Let $y(t)=\int_{-\infty}^0x(t+\tau)e^{\lambda\tau}\,{\rm d}\tau$, where $x(t)$ is an integrable function. Then how can we prove that $\frac{\rm d}{{\rm d}t}y(t)=x-\lambda y$?
Differentiation of a function including an integral
0
$\begingroup$
derivatives
-
0How can $\tau$ be a constant if we integrate with respect to it? – 2017-02-01
-
0Sorry. It was a mistake. – 2017-02-05
-
0Google Differentiation Under the Integral Sign. – 2017-02-05