integral_02

Let $I\left( a \right) =\int _{ -\infty }^{ 0 }{ \frac { { e }^{ ax }\cdot \sin { \left( x \right) } }{ x } } dx$

$\frac { dI }{ da } =\int _{ -\infty }^{ 0 }{ \frac { x{ e }^{ ax }\cdot \sin { \left( x \right) } }{ x } } dx=\int _{ -\infty }^{ 0 }{ { e }^{ ax }\cdot \sin { \left( x \right) } dx }$

$\frac { dI }{ da } =\frac { { e }^{ ax }\left( a\cdot \sin { \left( x \right) } -\cos { \left( x \right) } \right) }{ { a }^{ 2 }+1 } { | }_{ -\infty }^{ 0 }=\frac { -1 }{ { a }^{ 2 }+1 }$ //integration by parts

$I\left( a \right) =\int { \frac { -1 }{ { a }^{ 2 }+1 } da=-\tan ^{ -1 }{ \left( a \right) } } +C$

$when\quad a\longrightarrow \infty \qquad I\left( a \right) =\int _{ -\infty }^{ 0 }{ \frac { { e }^{ ax }\cdot \sin { \left( x \right) } }{ x } } dx=0$ // since x<0 ${ \Rightarrow e }^{ ax }={ e }^{ -\infty }=0$

$I\left( \infty \right) =-\tan ^{ -1 }{ \left( \infty \right) } +C=-\frac { \pi }{ 2 } +C=0\qquad \Rightarrow C=\frac { \pi }{ 2 }$

$I\left( a \right) =\frac { \pi }{ 2 } -\tan ^{ -1 }{ \left( a \right) }$

$\int _{ -\infty }^{ 0 }{ \frac { { e }^{ \frac { x }{ \sqrt { 3 } } }\cdot \sin { \left( x \right) } }{ x } } =I\left( \frac { 1 }{ \sqrt { 3 } } \right) =\frac { \pi }{ 2 } -\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { 3 } } \right) } =\frac { \pi }{ 3 } =\frac { \sqrt { 1 } \pi }{ 3 }$

a+b=4

Similar solution with @Hassan Abdulla's

$\begin{aligned} I(a) & = \int_{-\infty}^0 \frac {e^{ax}\sin x}x \ dx \\ \frac {\partial I(a)}{\partial a} & = \int_{-\infty}^0 e^{ax}\sin x \ dx & \small \color{#3D99F6} \text{By integration of parts} \\ & = - e^{ax} \cos x \bigg|_{-\infty}^0 + \int_{-\infty}^0 ae^{ax} \cos x \ dx \\ & = - 1 + ae^{ax} \sin x \bigg|_{-\infty}^0 - a^2 \color{#3D99F6} \int_{-\infty}^0 e^{ax} \sin x \ dx \\ & = - 1 + 0 - a^2 \color{#3D99F6} \frac {\partial I(a)}{\partial a} \\ & = - \frac 1{a^2+1} \end{aligned}$

$\begin{aligned} \implies I(a) & = - \int \frac 1{a^2+1} \ da \\ & = - \tan^{-1} a + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \end{aligned}$

We note that when $a=0$

$\begin{aligned} I (0) & = \int_{-\infty}^0 \frac {\sin x}x \ dx & \small \color{#3D99F6} \text{Replace }x \text{ with }-x \\ & = - \int^\infty_0 \frac {\sin x}x \ dx \\ & = {\color{#3D99F6} \text{Si } (0)} = \frac \pi 2 & \small \color{#3D99F6} \text{where } \text{Si } (z) \text{ is the sine integral.} \\ \implies C - \tan{0} & = \frac \pi 2 \\ C & = \frac \pi 2 \\ \implies I(a) & = \frac \pi 2 - \tan^{-1} a \\ I \left(\frac 1{\sqrt 3} \right) & = \frac \pi 2 - \tan^{-1} \frac 1{\sqrt 3} \\ & = \frac \pi 2 - \frac \pi 6 \\ & = \frac \pi 3 = \frac {\sqrt 1 \pi}3 \end{aligned}$

$\implies a + b = 1 + 3 = \boxed{4}$