I am not able to obtain the value of:$$ \int_{0}^{\infty}\dfrac{\sin(nx)}{x}dx ,$$
By changing the order of integration of: $$\int_{0}^{\infty}\int_{0}^{\infty}e^{-xy}\sin(nx) dx dy.$$ And by using differentiation under the integral sign.
If you can suggest some other method then please tell me. Thanks for help in advance.
Since you asked for a different method, start by substituting $u=nx$. Now, we must consider different cases:
For $n>0$, we obtain: $$\int_0^{\infty} \frac{\sin(nx)}{x}~dx=\int_0^{\infty} \frac{\sin(u)}{u}~du$$ And for $n<0$, we have: $$\int_0^{\infty} \frac{\sin(nx)}{x}~dx=\int_{0}^{-\infty} \frac{\sin(u)}{u}~du=-\int_{-\infty}^{0} \frac{\sin(u)}{u}~du=-\int_0^{\infty} \frac{\sin(u)}{u}~du$$ Where on the last step, we used the fact that $\dfrac{\sin(u)}{u}$ is an even function. The case where $n=0$ is extremely easy to do, since the integrand is just $0$.
Putting it all together, we have for all $n\in \mathbb{R}$: $$\int_0^{\infty} \frac{\sin(nx)}{x}~dx=\operatorname*{sgn}(n)\cdot \int_0^{\infty} \frac{\sin(u)}{u}~du$$ Where we have made use of the Sign function.
One can then make use of the following question on MSE: Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?
There are many different methods listed there without using the change of order in integration and differentiation under the integral sign.