calculate $\int_0^\infty \int_0^\infty e^{-xy} \sin(x) \,dx\, dy$ I want to prove $\int_0^\infty \sin(x) / x \,dx = \frac{\pi}{2}$ using
$$\int_0^\infty \int_0^\infty e^{-xy} \sin(x) \,dx\, dy= \int_0^\infty \int_0^\infty e^{-xy} \sin(x) \,dy\, dx $$
assuming I've proved the integral converges so I can switch integral.
but my trouble is how to calculate $$\tag1\int_0^\infty \int_0^\infty e^{-xy} \sin(x) \,dx\, dy$$I just wrote it in the below form
$$\int_0^\infty \int_0^\infty e^{-xy}\sin x\,dx\,dy = \int_0^\infty \operatorname{Im} \int_0^\infty e^{-xy + ix}\,dx\,dy = \int_0^\infty \operatorname{Im} \frac{1}{y -i}\,dy\\ =\int_0^\infty  \frac{1}{y^2 +1 }\,dy = \frac{\pi}2.
$$
but I'm also looking for another way to calculate this integral $(1)$ , any help is appreciated , thanks!
 A: \begin{align}
& \int_0^\infty \frac{\sin x} x \, dx = \int_0^\infty \left( \sin x \int_0^\infty e^{-xy}\, dy \right) \, dx\\[12pt]
= {} & \int_0^\infty \left( \int_0^\infty e^{-xy} \sin x \,dy\right) \, dx \\[12pt]
= {} & \iint\limits_{x,y\,:\,x\,>\,0,\,y\,>\,0} e^{-xy}\sin x \, d(x,y) \text{ ??} \tag 1
\end{align}
Is the “$=$” on line (1) above true? It is true if
$$
\iint\limits_{x,y\,:\,x\,>\,0,\,y\,>\,0} e^{-xy} \left| \sin x \right| \, d(x,y) <+\infty. \qquad \text{(?)}
$$
That seems less than clear, and questionable since $\displaystyle\int_0^\infty \frac{\left|\sin x\right|} x \, dx = +\infty.$
So we take a somewhat different tack:
\begin{align}
& \int_0^b \frac{\sin x} x \, dx = \int_0^b \left( \sin x \int_0^\infty e^{-xy}\, dy \right) \, dx\\[12pt]
= {} & \int_0^b \left( \int_0^\infty e^{-xy} \sin x \,dy\right) \, dx \\[12pt]
= {} & \iint\limits_{x,y\,:\, x\,>\,0,\ \& \\ 0\,<\,y\,<\,b} e^{-xy}\sin x \, d(x,y) \\
& \text{(For the moment I'll leave it as an exercise} \\
& \phantom{\text{(}} \text{that the integral of the corresponding} \\
& \phantom{\text{(}} \text{absolute value is finite.)} \\[12pt]
= {} & \int_0^\infty \left( \int_0^b e^{-xy} \sin x \, dx \right) \,dy
\end{align}
The inside integral calls for integration by parts twice, leading back to the same integral, thus yielding an algebraic equation in which the unknown to be solved for is the integral.
$$
\int_0^b e^{-xy} \Big( \sin x \, dx \Big) = \int u\,dv = \cdots
$$
If I'm not mistaken, integrating by parts twice yields
$$
\int_0^b e^{-xy} \sin x \, dx = \Big(\cdots\cdots\Big) - y^2 \int_0^b e^{-xy} \sin x \,dx
$$
where $\displaystyle\Big(\cdots\cdots\Big)\to0$ as $b\to+\infty,$
so that
$$
\int_0^b e^{-xy} \sin x \, dx \to \frac 1 {1+y^2} \text{ as } b\to+\infty.
$$
So finally we evaluate $\displaystyle \int_0^\infty \frac{dy}{1+y^2} = \frac \pi 2.$
A: For the inner integral, there is a different approach.
Let's find a general formula for integrals of this form.
$$I= \int e^{ax}\sin(bx)\ \mathrm{d} x $$
Integrating by parts two times,
$$I=\dfrac{ e^{ax}\sin(bx)}{a}-\dfrac{b\cos(bx)e^{ax}}{a^2}-\dfrac{b^2}{a^2}\underbrace{\int e^{ax}\sin(bx)\ \mathrm{d}x}_{I} \\ I= \dfrac{e^{ax}(a\sin(bx)-b\cos(bx))}{a^2+b^2}$$
So,
$$\begin{align} \int_0^{\infty} e^{-yx}\sin(x)\ \mathrm{d}x &= \left. \dfrac{e^{-yx}(-y\sin(x)-\cos(x))}{y^2+1}\right|_0^{\infty} \\ &= \dfrac{1}{y^2+1}\end{align} $$
The rest can be evaluated just like you did.
A: I liked your approach. You can write, alternatively
\begin{align}\displaystyle\int_0^\infty\displaystyle\int_0^\infty e^{-xy}\sin x\,\mathrm dx\mathrm dy&=\displaystyle\int_0^\infty \frac{1}{x^2+1}\mathrm dx\quad{(1)}\\&=\dfrac{\pi}{2}\end{align}
$(1)-$ We have used the known result to evaluate the inner integral that the Laplace transform of $\sin(\omega t)$ is $\frac{\omega}{s^2+\omega^2}$, therefore $\int_0^\infty \sin(t)e^{-at}\mathrm dt=\frac{1}{a^2+1}$.
