Which rule to apply?

$\large{I(a)=\displaystyle \int^{\infty}_{0} \frac{e^{-ax} \sin nx}{x} dx}$

Differentiating wrt $a$

$\large{I^{\prime}(a)=\displaystyle \int^{\infty}_{0} \frac{-\not{x}e^{-ax} \sin nx}{\not{x}} dx}$

$\large{I^{\prime}(a)=-\frac { 1 }{ \sqrt { { n }^{ 2 }+{ a }^{ 2 } } } { e }^{ -ax }\left( -a\sin { nx } -n\cos { nx } \right) \overset { \infty }{ \underset { 0 }{ | } } }$

$\large{\left(\because \quad \int { { e }^{ ax }\sin { \left( bx \right) dx } } =\frac { 1 }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \left( a\sin { \left( bx \right) -b\cos { \left( bx \right) } } \right) \right)}$

$\large{I^{\prime} (a)=-\frac{n}{\sqrt{n^2+a^2}}}$

$\large{I(a)=\int -\frac{n}{\sqrt{n^2+a^2}} da}$

$\large{I(a)=-\arctan \left (\frac{a}{n} \right)+c}$

We notice that $I(\infty)=0$ thus $\large{c=\frac{\pi}{2}}$

$\large{I(a)=\frac{\pi}{2}-\arctan \left (\frac{a}{n} \right)}$

$\large{I(a)=\cot^{-1} { \left( \frac { a }{ n } \right) } }$

$\large{\rightarrow I(a)=\boxed{\tan^{-1} \left (\frac{n}{a} \right)}}$

Property of Laplace Transforms Differentiate it wrt a and then integrate from a to infinity

we can use laplace transformation of ((sinnx)/x) first find L[sinnx] then integrate it with limits a to infinity and we get our solution