$\int_0^{\infty } e^{-t y} \text{sinc}(d t) \cos (t x) \, dt$ for $y>0$
$$\int_0^{\infty } e^{-t y} \text{sinc}(d t) \cos (t x) \, dt=\frac{\tan ^{-1}\left(\frac{d-x}{y}\right)+\tan ^{-1}\left(\frac{d+x}{y}\right)}{2 d}$$
I rewrote it as (omitting d)
$$\int_0^{\infty } \text{sinc}( t) \frac{e^{itx}+e^{-itx}}{2}e^{-t y}  \, dt$$
$$\frac{1}{2}\int_0^{\infty } \text{sinc}( t) \big(e^{-t(y-ix)}+e^{-t(y+ix)}\big) \, dt$$
And using Laplace transform properties I found it to be
$$\frac{1}{2}\big( \arctan(y-ix)+\arctan(y+ix)  \big)$$
Which is not correct.
So I want to learn why I am wrong. And I'd like to know how the complex arctan can be simplified to a real form.
 A: The Laplace transform of ${\rm sinc}(t)$ is $\arctan\dfrac{1}{s}$ if ${\rm Re}(s)>0$. Check e.g. Maple. Consequently
$$
\int_{0}^{\infty}e^{-ty}{\rm sinc}(t)\cos(xt)\, dt = \dfrac{1}{2}\left(\arctan\dfrac{1}{y-ix}+\arctan\dfrac{1}{y+ix}\right).
$$
But
$$
\arctan(s) = \dfrac{1}{2i}\log\dfrac{1+is}{1-is}.
$$
where $\log$ is the principal branch of the logarithm function and $s$ could be any complex number except those who have ${\rm Re}(s) = 0$ and $|s| \ge 1$.
Then we have that
$$
\int_{0}^{\infty}e^{-ty}{\rm sinc}(t)\cos(xt)\, dt = \dfrac{1}{2}\left(\arctan\frac{x+1}{y}-\arctan\dfrac{x-1}{y}\right) \quad \text{if }y>0.
$$
(Maple 17 calculated the integral to 
$$
\dfrac{1}{2}\left(\arctan\dfrac{2y}{x^{2}+y^{2}-1}\right)
$$
which isn't true for all our $x$ and $y$.)
If we instead use Fourier transformation and Parseval's formula we could almost completely avoid complex numbers.
\begin{gather*}
\int_{0}^{\infty}e^{-ty}{\rm sinc}(t)\cos(xt)\, dt = \dfrac{1}{2}\int_{-\infty}^{\infty}e^{-|ty|}\,\overline{{\rm sinc}(t)e^{-ixt}}\, dt \\[2ex]= \dfrac{1}{4\pi}\int_{-\infty}^{\infty}\dfrac{1}{y}\dfrac{2}{1+(\omega/y)^{2}}\pi({\rm H}(\omega+x+1)-{\rm H}(\omega+x-1))\,d\omega = \dfrac{1}{2}\int_{-1-x}^{1-x}\dfrac{1}{y}\dfrac{1}{1+(\omega/y)^{2}}\, d\omega \\[2ex] = \dfrac{1}{2}\left[\arctan\dfrac{\omega}{y}\right]_{-1-x}^{1-x} = \dfrac{1}{2}\left(\arctan\frac{x+1}{y}-\arctan\dfrac{x-1}{y}\right).
\end{gather*}
Here ${\rm H}$ is the Heaviside function.
