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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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    @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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    @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