ \left( { -q };{ q }^{ 6 } \right) }}$ can be found out by substituting $a=q^{5}$ and $b=q$ in Ramanujan Theta functions which states that $f\left( a,b \right) =\sum _{ n=-\infty }^{ \infty }{ { a }^{ \frac { n(n+1) }{ 2 } }{ b }^{ \frac { n(n-1) }{ 2 } } } ={ \left( -a;ab \right) }_{ \infty }{ \left( -b;ab \right) }_{ \infty }{ \left( ab;ab \right) }_{ \infty }$