- #1
yungman
- 5,755
- 292
I want to proof (1)##\delta(x)=\delta(-x)## and (2) ## \delta(kx)=\frac{1}{|k|}\delta(x)##
(1) let ##u=-x\Rightarrow\;du=-dx##
[tex]\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)[/tex]
but
[tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)[/tex]
I cannot proof (1) is equal as I don't know if ##f(u)=-f(-u)##. Please help.
(2)Let ##u=kx\Rightarrow\;\frac{du}{k}=dx##. The sign of the limit reverse depends on the polarity of k.
[tex]\int_{-\infty}^{\infty}f(x)\delta(kx)dx=\frac{1}{k}\int_{-\infty}^{\infty}f(\frac{u}{k})\delta(u)du=^+_-\frac{1}{k}f(0)[/tex]
Since sign is +ve if k is positive. Sign is -ve if k is -ve. This means the result is always ##\frac{1}{|k|}f(0)##. Am I correct?
Thanks
(1) let ##u=-x\Rightarrow\;du=-dx##
[tex]\int_{-\infty}^{\infty}f(x)\delta(x)dx=(0)[/tex]
but
[tex]\int_{-\infty}^{\infty}f(x)\delta(-x)dx=-\int_{-\infty}^{\infty}f(-u)\delta(u)du=-f(0)[/tex]
I cannot proof (1) is equal as I don't know if ##f(u)=-f(-u)##. Please help.
(2)Let ##u=kx\Rightarrow\;\frac{du}{k}=dx##. The sign of the limit reverse depends on the polarity of k.
[tex]\int_{-\infty}^{\infty}f(x)\delta(kx)dx=\frac{1}{k}\int_{-\infty}^{\infty}f(\frac{u}{k})\delta(u)du=^+_-\frac{1}{k}f(0)[/tex]
Since sign is +ve if k is positive. Sign is -ve if k is -ve. This means the result is always ##\frac{1}{|k|}f(0)##. Am I correct?
Thanks