A q double product integral

Although my solution is same as that of sir Mark Hennings , for the sake of remembrance of Ramanujan I'll make use of his theta function. 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 $a={ q }^{ 2 }$ and $b=1$ we have $f\left( { q }^{ 2 },1 \right) =\sum _{ n=-\infty }^{ \infty }{ { q }^{ { n }^{ 2 }+n }={ \left( -{ q }^{ 2 };{ q }^{ 2 } \right) }_{ \infty }{ \left( -1;{ q }^{ 2 } \right) }_{ \infty }{ \left( { q }^{ 2 };{ q }^{ 2 } \right) }_{ \infty } }$ where ${ \left( a;q \right) }_{ \infty }=\prod _{ k=0 }^{ \infty }{ \left( 1-{ aq }^{ k } \right) }$ is a pochammer notation. we came to see that the following product equates to an infinite bilateral series.Therefore we have the result as $\int _{ 0 }^{ 1 }{ \sum _{ n=-\infty }^{ \infty }{ q^{ { n }^{ 2 }+n }dq=\sum _{ n=-\infty }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }+n+1 } } =\frac { -\pi }{ i\sqrt { 3 } } } \left( \cot\left( \pi \left( \frac { 1+i\sqrt { 3 } }{ 2 } \right) \right) -\cot\left( \pi \left( \frac { 1-i\sqrt { 3 } }{ 2 } \right) \right) \right) =3.5962945.. }$

By the Jacobi Triple Product formula, we have $\begin{aligned} \prod_{k=0}^\infty \big(1 - q^{4(k+1)}\big)\big(1 + q^{2k}\big) & = \; \prod_{k=1}^\infty \big(1 - q^{4k}\big)\big(1 + q^{2k-2}\big) \\ &= \; \prod_{k=1}^\infty \big(1 - q^{2k}\big)\big(1 + q^{2k-2}\big)\big(1 + q^{2k}\big) \; = \; \sum_{m \in \mathbb{Z}} q^{m^2 + m} \end{aligned}$ for $0 < q < 1$ , and hence $\int_0^1 \prod_{k=0}^\infty \big(1 - q^{4(k+1)}\big)\big(1 + q^{2k}\big)\,dq \; = \; \sum_{m \in \mathbb{Z}} \frac{1}{m^2 + m +1} \; = \; \frac{2\pi}{\sqrt{3}}\tanh\left(\frac{\pi\sqrt{3}}{2}\right)$ making the answer $\boxed{3.596294561}$ .