Partial Integration

• January 4th 2011, 01:21 PM
bram kierkels
Partial Integration
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
• January 4th 2011, 02:50 PM
Random Variable
let $u = \alpha (s)$ and $dv = e^{\beta (t-s)} ds$

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

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