How to express the integral $\int_0^a\frac{\sin x}{x} dx$ using other integrals For $a >0$, I need to express the integral $$\int_0^a \frac{\sin x} x \, dx$$ according to the following integrals:
$$\int_0^\infty \frac{e^{-ay}}{1+y^2} \,dy ~~~\text{ and }~~~~ \int_0^\infty \frac{ye^{-ay}}{1+y^2} \,dx .$$
I know that $\int_0^a \frac{\sin x}{x} \,dx $ is the sine integral equals $\operatorname{si}(a)$ and $\frac{\pi}{2}$ when $a \rightarrow \infty$, but how to express the integral according to the other integrals?
 A: Sometimes it is useful to recall that
$$
\int_0^\infty e^{-xy} \, dy = \frac 1 x \quad \text{if } x>0.
$$
$$
{}
$$
\begin{align}
& \int_0^a \frac{\sin x} x \, dx \\[10pt]
= {} & \int_0^a \left( \int_0^\infty e^{-yx} \, dy  \right) \sin x \, dx \\[10pt]
= {} & \int_0^\infty \left( \int_0^a e^{-yx} \sin x \, dx \right) \, dy. \tag 1
\end{align}
Line $(1)$ is valid because
\begin{align}
& \iint\limits_{(0,a)\times(0,\infty)} \left| e^{-yx} \sin x \right| \, d(x,y) \\[10pt]
\le {} & \iint\limits_{(0,a)\times(0,\infty)} e^{-yx} \, d(x,y) = \frac a x < +\infty.
\end{align}
Now start with line $(1)$ and integrate by parts twice, getting
\begin{align}
& \text{integral} \\[8pt]
= {} & (\text{some expression}) + \Big(\text{something} \times \text{same integral}\Big)
\end{align}
and do a bit of algebra to get
\begin{align}
& \int_0^a e^{-yx} \sin x \, dx \\[10pt]
= {} & \frac{1 - (\cos a)e^{-ya} - y(\sin a) e^{-ya}}{1+y^2}.
\end{align}
Therefore
\begin{align}
& \int_0^a \frac{\sin x} x \, dx \\[10pt]
= {} & \int_0^\infty \frac{1 - (\cos a)e^{-ya} - y(\sin a) e^{-ya}}{1+y^2} \, dy \\[10pt]
= {} & \frac \pi 2 - (\cos a)\int_0^\infty \frac{e^{-ya}}{1+y^2} \, dy - (\sin a) \int_0^\infty \frac{ye^{-ya}}{1+y^2} \, dy.
\end{align}
