$$\begin {aligned} \int _{0}^{\infty }\frac {\lfloor x\rfloor}{1+e^{2\pi x}}dx &=\sum _{n=1}^\infty \int _{n}^{n+1}\frac {n}{1+e^{2\pi x}}dx\\ &=\sum _{n=1}^\infty \frac {n}{2\pi }\left [-\ln (1+e^{-2\pi x})\right ] _n^{n+1}\\ &=\frac {1}{2\pi }\sum _{n=1}^\infty n\ln \frac {1+e^{-2\pi n}}{1+e^{-2\pi (n+1)}} \end {aligned}$$

$$\begin {aligned} \sum _{n=1}^\infty n\ln \frac {1+e^{-2\pi n}}{1+e^{-2\pi (n+1)}} &=\sum _{n=1}^\infty n\left (a_n-a_{n+1}\right )\\ &=1(a_1-a_2)+2(a_2-a_3)+3(a_3-a_4)+\cdots \\ &=a_1+a_2+a_3+\cdots \quad \left (\because \lim _{n\to \infty }na_{n+1}=0\right )\\ &=\sum _{n=1}^\infty \ln (1+e^{-2\pi n})\\ &=\ln \prod _{n=1}^\infty \left (1+e^{-2\pi n}\right )\\ &=\ln \prod _{n=1}^\infty \left (1+q^{2n}\right )\quad \left (k=\frac {1}{\sqrt 2},q=e^{-\pi K'/K}\right )\\ &=\frac {1}6\ln \left (\frac {1}4e^{\frac {\pi }2}\left (\frac {1}{\sqrt 2}\right )\left (\frac {1}{\sqrt 2}\right )^{-1/2}\right )\\ &=\frac {\pi }{12}-\frac {3\ln 2}8 \end {aligned}$$

$$\begin {aligned} \int _{0}^{\infty }\frac {\lfloor x\rfloor}{1+e^{2\pi x}}dx&=\frac {1}{2\pi }\left (\frac {\pi }{12}-\frac {3\ln 2}8\right )\\ &=\textcolor {blue}{\frac {1}{24}-\frac {3\ln 2}{16\pi }}. \end {aligned}$$

$$k=\frac{\vartheta _2^2(q)}{\vartheta _3^2(q)},k'=\frac{\vartheta _4^2(q)}{\vartheta _3^2(q)}$$

$$\vartheta _3(q)\vartheta _4(q)=\vartheta _4(q^2)$$

$$\begin {aligned} \frac {\vartheta _2(q^2)\vartheta _3(q^2)}{\vartheta _2^2(q)} &=\frac {q^{1/2}\displaystyle \sum_{n ,m\in \mathbb Z}q^{2(n^2+n+m^2)}}{q^{1/2}\displaystyle \left (\sum_{n \in \mathbb Z}q^{n^2+n}\right )^2}\\ &=\prod _{n=1}^\infty \frac {\left (1-q^{4n}\right )\left (1+q^{4n}\right )\left (1+q^{4n-4}\right )\cdot \left (1-q^{4n}\right )\left (1+q^{4n-2}\right )^2}{\left (1-q^{2n}\right )^2\left (1+q^{2n}\right )^2\left (1+q^{2n-2}\right )^2}\\ &=\frac {1}2\prod _{n=1}^\infty \frac {\left (1-q^{4n}\right )^2\cdot \left (1+q^{4n}\right )^2\left (1+q^{4n-2}\right )^2}{\left (1-q^{2n}\right )^2\left (1+q^{2n}\right )^2\cdot \left (1+q^{2n}\right )^2}\\ &=\frac {1}2\prod _{n=1}^\infty \frac {\left (1+q^{2n}\right )^2}{\left (1+q^{2n}\right )^2}\\ &=\frac {1}2 \end {aligned}$$

$$\begin {aligned} \prod _{n=1}^\infty (1+q^{2n})^6&=\left (\prod _{n=1}^\infty \frac {1-q^{4n}}{1-q^{2n}}\right )^6\\ &=q^{-1/2}\frac {\eta^6(2\tau )}{\eta^6(\tau )}\quad \left (q=e^{\pi i\tau }\right )\\ &=q^{-1/2}\frac {\vartheta _2^2(q^2)\vartheta _3^2(q^2)\vartheta _4^2(q^2)}{\vartheta _2^2(q)\vartheta _3^2(q)\vartheta _4^2(q)}\\ &=q^{-1/2}\frac {\vartheta _2^2(q)}{\vartheta _3^2(q)}\frac {\vartheta _3(q)}{\vartheta _4(q)}\frac {\vartheta _2^2(q^2)\vartheta _3^2(q^2)\vartheta _4^2(q^2)}{\vartheta _2^4(q)\vartheta _3(q)\vartheta _4(q)}\\ &=q^{-1/2}kk'^{-1/2}\frac {\vartheta _2^2(q^2)\vartheta _3^2(q^2)}{\vartheta _2^4(q)}\\ &=\frac {1}4q^{-1/2}kk'^{-1/2}. \end {aligned}$$