derivatives - Evaluating $int_0^infty e^{-ax}frac{sin{(bx)}}{x}dx $

From the formula $\displaystyle\int_0^\infty e^{-tx}\frac{\sin{(x)}}{x}dx=\frac{\pi}{2}-\arctan{(t)}$ for $t>0$, how to use change of variables to obtain a formula for $\displaystyle\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}dx$, when $a$ and $b$ are positive?

Then how to use differentiation under integral sign with respect to b to find a formula for $\displaystyle\int_0^\infty e^{-ax}\cos{(bx)}dx$ when a and b are positive.

My attempt:
I know the result $\displaystyle\int_0^\infty e^{ax}\cos{(bx)}dx= \frac{e^{ax}}{a^2+b^2}(a\cos{(bx)}+b\sin{(bx)})$. Is this result useful here?
Secondly integral calculator gives me this answer
$$-\frac{i(Ei((ib-a)x)-Ei(-(ib+a)x))}{2}.$$

How to evaluate this answer involving exponential integrals if a=4 and b=5?
Note=It was assumed that $(ib+a)\not=0$

Answer

Let $t=bx$ and use the given integral,

$$I(b)=\displaystyle\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}dx
=\displaystyle\int_0^\infty e^{-\frac ab x}\frac{\sin{t}}{t}dt=\frac\pi2 - \arctan\frac ab$$

Alternatively, apply integration-by-parts twice to its derivative,

$$I'(b)=\displaystyle\int_0^\infty e^{-ax}\cos(bx)dx$$
$$=-\frac1a \cos (bx)e^{-ax} |_0^\infty - \frac ba \int_0^\infty e^{-ax}\sin(bx)dx $$
$$=\frac1a +\frac {b}{a^2}\int_0^\infty \sin(bx)d(e^{-ax})
=\frac1a -\frac {b^2}{a^2} I'(b)$$

which leads to,

$$I'(b) = \frac a{a^2+b^2}$$

Thus,

$$I(b) = \int_0^b I'(t)dt= \int_0^b \frac a{a^2+t^2}dt = \arctan\frac ba
= \frac\pi2 -\arctan\frac ab $$

So, both methods produce the same result as expected.