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 $(ib+a)\not=0$
 A: 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,
$$I'(b)=\displaystyle\int_0^\infty e^{-ax}\cos(bx)dx
=\frac1a +\frac {b}{a^2}\int_0^\infty \sin(bx)d(e^{-ax})
=\frac1a -\frac {b^2}{a^2} I'(b)$$
which leads to $I'(b) = \frac a{a^2+b^2}$ and
$$I(b) = \int_0^b I'(t)dt=  \int_0^b \frac a{a^2+t^2}dt = \arctan\frac ba 
$$
A: Assuming that the conditions are fulfilled,
$$\frac{\partial}{\partial a}\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}dx =-\int_0^\infty e^{-ax}\sin{(bx)}dx $$
and also 
$$\frac{\partial}{\partial b}\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}dx =\int_0^\infty e^{-ax}\cos{(bx)}dx.$$
From
$$\int_0^\infty e^{-ax}e^{-ibx}dx=\left.\frac{e^{(-a+ib)x}}{-a-ib}\right|_0^\infty=\frac{a-ib}{a^2+b^2},$$ we get the sine and cosine integrals,
$$-\frac b{a^2+b^2},\frac a{a^2+b^2}$$ and their respective antiderivatives
$$\log\sqrt{a^2+b^2}+C$$ which are identical.
