Calculate integral
$\int_{-\infty }^{\infty } \! x(a\lambda +\lambda _{0} ) \delta (b\lambda +\lambda _{1} ) \, d\lambda $
for $b>0$ and $b<0$.
$\lambda _{0}$, $\lambda _{1}$, and $b\neq 0$ are constant. $\delta$ is dirac comb and $x$ is steady signal.
Solution:
$b>0$
$I=\frac{1}{b}x(\lambda _{0}-\lambda _{1}\frac{a}{b} )$
$b<0$
$I=-\frac{1}{b}x(\lambda _{0}-\lambda _{1}\frac{a}{b} )$
I tried to solve this integral, but didnt have any success. Can someone tell me how to get these solutions? How much is important to understand "dirac comb" to solve this?
By definition, $\int_{-\infty}^{+\infty}f(x)\delta(x)dx=f(0)$
So,
$\int_{-\infty }^{\infty } \! x(a\lambda +\lambda _{0} ) \delta (b\lambda +\lambda _{1} ) \, d\lambda = \int_{-\infty }^{\infty }x(a\frac{u-\lambda_1}{b}+\lambda_0)\delta(u)\frac{du}{b}=\frac{1}{b}x(\lambda_0-\lambda_1\frac{a}{b}) $
for positive $b$ with change of variables $u=b\lambda+\lambda_1$.
If $b$ is negative, the integral bounds are reversed when doing the change of variables and a minus sign appears.
Keep in mind Dirac's delta "function" isn't really a function but a distribution, so the integral notation is a bit sloppy for a non-physicist.