Step Functions

**exphate** · August 29th 2009, 04:50 AM

Another stupid question!

Now, I know that we use "h(c-a)(F(c)-F(a)) - h(b-a)((F(b)-F(a))"

But, from what I can tell, that only applys for when there is only one step function. Is someone able to direct me on how to do this/post up the first couple of lines to help me p[lease? As I said, it's the inclusion of that second step function that has me in a twist

**chisigma** · August 29th 2009, 05:20 AM

First it is useful remember that $\sin (t-\frac{\pi}{2}) = - \cos t$ ... and after that you can compute the integral as sum of two integrals ...

$\int _{\frac{\pi}{2} } ^{\infty} \cos t \cdot e^{-pt}\cdot dt - \int _{\frac{\pi}{4} } ^{\infty} \cos t \cdot e^{-pt}\cdot dt$

Kind regards

$\chi$ $\sigma$

**exphate** · August 29th 2009, 12:38 PM

Being particularly poor at the multiple integration by parts, I've got no idea where to go.

I've done it a couple of times, but end up back where I started (that is, with an integral of $e^{-pt}cos(t)dt$ on both sides.

Im so stuck

**luobo** · August 29th 2009, 07:58 PM

Originally Posted by exphate

Being particularly poor at the multiple integration by parts, I've got no idea where to go.

I've done it a couple of times, but end up back where I started (that is, with an integral of $e^{-pt}cos(t)dt$ on both sides.

Im so stuck

$<br /> \int \cos t\; e^{-pt} \ dt = \frac{e^{-pt}\;(\sin t - p \cos t)}{1+p^2}<br />$

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