I am trying to compute this for a telecommunication formula.

$\displaystyle\int^{\infty}_{-\infty} e^{-\tau}u(\tau)(1 + \frac{1}{2}\cos(400\pi t - 400\pi\tau))dτ$

$u(t)$ is $1$ for $t>0$, $1/2$ for $t=0$, $0$ for $t<0$.

$t$ is different from $\tau$! $\tau$ is like a constant.

  • $\begingroup$ nop i mean τ.. propably it can go out from integral like normal ones? infinity confuses me. I am not sure how to break the infinity. $\endgroup$ – GorillaApe Feb 5 '12 at 12:04
  • $\begingroup$ i was wrong it wasnt dt it is dτ... $\endgroup$ – GorillaApe Feb 5 '12 at 12:37
  • 4
    $\begingroup$ Notice that you are really integrating on $[0, +\infty)$ and use Euler's formula (or write $\cos(x) = \Re(e^{ix})$) to compute the second part of the integral (I hope the first part is easy enough to compute). $\endgroup$ – Najib Idrissi Feb 5 '12 at 13:01
  • $\begingroup$ @zulon i though of writing cos(x) in e notation but i am confused why is [0,infity]. u(0) is 1/2. It is hard for me to compute these improper integrals however i am trying. $\endgroup$ – GorillaApe Feb 5 '12 at 13:49
  • $\begingroup$ as $\tau\to-\infty$ the integrand blows up ($e^{-\tau}$ grows exponentially). As written, the integral diverges. Is there another modification, either $e^{-|\tau|}$ or $e^{-\tau^2}$ or $\int_0^\infty$? $\endgroup$ – robjohn Feb 5 '12 at 14:12

Let's rewrite this as : $\displaystyle\int_0^{\infty} e^{-\tau}\left(1 + \frac{1}{2}\cos\left(400\pi (t-\tau)\right)\right)d\tau$

In exponential form this becomes : $\displaystyle\int_0^{\infty} e^{-\tau} + \frac14 e^{i 400\pi (t -\tau)-\tau}+ \frac14 e^{-i 400\pi (t -\tau)-\tau}d\tau$

$\displaystyle =\left[e^{-\tau} - \frac{e^{i 400\pi t -\tau(1+i400\pi)}}{4(1+i400\pi)}- \frac{e^{-i 400\pi t -\tau(1-i400\pi)}}{4(1-i400\pi)}\right]_0^{\infty}$ $\displaystyle =1+\frac{e^{i 400\pi t}}{4(1+i400\pi)}+\frac{e^{-i 400\pi t}}{4(1-i400\pi)}$

EDIT Let's rewrite this in standard trigonometric form :

$\displaystyle =1+\frac{(1-i400\pi)e^{i 400\pi t}+ (1+i400\pi)e^{-i 400\pi t}}{4(1+(400\pi)^2)}$

$\displaystyle =1+\frac{\cos(400\pi t) + 400\pi\sin(400\pi t)}{2(1+(400\pi)^2)}$

  • $\begingroup$ How do you cope with the fact that $u(0)=1/2$? Are you taking always it being 1 on all the integration range? $\endgroup$ – Jon Feb 5 '12 at 15:39
  • $\begingroup$ thank you :) The formulas i had for converting to exp form were different that confused me.. i'll try in trigonometric though.. $\endgroup$ – GorillaApe Feb 5 '12 at 15:44
  • $\begingroup$ @Jon: Well there is no singularity for $\tau=0$, nothing special happens for $t=\tau$ and the integral shouldn't depend on the value at one point. What are you fearing? $\endgroup$ – Raymond Manzoni Feb 5 '12 at 15:51
  • $\begingroup$ @RaymondManzoni: Ok, thanks. $\endgroup$ – Jon Feb 5 '12 at 15:53
  • $\begingroup$ @Parhs: to get the trigonometric form you only have to reduce the two terms at the right to the same denominator. Another way to solve this is to use change of variable two times to get the $cos$ a second time at the right. $\endgroup$ – Raymond Manzoni Feb 5 '12 at 15:58

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