<s>[INST] Seeking methods to solve $ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$ I am currently working on an definite integral that requires the following definite integral to be evaluated. 
$$ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$$
Where $k \in \mathbb{R}^{+}$
I was wondering what methods can be employed to solve this integral?
 [/INST] The method I took was:
Let 
$$ I(t) = \int_{0}^{\infty} \frac{\sin(kxt)}{x\left(x^2 + 1\right)} \:dx$$
Take the Laplace Transform 
\begin{align}
\mathscr{L} \left[I(t) \right]&=  \int_{0}^{\infty} \frac{\mathscr{L} \left[\sin(kxt)\right]}{x\left(x^2 + 1\right)} \:dx \\
&= \int_{0}^{\infty} \frac{kx}{\left(k^2x^2 + s^2\right)x\left(x^2 + 1\right)} \:dx \\
&= \int_{0}^{\infty} \frac{k}{\left(k^2x^2 + s^2\right)\left(x^2 + 1\right)} \:dx \\
&= \int_{0}^{\infty} \left[\frac{k^3}{\left(k^2 - s^2\right)\left(k^2x^2 + s^2 \right)}  - \frac{k}{\left(k^2 - s^2\right)\left(x^2 + 1\right)}\right] \:dx \\
&= \left[\frac{k^3}{\left(k^2 - s^2\right)}\frac{\arctan\left(kx\right)}{ks} -  \frac{k}{\left(k^2 - s^2\right)}\arctan(x)\right]_{0}^{\infty} \\
&= \frac{k^2}{s\left(k^2 - s^2\right)}\frac{\pi}{2} - \frac{k}{\left(k^2 - s^2\right)}\frac{\pi}{2} \\
&= \frac{k}{k^2 - s^2}\left[ \frac{k}{s} - 1 \right]\frac{\pi}{2} \\
&= \frac{k}{s\left(k + s\right)}\frac{\pi}{2}
\end{align}
And thus, 
$$ I(t) = \mathscr{L}^{-1}\left[\frac{k}{s\left(k + s\right)}\frac{\pi}{2} \right] =  \left[1 - e^{-kt} \right]\frac{\pi}{2}$$
Lastly,
$$ I = I(1) =  \left[1 - e^{-k} \right]\frac{\pi}{2}$$
</s>