A q product integral

The integrand is famously equal to $\sum\limits_{n=-\infty}^\infty (-1)^n q^{(3n^2-n)/2}.$ (You can find a proof here .)

Integrating gives $1 + \sum_{n \ne 0} \frac2{3n^2-n+2} (-1)^n = 1 + \sum_{n=1}^\infty (-1)^n \left( \frac2{3n^2-n+2} + \frac2{3n^2+n+2} \right)$ and this is easily evaluated by your favorite computer package to be approximately $\fbox{0.368413}.$

The following infinite product can be represented in a q pochammer symbol form i.e $\prod _{ k=1 }^{ \infty }{ \left( 1-{ q }^{ k } \right) } ={ \left( q;q \right) }_{ \infty }$ we know that the ramanujan theta function is defined as $f\left( a,b \right) =\sum _{ n=-\infty }^{ \infty }{ { a }^{ \frac { n(n+1) }{ 2 } }{ b }^{ \frac { n(n-1) }{ 2 } } } ={ \left( -a;ab \right) }_{ \infty }{ \left( -b;ab \right) }_{ \infty }{ \left( ab;ab \right) }_{ \infty }$ when $a =-q$ and $b={ -q }^{ 2 }$ we have $f\left( -q,{ -q }^{ 2 } \right) =\sum _{ n=-\infty }^{ \infty }{ { \left( -1 \right) }^{ n } } { q }^{ \frac { n(3n-1) }{ 2 } }={ \left( q;{ q }^{ 3 } \right) }_{ \infty }{ \left( { q }^{ 2 };{ q }^{ 3 } \right) }_{ \infty }{ \left( { q }^{ 3 };{ q }^{ 3 } \right) }_{ \infty }$ here the triple infinite product is ${ \left( q;{ q }^{ 3 } \right) }_{ \infty }{ \left( { q }^{ 2 };{ q }^{ 3 } \right) }_{ \infty }{ \left( { q }^{ 3 };{ q }^{ 3 } \right) }_{ \infty }=\prod _{ k=0 }^{ \infty }{ \left( 1-{ q }^{ 3k+1 } \right) \left( 1-{ q }^{ 3k+2 } \right) \left( 1-{ q }^{ 3k+3 } \right) =\left( 1-{ q }^{ 1 } \right) \left( 1-{ q }^{ 2 } \right) \left( { 1-q }^{ 3 } \right) \left( { 1-q }^{ 4 } \right) \left( { 1-q }^{ 5 } \right) ......={ \left( q;q \right) }_{ \infty } }$ therefore we can write the infinite product as the bilateral infinite series i.e $\int _{ 0 }^{ 1 }{ { \sum _{ n=-\infty }^{ \infty }{ { \left( -1 \right) }^{ n } } { q }^{ \frac { n(3n-1) }{ 2 } } }dq } =\sum _{ n=-\infty }^{ \infty } \frac { { \left( -1 \right) }^{ n }2 }{ { 3n }^{ 2 }-n+2 }$ by applying residue theorem for infinite series we can evaluate the sum to be $\frac { 2 }{ 3 } \sum _{ n=-\infty }^{ \infty } \frac { { \left( -1 \right) }^{ n } }{ \left( n-\left( \frac { 1+i\sqrt { 23 } }{ 6 } \right) \right) \left( n-\left( \frac { 1-i\sqrt { 23 } }{ 6 } \right) \right) } =\frac { -2\pi }{ i\sqrt { 23 } } \left( \cosec\left( \pi \left( \frac { 1+i\sqrt { 23 } }{ 6 } \right) -\cosec\left( \pi \left( \frac { 1-i\sqrt { 23 } }{ 6 } \right) \right) \right) \right)=0.36841...$

Start with approximating the product over the range $0\leq x \leq 1$ . Use the fact that, for all $0 \leq x \leq 1$ , we have $1-\frac{x}{2} \geq e^{-x} \geq 1-x.$ Hence, over the range of integration, the integrand can be bounded as $e^{-2\sum_{k=1}^{\infty} q^k} \leq \prod_{k=1}^{\infty}\big(1-q^k\big) \leq e^{-\sum_{k=1}^{\infty} q^k }.$ Summing up the geometric series within the exponent, we have $e^{-\frac{2q}{1-q}} \leq \prod_{k=1}^{\infty}\big(1-q^k\big) \leq e^{-\frac{q}{1-q}}.$ Thus, the integral can be upper and lower bounded by integrating the corresponding functions on the right and left. Numerical integration yields that the integral is at least $0.277343$ and is at most $0.403653$ . Better approximations yield better bounds.