Pi but why?

$\begin{aligned} \int _{ 0 }^{ \infty }{ \frac { \sin \left( 2019x \right) }{ { e }^{ 2\pi x }-1 } } dx&=\int _{ 0 }^{ \infty }{ \frac { 1 }{ { e }^{ 2\pi x }-1 } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( 2019 \right) }^{ 2n+1 } }{ \left( 2n+1 \right) ! }dx }&&&&&&&&&&&&&&\textcolor{#E81990}{\textrm{Sine Series}\quad\sin \left( 2019x \right) =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( 2019 \right) }^{ 2n+1 } }{ \left( 2n+1 \right) ! } } } \\&=\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( 2019 \right) }^{ 2n+1 } }{ \left( 2n+1 \right) ! } } \int _{ 0 }^{ \infty }{ \frac { { x }^{ 2n+1 } }{ { e }^{ 2\pi x }-1 }dx } &&&&&&&&&&&&&&\textcolor{#20A900}{\textrm{Gamma Zeta Relation}\quad\int _{ 0 }^{ \infty }{ \frac { { x }^{ 2n+1 } }{ { e }^{ 2\pi x }-1 } } dx=\frac { 1 }{ { \left( 2\pi \right) }^{ 2n+2 } } \Gamma \left( 2n+2 \right) \zeta \left( 2n+2 \right) } \\&=\frac { 1 }{ { 4\pi }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( 2019 \right) }^{ 2n+1 } }{ { \left( { 4\pi }^{ 2 } \right) }^{ n } } } \zeta \left( 2n+2 \right)&&&&&&&&&&&&&&\textcolor{#20A900}{\zeta \left( 2n+2 \right) =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2n+2 } } } }\\&=\frac { 2019 }{ 4{ \pi }^{ 2 } } \sum _{ k=1 }^{ \infty }{ \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( { 2019 }^{ 2 } \right) }^{ n } }{ { \left( 4{ \pi }^{ 2 } \right) }^{ n }{ \left( { k } \right) }^{ 2n+2 } } } }\\&=\frac { 2019 }{ 4{ \pi }^{ 2 } } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( { 2019 }^{ 2 } \right) }^{ n } }{ { \left( 4{ \pi }^{ 2 } \right) }^{ n }{ \left( { { k }^{ 2 } } \right) }^{ n } } } } &&&&&&&&&&&&&&\textcolor{#E81990} {\textrm{Geometric series}\quad{ \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n }{ \left( { 2019 }^{ 2 } \right) }^{ n } }{ { \left( 4{ \pi }^{ 2 } \right) }^{ n }{ \left( { { k }^{ 2 } } \right) }^{ n } } =\frac { { 4\pi }^{ 2 }{ k }^{ 2 } }{ { 4\pi }^{ 2 }{ k }^{ 2 }+2019^{2} } } }}\\&=\frac { 2019 }{ 4{ \pi }^{ 2 } } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 }+\frac { 2019^{2} }{ { 4\pi }^{ 2 } } } } &&&&&&&&&&&&&&\textcolor{#20A900}{\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 }+\frac { { 2019 }^{ 2 } }{ { 4\pi }^{ 2 } } } } =\left( \frac { \pi ^{ 2 } }{ 2019 } \coth \left( \frac { 2019 }{ 2 } \right) -\frac { 2{ \pi }^{ 2 } }{ { 2019 }^{ 2 } } \right) }\\&=\frac { 1 }{ 4 } \coth \left( \frac { 2019 }{ 2 } \right) -\frac { 1 }{ 4038 } \end{aligned}$ so we get $a=4,b=2019,c=2$ $\boxed{\sqrt { a+b+c } =\sqrt { 2025 } =45}$

This solution is base on another identity

$\sum_{n=-\infty}^\infty \dfrac1{n^2+a^2}=\dfrac\pi{a}\coth(\pi a)$

and you can find the proof in here: Calculate a special infinite series .

Since for $x\geq 1$ , we always have $e^{2\pi x}> 2$ , so $e^{2\pi x}-1\geq \dfrac12e^{2\pi x}\quad\text{ for all }x\geq 1.$

and since $|\sin(2019x)|\leq 1$ , we see

$\bigg|\int_1^M \dfrac{\sin(2019x)}{e^{2\pi x}-1}dx\bigg|\leq \int_1^M 2e^{-2\pi x}dx=\dfrac1\pi(e^{-2\pi }-e^{-2\pi M})$

which saying that the improper integral converges, therefore we can using a geometric series to expand $(e^{2\pi x}-1)^{-1}$ , which leads to

$I=\int_0^\infty \dfrac{\sin(2019x)}{1-e^{-2\pi x}}\cdot e^{-2\pi x}dx=\int_0^\infty \sin(2019x)(e^{-2\pi x}+e^{-4\pi x}+e^{-6\pi x}+\cdots)dx=\int_0^\infty \sum_{r=1}^\infty e^{-2\pi rx}\sin(2019x)dx.$

Note that the Laplace transform of $\sin bx$ is given by $G(b,s)=\mathfrak L\{\sin(bx)\}=\int_0^\infty e^{-sx}\sin(bx)dx=\dfrac{b}{s^2+b^2},$

Since $I$ converges, we can exchange the integral sign and summation sign, we have

$\begin{aligned}I&=\sum_{r=1}^\infty \int_0^\infty e^{-2\pi rx}\sin(2019x)dx\\ &=\sum_{r=1}^\infty G(2019,2\pi r)\\ &=\sum_{r=1}^\infty \dfrac{2019}{(2\pi r)^2+2019^2}\\ &=\dfrac{2019}{4\pi^2}\sum_{r=1}^\infty \dfrac1{r^2+(2019/2\pi)^2}\\ &=\dfrac{2019}{8\pi^2}\bigg[\sum_{r=-\infty}^\infty \dfrac1{r^2+(2019/2\pi)^2}-\dfrac{4\pi^2}{2019^2}\bigg]\\ &=\dfrac{2019}{8\pi^2}\bigg[\dfrac{2\pi^2}{2019}\coth \bigg(\dfrac{2019}2\bigg)-\dfrac{4\pi^2}{2019^2}\bigg]\\ &=\dfrac14\coth\bigg(\dfrac{2019}2\bigg)-\dfrac1{4038}\end{aligned}$

which gives that $a=4,b=2019,c=2$ , hence $\sqrt{4+2019+2}=\sqrt{2025}=\boxed{45}$ .

I hope there is a more elegant approach, something like a direct application of contour integral will be the best!