# $\int_0^{\infty } e^{-t y} \text{sinc}(d t) \cos (t x) \, dt$

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.

• arctan is an odd function. Does that help? – user3697176 Oct 5 '15 at 20:28
• Not really?$\phantom{}$ – grdgfgr Oct 5 '15 at 21:02
• Because arctan(z) is odd, you seem to be only off by a minussign. – tired Oct 6 '15 at 10:42
• I realized that both of those were wrong (the one at the top is correct though.) – grdgfgr Oct 6 '15 at 11:00
• i know, and it is not difficult to obtain. if you want i can write down a solution. – tired Oct 6 '15 at 11:02

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.