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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.