Partial Integration

**bram kierkels** · January 4th 2011, 12:21 PM

I have $\alpha:[0,T]\rightarrow \mathbb{R}$ is continuous and $\beta>0$
How do I apply partial integration on $\alpha(t)+\beta\int_0^t e^{\beta(t-s)}\alpha(s)ds$ such that it equals
$e^{\beta t}\alpha(0)+\int_0^te^{\beta(t-s)}\alpha'(s)ds$
Thanks

**Random Variable** · January 4th 2011, 01:50 PM

let $u = \alpha (s)$ and $dv = e^{\beta (t-s)} ds$

$\displaystyle = \alpha (t) - \alpha(s) e^{\beta(t-s)} \Big|^{t}_{0} + \int^{t}_{0} e^{\beta(t-s)} \alpha^{'}(s) \ ds$

$\displaystyle = \alpha (t) - \alpha(t) + \alpha(0) e^{\beta t} + \int^{t}_{0} e^{\beta(t-s)} \alpha^{'}(s) \ ds$

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