1. ## Step Functions

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. 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 $\displaystyle e^{-pt}cos(t)dt$ on both sides.

Im so stuck

4. 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 $\displaystyle e^{-pt}cos(t)dt$ on both sides.

Im so stuck
$\displaystyle \int \cos t\; e^{-pt} \ dt = \frac{e^{-pt}\;(\sin t - p \cos t)}{1+p^2}$