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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.

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    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.2012-02-05
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    i was wrong it wasnt dt it is dτ...2012-02-05
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    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).2012-02-05
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    @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.2012-02-05
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    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 $\int_0^\infty$?2012-02-05
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    @robjohn: Since for $\tau <0$ the function $u(\tau)=0$, so is the integrand, isn't it?2012-02-05
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    in case its τ>0 its different..2012-02-05
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    @Américo: You are right. I misread the definition for $u(\tau)$ to be $\operatorname{signum}(\tau)$. $u(\tau)=\frac12(1+\operatorname{signum}(\tau))$. Ignore my comment.2012-02-05

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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 :

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

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

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    How do you cope with the fact that $u(0)=1/2$? Are you taking always it being 1 on all the integration range?2012-02-05
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    thank you :) The formulas i had for converting to exp form were different that confused me.. i'll try in trigonometric though..2012-02-05
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    @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?2012-02-05
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    @RaymondManzoni: Ok, thanks.2012-02-05
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    @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.2012-02-05
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    @RaymondManzoni i did it in standard and i got the same result i think.I used the type for cos(A-B) firstly , i dont know whether i could have avoided thatr and then the integral for e*cos and e*sin2012-02-06
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    @Parhs: If you mean using [Laplace tables](http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms) of course this works too!2012-02-06